Nuclear and Particle Physics 9789391029135

Table of contents :
front cover
Title
Copyright
Preface
Contents
Chapter 1: Nuclear Structure and General Properties of Nuclei
1.1 Introduction
1.2 Constituents of the Nucleus
1.3 Basic Properties of Atomic Nucleus
1.4 Nuclear Forces
1.5 Yukawa Theory of Nuclear Forces
1.6 Nuclear Stability
1.7 Nuclear Models
1.8 Liquid Drop Model (Semi-Empirical Mass Formula)— C.F. Weizsacker Formula
1.9 Magic Number and Its Experimental Evidence
1.10 The Nuclear Shell Model
1.11 Fermi Gas Model
Summary
Short Answer Questions
Long Answer Questions
Multiple-Choice Questions
Chapter 2: Nuclear Reactions
2.1 Introduction
2.2 Types of Nuclear Reaction
2.3 Conservation Laws During Nuclear Reaction
2.4 Kinematics of Nuclear Reaction: Q-value of a Reaction
2.5 Physical Significance of the Q-value of a Reaction
2.6 Nuclear Cross Section
2.7 The Concept of Compound Nucleus
2.8 Energy Levels of Compound and Product Nucleus
2.9 Nuclear Transmutation
2.10 Discovery of Neutron (A Useful Probe for Nuclear Reaction)
2.11 Rutherford Scattering (Coulomb’s Scattering)
2.12 Nuclear Fission
2.13 Bohr and Wheeler Theory of Nuclear Fission
2.14 Source of Energy in Nuclear Fission
2.15 Chain Reaction
2.16 Nuclear Fusion
2.17 Thermonuclear Reactions and the Stellar Energy
Summary
Short Answer Questions
Long Answer Questions
Multiple-Choice Questions
Chapter 3: Radioactivity
3.1 Introduction
3.2 Radioactivity
3.3 Modes of Radioactive Decay
3.4 Nature and Properties of Nuclear Radiation
3.5 Law of Radioactive Disintegration
3.6 Disintegration Constant (l)
3.7 Half-life (T)
3.8 Activity and Its Units
3.9 Average Life or Mean Life
3.10 Law of Successive Disintegration (Radioactive Growth and Decay)
3.11 Radioactive Series
3.12 Geiger-Nuttal Law
3.13 Uses of Radioactivity
3.14 Radioactive Tracer and Its Uses
3.15 Radioactive Dating
3.16 Nuclear Potential Barrier
3.17 Theory of a-Decay
3.18 Theory of b-decay: The Neutrino Theory
3.19 Theory of Gamma Decay
Summary
Short Answer Questions
Long Answer Questions
Multiple-Choice Questions
Chapter 4: Interaction of Radiation with Matter
4.1 Introduction
4.2 Energy Loss Due to Interaction of Heavy Charged Particle with Matter
4.3 Bethe Bloch Relation
4.4 Range of Charged Particle
4.5 Bremsstrahlung
4.6 Range Straggling
4.7 Cherenkov Radiation
4.8 Interaction of g-ray with Matter
Summary
Short Answer Questions
Long Answer Questions
Multiple-Choice Questions
Chapter 5: Radiation Detection and Monitoring Devices
5.1 Introduction
5.2 Ionization Chamber
5.3 Geiger-Muller Counter (G-M Counter)
5.4 Proportional Counter
5.5 Wilson’s Cloud Chamber
5.6 Scintillation Counter
5.7 Bubble Chamber
5.8 Semiconductor Detector
5.9 Cherenkov Detector
Summary
Short Answer Questions
Short Questions
Long Answer Questions
Multiple-Choice Questions
Chapter 6: Particle Accelerators
6.1 Introduction
6.2 Van de Graaff Electrostatic Generator
6.3 The Cyclotron
6.4 Linear Accelerator
6.5 Synchrotrons
6.6 Betatron
Summary
Short Answer Questions
Long Answer Questions
Multiple-Choice Questions
Chapter 7: Elementary Particle Physics
7.1 Introduction
7.2 Classification of Elementary Particles
7.3 Interaction between Elementary Particles
7.4 Quantum Numbers Associated with Elementary Particles
7.5 Conservation Laws Obeyed
7.6 CPT Theorem
7.7 Quark Model
Summary
Short Answer Questions
Long Answer Questions
Multiple-Choice Questions
Chapter 8: Cosmic Rays
8.1 Introduction
8.2 Cosmic Rays
8.3 Types of Cosmic Rays
8.4 Nature of Cosmic Rays
8.5 Cosmic Ray Shower
8.6 Origin of Cosmic Rays
Summary
Short Answer Questions
Long Answer Questions
Multiple-Choice Questions
Chapter 9: Experiments of Nuclear Lab
Experiment 1: To determine the optimal voltage and plateau of a G-M counter
Experiment 2: To determine the dead time and recovery time in G-M counter
Experiment 3: To determine the half-life of a radioisotope using G-M counter
Experiment 4: To prove inverse square law using G-M counter
Experiment 5: To estimate the efficiency of gamma source
Experiment 6: To calibrate radiation survey meter
University Question Papers
Index
back cover

Citation preview

This book includes to-the-point description of titles, related theories, labelled diagrams, tables, examples, etc. This book is beneficial for B.Tech. Physics, B.Sc. Physics and for M.Sc. Applied Physics students. It helps students to prepare for JAM, NET SET exams. This book has been divided into nine chapters. Topics like nuclear structure, nuclear reactions, radioactivity, elementary particles, accelerators, radiation detection, interaction of radiation, cosmic rays are covered in detail. The discussion on all the topics is simple and mathematical derivations are less complex. Data, graphs and figures are mentioned wherever required. It also includes the basic lab experiments required in the nuclear lab for both B.Sc. as well as M.Sc. students. Salient Features ? Discussion on all the topics is simple and mathematical derivations are less complex. ? Data, graph and figures are mentioned wherever required. ? Short and long questions from various examinations of different universities. ? All concepts of nuclear physics are supported with relevant diagrams. Shefali Kanwar is Assistant Professor, Department of Applied Physics, Amity University, Noida, U.P. She obtained her Ph.D. in Nuclear Physics from Thapar University, Patiala. She is a lifetime member of Indian Association for Engineers (IAENG). She has more than 12 years of teaching experience.

Shivani is Assistant Professor, Department of Applied Physics, Amity Institute of Applied Sciences, Amity University, Noida. She has more than 12 years of teaching experience. She is a Gold Medallist in M.Sc. from the University of Allahabad in 2002. She obtained her Ph.D. in 2009 from the University of Allahabad. She has published more than 15 research papers in national and international journals.

978-93-91029-13-5

Nuclear and Particle Physics Shefali Kanwar Pramila Shukla Shivani

• Shefali Kanwar • Pramila Shukla • Shivani

Pramila Shukla is Assistant Professor, Department of Applied Physics, Amity University, Noida, U.P. She obtained her Ph.D. in Physics from the University of Allahabad. She is editor of International Journal of Scientific Research Letters and member of editorial board of International Journal of Engineering Science & Technology Letters. She is a lifetime member of Indian Association for Physics Teachers (IAPT) and has been a member of Optical Society of America. She has more than 14 years of teaching experience.

Nuclear and Particle Physics

Nuclear and Particle Physics

TM

Distributed by: 9 789391 029135 TM

Nuclear and Particle Physics

Shefali Kanwar Assistant Professor Department of Applied Physics Amity Institute of Applied Sciences Amity University, Noida (U.P.)

Pramila Shukla Assistant Professor Department of Applied Physics Amity Institute of Applied Sciences Amity University, Noida (U.P.)

Shivani Assistant Professor Department of Applied Physics Amity Institute of Applied Sciences Amity University, Noida (U.P.)

Nuclear and Particle Physics Authors: Shefali Kanwar Published by I.K. International Pvt. Ltd. 4435, 36/7, Ansari Rd, Daryaganj, New Delhi, Delhi 110002 ISBN: 978-93-91029-13-5 EISBN: 978-93-91029-21-0 ©Copyright 2021 I.K. International Pvt. Ltd., New Delhi-110002. This book may not be duplicated in any way without the express written consent of the publisher, except in the form of brief excerpts or quotations for the purposes of review. The information contained herein is for the personal use of the reader and may not be incorporated in any commercial programs, other books, databases, or any kind of software without written consent of the publisher. Making copies of this book or any portion for any purpose other than your own is a violation of copyright laws. Limits of Liability/disclaimer of Warranty: The author and publisher have used their best efforts in preparing this book. The author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness of any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. The accuracy and completeness of the information provided herein and the opinions stated herein are not guaranteed or warranted to produce any particulars results, and the advice and strategies contained herein may not be suitable for every individual. Neither Dreamtech Press nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Dreamtech Press is not associated with any product or vendor mentioned in this book.

Preface Nuclear and Particle Physics is an emerging and most researched branch of physics. It has found its applications in the field of physics, i.e., medical physics, like interaction of radiation with matter and use of different rays in radiotherapy. Particle physics has huge scope in modern-day research. This book is designed for B.Sc., and B.Tech. students of Indian universities. It can be used as an introductory book for postgraduate physics students. This book includes to-the-point description of titles, related theories, labelled diagrams, tables, examples and whatever is required. It also contains solved numerical problems, which cover almost all topics and numericals for practice also. It provides detailed theoretical explanation of topics, like Nuclear Potential Barrier; Bohr Wheeler Theory; etc., which are generally not covered in general books. This book can also be beneficial for B.Tech. Physics (Engineering Physics) and for M.Sc. Applied Physics students. It covers all the topics of UPTU and DU syllabus. Chapter 1 deals with basic concepts of nuclear structure and general properties of matter. Chapter 2 deals with nuclear reactions. Chapter 3 covers radioactivity, which explains a-decay, b-decay and g-decay reactions in detail. Chapter 4 covers interaction of radiation with matter, which is useful for medical physicists. Chapter 5 covers radiation detection and monitoring devices, which is a basic topic for M.Sc. physics students. Chapter 6 covers particle accelerators, Van de Graaff accelerator, cyclotron and different types of synchrotron. Chapter 7 covers elementary particle physics, which is an emerging scientific branch. Chapter 8 covers cosmic rays. Chapter 9 covers all the basic lab experiments required in the nuclear lab for both B.Sc. as well as M.Sc. students. The discussion on all the topics is simple and mathematical derivations are less complex. Data, graph and figures are mentioned wherever required. It also contains short and long questions, which are taken from various examinations of different Indian universities. We express our sincere gratitude to respected Dr. Ashok K. Chauhan, Founder President, Amity University; Atul Chauhan, Chancellor, Amity University; Prof. Balvinder Shukla, V.C., Amity University, Noida; Prof. Sunita Rattan, Director, AIAS, Amity University, Noida and Prof. R. Pandey, Head, Department of Physics, AIAS, Amity University, Noida.

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We are grateful to our family members for their blessings and motivation in bringing out this book. We owe a special thanks to Dr. Satish Kanwar, whose continuous input for this book helped in improving the quality. We are thankful to all our colleagues and students for their encouragement and constant support. Shefali Kanwar Pramila Shukla Shivani

Contents

Preface

v

1. Nuclear Structure and General Properties of Nuclei .................................. 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Introduction Constituents of the Nucleus Basic Properties of Atomic Nucleus Nuclear Forces Yukawa Theory of Nuclear Forces Nuclear Stability Nuclear Models Liquid Drop Model (Semi-Empirical Mass Formula)— C.F. Weizsacker Formula 1.9 Magic Number and Its Experimental Evidence 1.10 The Nuclear Shell Model 1.11 Fermi Gas Model Summary Short Answer Questions Long Answer Questions Multiple-Choice Questions

1 2 6 17 19 20 21 22 26 27 31 37 40 42 43

2. Nuclear Reactions ............................................................................................ 48 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9

Introduction Types of Nuclear Reaction Conservation Laws During Nuclear Reaction Kinematics of Nuclear Reaction: Q-value of a Reaction Physical Significance of the Q-value of a Reaction Nuclear Cross Section The Concept of Compound Nucleus Energy Levels of Compound and Product Nucleus Nuclear Transmutation

48 48 51 52 54 57 60 61 63

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2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17

Discovery of Neutron (A Useful Probe for Nuclear Reaction) Rutherford Scattering (Coulomb’s Scattering) Nuclear Fission Bohr and Wheeler Theory of Nuclear Fission Source of Energy in Nuclear Fission Chain Reaction Nuclear Fusion Thermonuclear Reactions and the Stellar Energy Summary Short Answer Questions Long Answer Questions Multiple-Choice Questions

71 76 83 84 85 86 88 88 91 93 93 94

3. Radioactivity ..................................................................................................... 99 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19

Introduction Radioactivity Modes of Radioactive Decay Nature and Properties of Nuclear Radiation Law of Radioactive Disintegration Disintegration Constant (l) Half-life (T) Activity and Its Units Average Life or Mean Life Law of Successive Disintegration (Radioactive Growth and Decay) Radioactive Series Geiger-Nuttal Law Uses of Radioactivity Radioactive Tracer and Its Uses Radioactive Dating Nuclear Potential Barrier Theory of a-Decay Theory of b-decay: The Neutrino Theory Theory of Gamma Decay Summary Short Answer Questions Long Answer Questions Multiple-Choice Questions

99 99 100 101 103 105 105 107 107 109 112 113 114 114 116 121 123 128 130 139 140 145 147

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4. Interaction of Radiation with Matter ......................................................... 152 4.1 Introduction 4.2 Energy Loss Due to Interaction of Heavy Charged Particle with Matter 4.3 Bethe Bloch Relation 4.4 Range of Charged Particle 4.5 Bremsstrahlung 4.6 Range Straggling 4.7 Cherenkov Radiation 4.8 Interaction of g-ray with Matter Summary Short Answer Questions Long Answer Questions Multiple-Choice Questions

152 153 156 157 158 159 160 164 176 179 180 181

5. Radiation Detection and Monitoring Devices ......................................... 186 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

Introduction Ionization Chamber Geiger-Muller Counter (G-M Counter) Proportional Counter Wilson’s Cloud Chamber Scintillation Counter Bubble Chamber Semiconductor Detector Cherenkov Detector Summary Short Answer Questions Short Questions Long Answer Questions Multiple-Choice Questions

186 187 189 192 196 199 203 205 209 216 217 219 220 221

6. Particle Accelerators ...................................................................................... 226 6.1 6.2 6.3 6.4 6.5 6.6

Introduction Van de Graaff Electrostatic Generator The Cyclotron Linear Accelerator Synchrotrons Betatron Summary Short Answer Questions Long Answer Questions Multiple-Choice Questions

226 227 229 232 235 239 246 248 248 249

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7. Elementary Particle Physics ......................................................................... 253 7.1 7.2 7.3 7.4 7.5 7.6 7.7

Introduction Classification of Elementary Particles Interaction between Elementary Particles Quantum Numbers Associated with Elementary Particles Conservation Laws Obeyed CPT Theorem Quark Model Summary Short Answer Questions Long Answer Questions Multiple-Choice Questions

253 254 257 260 263 266 270 275 277 277 278

8. Cosmic Rays .................................................................................................... 284 8.1 8.2 8.3 8.4 8.5 8.6

Introduction Cosmic Rays Types of Cosmic Rays Nature of Cosmic Rays Cosmic Ray Shower Origin of Cosmic Rays Summary Short Answer Questions Long Answer Questions Multiple-Choice Questions

284 284 284 285 289 290 292 294 294 294

9. Experiments of Nuclear Lab......................................................................... 296 Experiment 1: To determine the optimal voltage and plateau of a G-M counter 296 Experiment 2: To determine the dead time and recovery time in G-M counter 298 Experiment 3: To determine the half-life of a radioisotope using G-M counter 300 Experiment 4: To prove inverse square law using G-M counter 301 Experiment 5: To estimate the efficiency of gamma source 302 Experiment 6: To calibrate radiation survey meter 303

University Question Papers ................................................................................ 305 Index ........................................................................................................................ 313

1 Nuclear Structure and General Properties of Nuclei

1.1

INTRODUCTION

As we know matter in all its manifestations—solid, liquid and gas (vapor)—consist of tiny discrete particles called atoms. The various investigations of chemical and electrical properties of materials support the atomic concept of matter. The atom in the normal state is always found to be electrically neutral which means that net positive charge in the atom is exactly equal to net negative charge. To account for the experimentally obtained spectroscopic data, various atom models have been proposed from time to time. For instance, Thomson Atomic Model (1897), Rutherford Nuclear Atom Model (1911), Bohr Atom Model (1912), Sommerfeld Vector Atom Model (1915) and so on are some of well-known models. These models step by step mark the improvement of in our understanding about the structure of atom. Thomson Atom Model explained why an atom was stable and how it could radiate certain spectral lines. But it failed to explain a-scattering as observed in 1911 by Rutherford. But this failure proved to be most profitable because it led Rutherford and co-workers to the concept of nuclear atom. And this concept has been accepted as a fundamental concept in atomic and nuclear physics. According to Rutherford model, there is a central part of the atom in which whole mass and a positive charge is enclosed. This central part occupying a very small region of space; Rutherford called this part nucleus. The nucleus is surrounded by negatively charged electrons. Almost the entire mass of the atom is contributed by the nucleus. Hence, density of the nucleus is very large. The space between the nucleus and electron is empty. This model was the result of experimental study of large angle scattering of a-particle. Nuclear Physics is basically a younger scientific discipline that concerns itself with the structure of nucleus and the force holding the constituents of nucleus. It is also called Nuclear Structure Physics. In order to differentiate it from the elementary particle physics, that deals with particles of physics, their interactions and their structure; the former is called low energy (nuclear) physics, while the latter is called high energy physics.

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Most of the concepts and expressions used in nuclear physics are borrowed from macroscopic physics. By macroscopic particles, we mean particle with diameter, of say more than 10–6 m. The atomic and molecular phenomena which involve systems with diameters of the order of 10–10 m, are successfully described by what is known as Wave Mechanics. While coming over for atomic dimensions to the realm of the nucleus, the linear dimensions decrease approximately by the factor of 10 4. Even more spectacular is the difference in the measurable energy density between atomic and nuclear system, e.g., the binding energy of all the electrons in heavy element atoms is about 1032 J/m3 of nuclear matter. We would then like to know whether or not wave mechanics can properly describe all important nuclear phenomena. Wave mechanics has been very successful in some cases, yet there are instances when the mathematical complexities of the problem make it difficult to decide whether it is a right approach. In this section, we will discuss the various properties including electric and magnetic of nucleus, nuclear forces and various nuclear models.

1.2

CONSTITUENTS OF THE NUCLEUS

Rutherford’s a-scattering experiments established that the bulk of atomic mass and all its positive charge concentrated in a very small central core of the atom called nucleus. Since then various methods have been developed for explaining the nuclear structure. Various theories or hypotheses have been put forward about the nuclear constituent. The basis of these theories is some experimental data. This data is based on the following facts: (a) The precise measurement of atomic masses, (b) radioactivity, (c) optical spectroscopy, (d) nuclear spin and magnetic moments, etc. The first one was the proton-electron hypothesis, but ultimately failed on the grounds of some experimental observation. After the discovery of neutron, proton-neutron hypothesis was given for nuclear constituents.

1.2.1

The Electron-Proton Hypothesis

Until the discovery of neutron, it was assumed that an atom only contains electrons and protons. The concept of proton-electron constituent of the nucleus was based upon the following experimental observations: 1. Some radioactive elements emit b-particle (stream of fast moving electrons). 2. The discovery of isotopes and their whole number atomic weight suggested that all elements were built out of hydrogen atoms. The fractioned atomic weights are due to the presence of two or more isotopes each of which has an integral mass. The experimental results show that nuclei of different elements can be regarded as a collection of an integral number of hydrogen nuclei. The nucleus of hydrogen has positive charge equal in magnitude to that of the electron and is known as a proton.

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According to electron-proton model, a nucleus of atomic mass number A and atomic charge Z consisted of A protons and (A – Z) electrons, thus, giving it a net charge +Ze. As the electrons have a very small mass, hence, they hardly contribute to the mass of the nucleus. Since the atom as a whole is electrically neutral, the nucleus was supposed to be surrounded by Z extra nuclear electrons. For example, an atom of carbon of atomic weight 12 and atomic number 6 was supposed to contain 12 protons and 12 electrons. 12 protons and 6 electrons are packed together in the small space of the nucleus and remaining 6 revolve around the nucleus in a different orbit.

1.2.2

Failures of Electron-Proton Hypothesis

Although electron-proton model had some satisfactory aspects, yet there were a number of objections against the model which are: 1. If the electrons really existed inside the nucleus, their de-Broglie wavelength should not exceed the radius of the nucleus (R). Thus, taking l @ 10–14 m (size of nucleus) and h (Plank’s constant) = 6.62 ¥ 10–34 Js. The de-Broglie equation of matter waves, l = h/p or

p = h/l =

6.62 ¥ 10 -34 = 6.62 ¥ 10–20 kg ms–1 10 -14

Thus, the electron energy should be almost equal to E = pc = 6.62 ¥ 10–20 ¥ 3 ¥ 108 @ 2 ¥ 10–11 J or if the electrons exist inside the nucleus, the minimum energy of the electron should be of the order of 2 ¥ 10–11 J = 100 MeV. But experimentally, b-particles with energies more than 2 MeV to 3 MeV have never been observed. In fact the de-Broglie wavelength of such low energy electron is very large (l >>>> R). 2. If the electrons really existed inside the nucleus, even Heisenberg’s uncertainty principle also demands that an electron momentum, Dp @

h 6.62 ¥ 10 -34 = = 6.62 ¥ 10–24 kg ms–1 Dx 10 -14

Dx @ l @ 10–10 m then the energy of the electron. E=

p 2 c 2 + m02 c 4 where p = Dp = 6.62 ¥ 10–24 kg ms–1

E=

(6.62 ¥ 10 -34 ¥ 3 ¥ 108 )2 + (9.1 ¥ 10 -31 )2 ¥ (3 ¥ 108 )4

@ 2 ¥ 10–11 J =

2 ¥ 10 -11 1.6 ¥ 10 -13

@ 100 MeV

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3. There are difficulties in nuclear spin also. For example, a nitrogen atom should be expected to have half integral spin, i.e., 14 protons and 7 electrons inside the nucleus of nitrogen atom. But actually, it is only an integral value. It violates the established fact that nuclei with odd mass number (A) should have half integral spin and those with even mass number, integral spin. 4. Nuclear magnetic moments are very small. Magnetic moment of an electron is much larger (about 200 times) than that of a proton. If the electrons do not exist inside the nucleus, then electronic magnetic moment must dominate the nuclear magnetic moment. Experiments show that nuclear magnetic moments are very small. We can safely say that electrons do not contribute to it or they simply do not exist inside the nucleus.

1.2.3

The Proton-Neutron Hypothesis

With the discovery of neutron the setup of the nucleus has changed. To overcome the difficulties experienced by proton-electron model, Heisenberg proposed a new hypothesis, i.e., a nucleus consists of protons and neutrons. According to protonneutron, a nucleus of atomic mass A and atomic number Z consists of Z protons and (A – Z) neutrons giving it a net charge +Ze and total atomic mass (N + Z) = A. For example, the carbon nucleus consists of 6 protons and 6 neutrons, thus, making atomic mass 12 and atomic number 6. These are 6 extra nuclear electrons revolving around the nucleus make the atom on the whole, neutral. As the nucleus consists of protons and neutrons, hence, it is called nucleon. The proton and neutron are the two charge states of the same particle—the nucleon. The proton in the protonic state with charge +e and the neutron is the neutronic state with zero charge. The proton-neutron theory overcomes the failure of proton-electron theory. The standard convention for denoting the nucleus of a given element is ZXA.

1.2.3.1

Acceptance of Proton-Neutron Hypothesis

1. The hypothesis removes the difficulty of having an electron in the nucleus and is based upon two actually existing fundamental particles, the proton and the neutron. 2. The presence of protons and neutrons in the nucleus is in accordance with the Heisenberg’s uncertainty principle, according to which Dx Dp =

h 2p

where Dp is the uncertainty in momentum. As Dx should be at least equal to the radius of the nucleus, i.e., Dx @ R @ 10–14 m the value of Dp is Dp =

h 6.62 ¥ 10 -34 = 1.05 ¥ 10–20 kg ms–1 = 2p ¥ Dx 2p ¥ 10 -14

The rest mass of proton (or neutron) of the order of mp = 1.67 ¥ 10–27 kg

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Hence, corresponding to the velocity, v = v=

Dp m

1.05 ¥ 10 -20 = 6.31 ¥ 106 ms–1 -27 1.67 ¥ 10

and the value of kinetic energy of the proton, EK =

p2 (1.05 ¥ 10 -20 )2 = = 3.30 ¥ 10–14 J = 0.206 MeV 2mp 2 ¥ 1.67 ¥ 10 -27

Since the energy carried by the proton and neutrons emitted by the nucleus is of the order of 8 MeV greater than the value 0.206 MeV, the proton and neutron can exist within the nucleus. 3. The process of b-emission from the nucleus is explained on the basis of conversion of a neutron into proton and electron according to the equation n Æ p + e –, the electron so produces will come out of the nucleus in the form of b -particle. 4. The Compton wavelength associated with the neutron (or proton) is given by ln =

h 6.62 ¥ 10 -34 = = 1.3 ¥ 10–15 m = 0.13 ¥ 10–14 m -27 8 m0 c 1.67 ¥ 10 ¥ 3 ¥ 10

= 0.13 ¥ Nuclear diameter Therefore, ln 180, f is positive. Also we find that f has minimum value for A = 56 which has significance as the most stable nuclide. Packing fraction has been a useful parameter in the study of isotropic masses.

1.3.8

Binding Energy per Nucleon

Binding energy per nucleon is defined as the average energy required to remove a nucleon from the nucleus. It is equal to the ratio of total binding energy (BE) to the mass number (A) of the nucleus. Total BE A It is BE per nucleon and not the total binding energy that gives stability to the nucleus. More the BE/A more be the stability of the nucleus. The variation of BE/A with mass number (A) is given in Fig. 1.3. From the curve we find that: BE per nucleon =

1. The average BE per nucleon is small for light nuclei (1H1, 1H2, 1H3). 2. BE/A first increases sharply up to the mass number 20 and then increases gradually between A = 20 and A = 40. There are certain peaks corresponding to 2He4, 6C12, 8O16 etc. The peaks indicate that these nuclei are relatively more stable than the other nuclei in their neighbourhood. 3. Between the mass number 40 and 120 B.E./A more or less constant corresponding to average value equal to 8.5 MeV. The peak value of BE/A in this region is 8.8 MeV for 26Fe56. 4. Beyond the mass number 120 BE/A decreases with increase of mass number and it decreases to 7.6 MeV for 92U235. The decrease in the BE/A for heavy nuclei is due to large Coulomb’s force of repulsion between large number of protons in a heavy nuclei.

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Fe

8.8 MeV 8.5 7.6 MeV

16 8O 6

S

C12

4 2He

B.E./A (MeV)

3

3

2

7N

14

D

Li6

H1

H1 A 20

40

120

235

‹‰ǤͳǤ͵౨Binding energy per nucleon curve

5. Since from the graph we find that B.E./A for light as well as for heavy nuclei is small as compared to the nuclei whose mass numbers lie between 40 and 120. Hence, light nuclei combine together to form heavy nucleus. The process is known as nuclear fusion. Whereas heavy nucleus disintegrates into light nuclei, the process is known as nuclear fission. In both the cases, BE/A will increase or energy will release and nucleus attain stable configuration.

Numerical Problems Problem 1 (a) What is the energy of electron at rest? m (b) Calculate the ratio for electron having kinetic energy 1 MeV. m0 where m0 = rest mass energy of electron = 9.1 ¥ 10–31 kg and m is the mass when in motion? Solution:

(a) Rest mass energy of electron, E = m0c2 = 9.1 ¥ 10–31 ¥ (3 ¥ 108)2 = 9.1 ¥ 10–31 ¥ 9 ¥ 1016 J = 81.9 ¥ 10–15 J =

81.9 ¥ 10 -15 = 0.51 MeV 1.6 ¥ 10 -13

(b) Let m be the mass of the electron when in motion and m0 be the rest mass then Kinetic energy = (m – m0)c2 = 1 MeV mc2 – m0c2 = 1 MeV but m0c2 = 0.51 MeV mc2 = 1 + 0.51 = 1.51 MeV

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mc 2 1.51 m = fi = 2.96 m0 c 2 0.51 m0

Therefore,

Problem 2 (a) The mass of a lithium atom is 7.01822 amu. Calculate the binding energy of 3Li7 nucleus. Given mass of proton = 1.00814 amu, mass of neutron = 1.00893 amu and mass of electron = 0.00055 amu. (b) Calculate the average binding energy per nucleon for 28Ni64 having 63.9280 amu. Given mp = 1.007825 amu and mp = 1.008665 amu. (a) Mass defect of 3Li7 atom,

Solution:

Dm = {3mp + 4mn + 3me} – M = 3 ¥ 1.00814 + 4 ¥ 1.00893 + 3 ¥ 0.00055 – 7.01822 = 0.04337 amu BE = Dm ¥ 931.48 = 40.58 MeV (b) Mass defect of

BE of

28Ni

28Ni

64

nucleus,

Dm = 28mp + 36mn – M = 28 ¥ 1.007825 + 36 ¥ 1.008665 – 63.9280 = 0.6030 amu nucleus = Dm(amu) ¥ 931.48

64

= 0.6030 ¥ 931.48 = 561.7 MeV BE energy per nucleon =

Total BE 561.7 = = 8.77 amu. 64 A

Problem 3 Calculate the binding energy per nucleon for 1.007825 amu, mn = 1.008665 amu? Solution: Mass defect of

17Cl

35

17Cl

35

. Given that mp =

,

Dm = 17mp + 18mn – M = 17 ¥ 1.007825 + 18 ¥ 1.008665 – 34.98000 = 0.309 amu Total BE of

17Cl

35

= Dm(amu) ¥ 931.48 = 0.309 ¥ 931.48 = 287.83 MeV

BE per nucleon = Problem 4

287.83 = 8.22 MeV/nucleon. 35

The binding energy of 10Ne20 is 160.64 MeV. Find the atomic mass. Given

that mp = 1.007825 amu, mn = 1.008665 amu?

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Solution: Binding energy of

10Ne

20

= 160.64 MeV BE = Dm ¥ 931.48

Mass defect, Dm =

BE 160.64 = 0.17 245 amu = 931.48 931.48

Dm = 10mp + 10mn – M M = 10 ¥ 1.007825 + 10 ¥ 1.008665 – 0.17 245 = 19.9979 amu = 19.998 amu Therefore, nuclear mass of 10Ne20 = 19.998 amu.

Since \

1.3.9

Nuclear Energy

Nuclear energy is due to nuclear fission or nuclear fusion. Nuclear fission is a process in which a heavy nucleus splits into two or more light nuclei either spontaneously or after the absorption of a certain particle such as neutron (n) or g -rays with the release of large amount of energy. Nuclear fusion is a process by which two or more light nuclei combine together to form a heavy nucleus with the release of energy. The release of energy in both the processes is due to the conversion of mass defect into energy hence, they are exothermic reactions.

1.3.10

Angular Momentum of the Nucleus

 The total angular momentum of the nucleus ( I ) for a particular nuclear state is the resultant of individual total angular momentum ofall constituent nucleons in the nucleus. The total angular momentum of a nucleon ( J ) is the vector sum of its orbital angular momentum and spin angular momentum, i.e.,    I = L + S  (a) Spin angular momentum, S: Each nucleon (p or n) in the nucleus has an intrinsic angular momentum due to the spinning motion of the nucleon about its own axis passing through its centre of mass. The magnitude of spin angular momentum is given as  |S|= s(s + 1) where s = 1/2 (spin angular momentum quantum number) (b) Orbital angular momentum: The nucleons are also revolving around the centre of the nucleus in circular orbits. Hence, angular momentum is also associated with each nucleon due to its orbital motion defined as orbital angular momentum. The magnitude of orbital angular momentum is given by  |L|= l(l + 1) where l, the orbital angular momentum quantum number, has only integral values 0, 1, 2, 3….

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1.3.11

Magnetic Moment

The magnetic dipole moment is associated with the proton and neutron due to their orbital and spin motions. It has been found the magnetic dipole moment associated with the proton and neutron is mp = 2.792 e/2 mp (with proton) mn = –1.913 e/2 mn (with neutron) Having positive charge the magnetic field around a proton is parallel to the direction of total angular momentum hence, magnetic dipole moment is positive. A negative magnetic moment is a clear indication that the neutron is a complex particle containing positive and negative charges in equal amount and that the an negative charge is on an average farther from the axis of rotation. Total magnetic moment of the nucleus is the sum of magnetic dipole moment of protons and neutrons.

1.3.12

Electric Quadruple Moment

It is found that nuclei do not have electric dipole moment but electric quadruple moment has been observed in a number of nuclei. The electric quadruple moment of a nuclear charge distribution which is symmetric about the Z axis is given by Q=

1 (3 z 2 - r 2 )r dv eÚ

where r is the volume charge density.

1.3.13

Wave Mechanical Properties

Nucleus has two wave mechanical properties. (1) Statistics: The quantum mechanical description of a system with the number of particles is given by either Bose-Einstein or Fermi-Dirac statistics. (a) Bose-Einstein statistics. All particles with integral spin or zero obey BoseEinstein statistics and are called bosons, e.g., photon, p-mesons, etc. All nuclei with even number A obey Bose-Einstein Statistics. The wave function of a system obeying Bose-Einstein statistics is symmetric, i.e., y (x1, x2, … xi, xj, …, xn) = y (x1, x2, … xj, xi, …, xn) (b) Fermi-Dirac statistics. All particles with half integral spin obey FermiDirac statistics and the particles are known as fermions, e.g., electron, proton, neutrons, etc. The wavelength of the system obeying Fermi-Dirac statistics is antisymmetric, i.e., y (x1, x2, … xi, xj, …, xn) = –y (x1, x2, … xj, xi, …, xn)

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(2) Parity: The parity of a system refers to the behaviour of the wave function (y) under inversion of coordinates through the origin, i.e., when x is replaced by –x, y by –y and z by –z. If the change of sign of x, y, z does not change the sign of wave function, i.e., Y(x, y, z) = Y(–x, –y, –z) Then the wave function is said to be having even or positive parity. If the change of sign of x, y, z changes the sign of wave function, i.e., Y(x, y, z) = –Y(–x, –y, –z) Then the wave function is said to be having odd or negative parity. In general, Y(x, y, z) = P Y(–x, –y, –z) where P = ± 1. P can be taken as quantum number and the property defined by it is called the parity of the system. P = +1 mean even or positive parity P = –1 mean odd or negative parity Parity related to orbital quantum number as P = (–1)l where l is orbital quantum number.

Numerical Questions 1. Assuming a proton to be a sphere of radius r = 1.5 ¥ 10–15 m and of h uniform density and has an angular momentum of s(s + 1) where, 2p 1 s = . Find its frequency of revolution? 2 h Ans. Since angular momentum, s = s(s + 1) 2p If m is the mass, r is the radius and w is the angular velocity of proton spin, then spin angular momentum, S = mr2w Therefore,

mr2w =

h s(s + 1) 2p

w = 2pn = Frequency of revolution, n =

h s(s + 1) 4p 2 mr 2

h s(s + 1) 2p

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1 3 ¥ 2 2 = -27 4 ¥ 3.142 ¥ 1.67 ¥ 10 ¥ (1.5 ¥ 10 -15 )2 6.62 ¥ 10 -34

= 0.38 ¥ 1027 rev/s. 2. The quadrupole moment (Q) is given by 1 r (3z 2 - r 2 )dv eÚ where r is the uniform density of nucleus in the volume element at point (z, r). Prove that it may also be written as:

Q=

Q = Z < 3Z2 – r2 > a? Ans. Let Z be the atomic number of a nucleus then r= Thus,

Q=

Ze Ê Charge ˆ Á= ˜ Ú dv Ë Volume ¯

1 r (3z 2 - r 2 )dv eÚ

r (3 z 2 - r 2 )dv (Because r is uniform) eÚ 2 2 Ze Ze Ú(3z - r )dv 2 2 ( 3 ) = = Z{·3Z2 – r2Ò} = z r dv Ú e e Ú dv Ú dv =

Q = Z{·3Z2 – r2Ò} 3. Calculate the binding energy per nucleon for 17Cl35. Given that mp = 1.007825 amu, mn = 1.008665 amu, M = 34.98000 amu. Ans. Mass defect of

17Cl

35

,

Dm = 17mp + 18mn – M = 17 ¥ 1.007825 + 18 ¥ 1.008665 – 34.9800 = 0.309 amu Total binding energy of

17Cl

35

= Dm(amu) ¥ 931.48 = 0.309 ¥ 931.48 = 287.83 MeV

287.83 = 8.22 MeV. 35 4. The binding energy of 10Ne20 is 160.64 MeV. Find the atomic mass. Given mp Therefore, binding energy per nucleon =

= 1.007825 amu, mn = 1.008665 amu. Ans. Binding energy of

10Ne

20

= 160.64 MeV BE = Dm ¥ 931.48

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Mass defect, Dm =

B.E. 160.64 = = 0.172459 amu. 931.48 931.48

Since Dm = 10Mp + 10Mn – M Therefore, M = 10 ¥ 1.007825 + 10 ¥ 1.008665 – 0.17245 = 19.9979 amu = 19.998 amu Therefore, nuclear mass of

1.4

10Ne

20

= 19.998 amu.

NUCLEAR FORCES

A nucleus consists of protons and neutrons. Protons carrying positive charge exert a strong electrostatic force of repulsion. There is also gravitational force of attraction among the nucleons. The magnitude of electrostatic force is about 10 36 times more than the gravitational force of attraction. If only two forces are acting among the nucleons, then nucleons should not remain in the small volume (nucleons) but should fly apart or same repulsive force should cause a disruption in the nucleus. But we know that nucleons are inside the nucleus. Thus, there must be some third force which will bind the nucleons together inside the nucleus. This third force is the nuclear force, which holds the nucleons together inside the nucleus. It is attractive in nature and its magnitude is 102 times more than the electrostatic force of repulsion.

Properties of Nuclear Force 1. Nuclear forces are short range forces: Nuclear force is effective only up to a distance @ 10–14 m. If the distance between two nucleons inside the nucleus is more than 10–14 m the nuclear force vanishes for all practical purposes. Due to the short range, a nucleon will attract only those nucleons which are in its immediate neighborbood and these alone will contribute to the binding energy of the nucleons making it almost proportional to A.

0.5 ¥ 10–15 m

F

r O

A

10–15 m

B(10–14 m)

F

‹‰ǤͳǤͶ౨Variation of inter-nuclear force

18പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

2. Nuclear force is a charge-independent force: The magnitude of nuclear force between two protons (p-p interaction) or between the neutrons (n-n interaction) or a proton and neutron is almost same. This fact also verifies that the total binding energy of a nucleus depends upon the total number (A) and not on their nature. Hence nuclear force does not depends upon the charge of nucleons. Therefore, F(p – p) = F(n – n) = F(p – n). This shows that nuclear force is a charge-independent force. 3. Nuclear force is a spin-dependent force: It has been found that the magnitude of nuclear force between two nucleons with parallel spin is more than between two nucleons with anti-parallel spin, i.e.,

F( ≠≠ ) > F( ≠Ø )

Hence, nuclear force is a spin-dependent force. 4. Nuclear force is the strongest force in the nature: Magnitude of nuclear force is 102 times more than electrostatic force and 1038 times more than gravitational force between two nucleons, i.e., FN = 102 Fe = 1038 Fg Hence, it is the strongest force in the nature. 5. Nuclear force is saturated in nature: Since due to short range nature a nucleon will interact with only a limited number of nucleons which lie within the attractive range ( @ 10–14) and will not form bonds with other nucleons. This effect is known as saturation of nuclear force. If the nuclear force is not saturated in nature, then each nucleon will interact with remaining (A – 1) nucleons. Hence, binding energy would be proportional to the A( A - 1) number of nucleon pairs, i.e., or binding energy will be proportional 2 to A2. But the BE is proportional to A. Hence, nuclear force is a saturated force. 6. Nuclear force is a non-central force: The nuclear force does not lie along the line joining the centre of two nucleons. Hence, nuclear force is not a central force. 7. Nuclear forces are exchange forces: Nuclear force is due to the fast exchange of p-mesons between two nucleons. 8. Nuclear force is an attractive force but it also has a repulsive component if the distance between the two nucleons is less than 0.5 ¥ 10–15 m.

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1.5

YUKAWA THEORY OF NUCLEAR FORCES

According to Yukawa nuclear force is due to fast exchange of particles having mass about 270 me. On theoretical grounds Yukawa predicted a new particle, the meson, which may have positive, or a negative charge or may be neutral. He assumed that this new particle is exchanged between the nucleons and the corresponding exchange force is responsible for the binding energy of the nucleons. Just as a photon is a quanta of electromagnetic field, graviton is a quanta of gravitation field, a meson is a quanta of nuclear field. Nuclear forces are thus produced by a meson field similar to the electromagnetic field but of a much shorter range. This is because the exchange of mesons between the nucleons can take place only when the nucleons are close together. The prediction of Yukawa was confirmed by the experimental discovery of p-mesons or pion which explained the observed nuclear binding energies. According to Yukawa, the exchange of pions between two nucleons is so fast that the nucleons remain in the bind state. Also due to the exchange of p-mesons, nucleons may lose their identity, but the total number of protons and neutrons at any instant always remains constant, for example: p+ (i)

(ii)

(iii)

p Æ p+ + n Ø p+ + n Æ p

n Æ p– + p Ø p– + p Æ n p1 Æ p 0 + p*1 Ø p 0 + p2 Æ p*2 p*2 Æ p 0 + p2 Ø p 0 + p*1 Æ p1

n(p)

p(n) p+ p–

p (n)

n(p)

p– p0 p2 ( p2* )

p1( p1* )

p0 p0

(iv)

n1 Æ p 0 + n*1 Ø p 0 + n2 Æ n*2 n*2 Æ p 0 + n2 Ø p 0 + n*1 Æ n1

n1(n1* )

n2 (n2* )

p0

‹‰ǤͳǤͷ౨Nuclear force due to exchange of mesons

Hence, the force between a proton and neutron is carried by p+ or p – meson whereas force between two protons and two neutrons is carried by p 0. The exchange of p+ and p – between proton and neutron, the charge is carried from one nucleon to another and the proton is changed to neutron and vice versa.

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1.6

NUCLEAR STABILITY

A nucleus is stable if its identity remains same with the passage of time, that is, it does not transform itself into another nucleus without the addition of some energy from outside. This aspect is discussed in relation with the binding energy of nucleons in a given nucleus. Greater the binding energy of a nucleus, more stable it is and vice versa. The binding energy arises on account of mass defect. Unless and until an amount at least equal to the binding energy is given to the nucleus, its components (nucleons) cannot just be separated as free. Thus, every nucleus is stable against breaking up into say, two fragments, if its binding energy is more than the combined binding energy of two (stable) fragments, otherwise the nucleus is unstable and will break. Let us consider examples: 1.

92U

238

Æ

82Pb

206

+ 101H1 + 220n1

This breakup is possible only if we supply 188.68 MeV energy from outside; otherwise the uranium nucleus is stable. It essentially follows from the following facts: Total BE of Total BE of

92U

238

82Pb

= 238 ¥ 7.57 MeV = 1801.66 MeV

206

= 206 ¥ 7.83 MeV = 1612.92 MeV

We find that BE of 92U238 > BE of 82Pb206 Hence, we find that uranium nucleus as such is stable. But to realize the above breakup, we must supply an energy (minimum) equal to the difference = 1801.66 – 1612.92 = 188.68 MeV. 2.

92U

238

Æ 82Pb206 + 82He4 + 6–1e0

Now the binding energy of the products: BE of

82Pb

206

+ BE of 8 a-particles = 206 ¥ 7.83 + 8 ¥ 28.3 = 1839.38 MeV

Since BE of the product is more than the BE of uranium (1839.38 > 1801.66) the breakup is possible by itself. No energy is needed to realize this breakup. Let us discuss the experimental observation regarding stable nuclides in regard to the number of protons or neutrons. The most common practice here is to look at the plot of N versus Z. The striking feature here is the tendency of all nuclei to fall (through various emissions) into a narrow band. In the graph the expected stability line at N = Z. The actual observed stability line is N > Z. The two lines run together up to Z @ 20 (A @ 40 ) thereafter, they bifurcate (separate in two lines). Also we find that slope of the line N > Z is more than 45° after Z = 20. The attaining of stability from unstable nuclei through negative b-decay (n Æ p + e– case a), positive b-decay (p Æ n + e+ case b) and a-decay (case c) has been plotted very nearly in Fig. 1.6.

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N>Z c

a

N=

b

z lm

120 90 60 30

45° 20

40

80 60 Proton number (z)

‹‰ǤͳǤ͸౨The N versus Z graph and nuclear stability

The unstable nuclei find to fall into stability zone up to Z = 20, we have the number of neutrons equal to the number of protons. Beyond that, the stable nuclei show an excess of neutron. More number of neutrons provide stability to the nucleus (by reducing the repulsion force between protons). An all-proton nucleus, obviously cannot be stable. It has also been found nuclei with even neutrons (even protons) are more than even proton-odd neutrons or odd proton-even neutrons. The odd protonodd neutrons are less stable.

1.7

NUCLEAR MODELS

We do not have precise and detailed theory of nuclear structure so attempts have been made to correlate nuclear data in terms of various nuclear models. Thus, in order to interpret nuclear properties like stability, spin, magnetic moment, binding energy, shape, radioactivity and parity, etc., various nuclear models have been proposed. It is just like we have various models proposed for the peripheral electrons in an atom. Each model is based upon certain simplifying assumptions and has limited success. No model is capable of explaining every observed nuclear fact. The acceptance of a model depends on how much it can explain the observed behaviour of atomic nuclei. Some of the models are: 1. 2. 3. 4.

Alpha particle model (Wheeler Weizsacker) Liquid drop model (Bohr Kalcker) Shell model (Mayer Jensen) Collective model (Bohr)

Since no single model predicts all the results that are exhibited by various nuclei, simple and complex. We can only say these models are complementary.

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1.8

LIQUID DROP MODEL (Semi-Empirical Mass Formula)—C.F. Weizsacker Formula

The liquid drop model of nucleus as proposed by Bohr uses the concept of nuclear potential barrier developed by Gamow. According to this model, the nuclei of all elements behave like a liquid drop of electrically charged, incompressible liquid of constant density but of varying mass. The liquid in reality consists of neutrons and protons. The main assumptions of liquid drop model are: 1. The nucleus is supposed to be spherical in its stable state due to nuclear force just as a liquid drop is spherical due to the symmetric force of surface tension. 2. The volume as well as mass of the nuclei is proportional to the atomic mass number A, the density of all nuclei is same and constant and independent of volume. Similarly, the density of a liquid drop is constant independent of its volume. 3. The nucleons are considered to move about within a spherical enclosure represented by the nuclear potential barrier just as the molecules move about within the spherical drop of the liquid, the shape of which is determined by the properties of surface tension. 4. The two properties of nuclear forces are their short range and saturation in nature. These can be deduced from the fact that (a) The BE per nucleon is almost a constant quantity of 40 < A < 120. (b) The volume of the nucleus is also proportional to the atomic mass. 5. Just like the latent heat of vaporization is a constant for a liquid, the binding energy per nucleon is also constant. 6. The molecules evaporate from a liquid drop on increasing the temperature due to their increased thermal agitation. Similarly, when energy is given to a nucleus by bombarding it with nuclear projectiles, a compound nucleus is formed which emits nuclear radiation almost immediately. 7. When a small drop of liquid allows to oscillate, it breaks up into two smaller drops of equal size. The process of nuclear fission is similar and the nucleus breaks in two nuclei of comparable mass. Since we find that properties of the nucleus are quite similar to the properties of liquid drop. Hence, this model of nucleus is called liquid drop model.

Expression for Binding Energy The mass M of a nucleus containing Z protons and (A – Z) neutrons then Mass defect, Dm = [Zmp + (A – Z)mn – M] Binding energy, Eb = Dmc2 = [ZMp + (A – Z)Mn – M] c2

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And mass of the nucleus on physical scale, M = ZMp + (A – Z)Mn – Eb/c2

(1.1)

The total binding energy (Eb) of the nucleus is contributed by the following factors.

(a) Volume Energy. The nucleons inside the nucleus interact only with the neighbours which lie within the range of nuclear force. Thus, the total binding energy will have a contribution proportional to the volume of the nucleus which, in turn, depends upon atomic number (A). This term of BE is defined as volume energy of the nucleus. Therefore,

Eb μ A or Eb = avA

(1.2)

where av is constant having value 14 MeV. This represents the largest term in total binding energy of nucleus.

(b) Surface Energy. Equation (1.2) has been derived assuming that all nucleons are surrounded by equal number of neighbours and are equally attracted on all sides. This is true only for interior nucleons. The nucleons on the surface or near the surface of the nucleus interact only with nucleons on one side. Hence, the nucleons near or on the surface are less attractive or in other words, the nucleons on or close to the nuclear surface will have less binding energy than interior nucleons. The number of such nucleons depends upon the surface area of nucleus. As surface area, 4pR2 = 4p(R0 A1/3)2 = 4p R02A2/3 Hence, reduction in the binding energy due to the surface of the nucleus, ES = –K4p R02A2/3 = –as A2/3 where as is constant and its value is 13 MeV. Negative sign indicates that this energy contribution is in the direction opposite to the volume energy. This energy term is more significant for lighter nuclei than for heavier ones as greater fraction of nucleons lies on the surface of light nuclei. For maximum stability of a nucleus, its binding energy should be large which is possible only if the contribution of surface energy is small. Because of this reason, in the absence of external forces, a nucleus should be spherical, since for a given volume, only a sphere has the least surface area.

(c) Coulomb’s Energy. Since protons inside the nucleus possess positive charge, over and above the nuclear force of attraction, they also exert a Coulomb’s force of repulsion. The repulsive force between them tends to decrease the nuclear binding energy (or its stability is lowered). The electrostatic potential energy of a pair of protons at a distance r, U=

1 e2 1 q1q2 = 4pe 0 r 4pe 0 r 2

24പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

For Z protons inside the nucleus, there are

Z(Z - 1) proton pairs in all. 2

Therefore, Coulomb binding energy = – Potential energy of one pair ¥ Number of proton pairs, EC = –

-e 2 Z(Z1 ) 4pe 0 r 2

As Z protons are uniformly distributed within the nuclear volume (uniform distribution of charge in the nucleus) of radius R. Therefore,

where ac =

r=R EC =

- e 2 Z(Z - 1) - e 2 Ê Z(Z - 1) ˆ = 8pe 0 ÁË R0 A1/3 ˜¯ 8pe 0 R

EC =

- e 2 Ê Z(Z - 1) ˆ Z(Z - 1) = – ac ÁË 1/3 ˜ ¯ 8pe 0 R0 A A1/3

(1.3)

+e 2 = 0.6 MeV is constant. Coulomb energy is also negative as 8pe 0 R0

Coulomb force tends to disrupt the nucleus.

(d) Asymmetry Energy. In general, in heavy nuclei, the number of neutrons is more than the number of protons (N > Z). In light nuclei N = Z and these nuclei are highly stable. As the number of neutrons increases, the nucleus acquires an asymmetrical character. As a consequence, a force comes into play which reduces the volume energy and the net binding energy as well. The reduction in binding energy is directly proportional to the square of the excess of neutrons over the protons and inversely proportional to the number of nucleons. Therefore,

Ea μ

( N - Z )2 A

( N - Z )2 ( A - 2 Z )2 = –aa (1.4) A A where aa constant and its value can be taken to be 19 MeV for most common cases. Ea = –aa

(e) Even-Odd Effect (or Pairing Energy). Nucleons have a tendency to exist in pairs. Hence the nuclei with even number of protons (Z) and even number of neutrons (N) are highly stable that even Z-odd N or odd Z-even N. Also nuclear energy levels inside the nucleus are filled with certain number of nucleons obeying relevant selection rules. Taking these two effects into consideration, the term contributing towards volume energy is given by Ep = ap A–3/4

(1.5)

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We take

ap = –34 MeV for even Z-even N = +34 MeV for even Z-odd N or odd Z-even N =0

for odd Z-odd N

Weizacker’s Semi Empirical Formula Combining all the terms which contribute towards the binding energy of the nucleus, we get Weizacker’s semi-empirical binding energy formula. Total binding energy, E = EV + ES + EC + Ea + Ep E = av A – as A2/3 – aC

Z(Z - 1) ( A - 2 Z )2 – a + ap A–3/4 a A A1/3

The binding energy per nucleon = E/A = av – as

A2/3 Z(Z - 1) ( A - 2 Z )2 A-3/4 – aC – a + a a p A A A4/3 A2

This formula is known as semi-empirical mass formula for the simple reason that the values of the parameters av, as, ac, aa and ap can be obtained empirically by fitting in it the known masses of some nuclei. Volume energy Ev/A Total energy E/A

BE/A

A Surface energy ES/A

O

Asymmetry energy Ea /A Coulomb energy Ec /A

‹‰ǤͳǤ͹౨BE/A curve for various forms

Success of Liquid Drop Model 1. Stable nucleus. The stability of the nucleus can be explained using the liquid drop model. As the stability of a liquid drop is due to the force of cohesion between the molecules, similarly, the stability of the nucleus is due to the binding energy of each nucleon. To remove a molecule from a liquid drop energy has to be supplied to the drop in the form of heat, similarly energy will have to be supplied to the nucleon equal to or greater than its binding energy to remove the nucleon. Hence, the stability of the nucleus.

26പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

2. Radioactivity. As we increase the temperature of the liquid drop, thermal agitation of its molecules becomes more and more rapid and at a particular temperature evaporation may take place. A molecule in a liquid drop evaporates by gaining energy from its neighbouring molecules during the process of collision. Similarly, a nucleon or group of nucleons may leave by gaining energy from the neighbouring nucleons and show the phenomenon of radioactivity. 3. Artificial radioactivity. When a fast moving projectile particle strikes a target nucleus, it is completely absorbed by it and a compound nucleus is formed. The energy of the incoming particles is shared by all the nucleons. If the energy of the incident particle is concentrated at a point, the nucleus disintegrates and hence showing the phenomenon of artificial radioactivity. 4. It can also explain the phenomena of nuclear fission. 5. It also helps to calculate the atomic masses and binding energy of nuclei quite accurately.

Failures 1. The liquid drop model fails to explain the high stability of nuclei with magic numbers. 2. Nuclear spin and magnetic moments are also not precisely calculated. 3. Semi-empirical mass formula is in better agreement for a heavy nucleus than for light nucleus.

1.9

MAGIC NUMBER AND ITS EXPERIMENTAL EVIDENCE

It has been observed that nuclei having either the number of proton (Z) or number of neutron (N) equal to one of the numbers 2, 8, 20, 50, 82, 126 have a very high stability as compared to their neighbours. These numbers are called magic numbers.

Evidences in Support of Magic Numbers 1. The strongest evidence comes from the study of stable nuclide 2He4, 8O16 (Z = N = 2 or 8) etc., are more stable than their neighbours. The fact can be seen in relation with the binding energy curve. Similar peaks of greater stability have been noticed to occur for Pb208 (Z = 82), Ce140 (N = 82), Sn120 (Z = 50). 2. Stability is usually related with high natural abundance. From the study of relative natural abundance of nuclei in the composition of the sun, the earth, stars, etc., pronounced peaks have been found for 8O16 (N = Z = 8), Y89 (N = 50), Ca40 (N = Z = 20), Pb208 (Z = 82). 3. Relative stability of different elements is also indicated by the number of stable isotopes they have, e.g., oxygen has two isotopes. Sulphur ( 16S36: N = 20) has

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four stable isotopes, five isotopes occur in Ca40 (N = Z = 20), seven in Mo92 (N = 50), and ten isotopes occurs in 50Sn118 (Z = 50). 4. The energies of the a-particles emitted by heavy nuclei provide strong evidence for the magic numbers Z = 82 and N = 126. The nuclei with Z = 82 or N = 126 emit exceptionally energetic a-particles, resulting the daughter nuclei with N = 126 a stable nuclei. 5. Similar relations have been observed in b-decay. The energies of the emitted b -particles are exceptionally large when the number of neutrons and protons in the daughter nucleus corresponds to a magic number. As a result the end product has low energy and higher stability.

1.10

THE NUCLEAR SHELL MODEL

The shell model of a nucleus is an attempt to explain and account for the existence of magic numbers. It also seeks to explain certain other properties in terms of nuclear behaviour in a common course field. Here we consider a nucleon as a particle moving in potential well produce all remaining nucleons. The average potential well seen by each nucleon is almost the same. This way we are in a position to treat each nucleon as moving independently of the other nucleon in the potential well. Various assumptions of potential well 1. Nucleons form a closed sub-shell within the nucleus in a similar manner as the electrons do in the case of atoms. Hence, it is named shell model. 2. The nucleons which constitute the nucleus are arranged in some type of shell structure. The shell gets closed with suitable number of protons and neutrons. 3. Each nucleon behaves like an independent particle and moves under the influence central potential produced by remaining (A – 1) nucleons. Each nucleon possessing intrinsic spin angular momentum of magnitude  |S|= S(S + 1) where S = 1/2 is the spin quantum number and orbital angular  momentum of magnitude |L|= l(l + 1) where l is the orbital quantum number having values 0, 1, 2, 3…. 4. The electrons revolve in the Coulomb electrostatic field of the nucleus which is supposed to be heavy and at a very large distance. The electrons revolve in specific permitted orbits.

Shell Model The nuclear shell model has been developed on the basis of above assumptions. The model is similar to Bohr model for electrons in the extra nuclear space. By analogy with the closed sub-shells and shells in the case of atoms it is assumed that nucleons also form similar closed sub-shells and shells within the nucleus. The neutrons and

28പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

protons constituting the nucleus are supposed to be arranged in some type of shell structure and those shells get close with suitable number of protons and neutrons. Each nucleon moves independently inside the nucleus in a fixed orbit under the influence of a central field of force or a central potential V(r) produced by the average interaction between all the remaining (A – 1) nucleons. It is also assumed that the potential field is fairly constant within the nucleus and changes rapidly near the edges. Various potential shapes have been suggested to derive the magic numbers and nuclear energy levels. The simplest one are: 1. Infinite deep square well potential of the form, V(r) = –V0 for r < r0 for r ≥ r0

=d

2. Harmonic oscillator potential of the form, 1 1 Kr2 = m w2 r2 V(r) = 2 2 where w is the angular frequency of the harmonic oscillator of mass m. In onedimensional case, the energy levels are given by 1ˆ Ê En = Á n + ˜ w Ë 2¯ and in the general three dimensional case 3ˆ Ê E = Á n1 + n2 + n3 + ˜ w Ë 2¯ 3ˆ Ê En = Á N + ˜ w Ë 2¯ where n1, n2, n3 are integers specifying the wave functions and N = n1 + n2 + n3 ≥ 0 is the oscillator quantum number (N = 0, 1, 2, 3…). The above equations show that all the energy states in the harmonic oscillator model are equally spaced.  Since each nucleon is supposed to have an orbital angular momentum, |L|= l(l + 1) where l is defined as the orbital quantum number having values 0, 1, 2, 3, … . Another quantum number, similar to principle quantum number of electronic orbit, characterizes the radial part of the nuclear wave function and is denoted by n. The values of n are 1, 2, 3, … . The oscillator quantum number related to orbital quantum number l and quantum number n, N = 2(n – 1) + l = (2n + l – 2) The energy levels corresponding to each value of l for the nucleon are represented by specific spectroscopic rotations as l=0 1 Spectroscopic rotation s p

2 d

3 f

4 g

5 h

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The energy level for a harmonic oscillator potential for N = 0 to 4 with the maximum number of nuclei is given below: N

n

l = (N + 2 – 2n)

State

Number of nucleons 2(2l + 1)

Energy

0

3 w 2 2...........……………. Not allowed state as l is negative ...........…………….

1

1

1

1p

6

1

5 w 2 2...........……………. Not allowed state as l is negative ...........…………….

2

1

2

1d

2

2

0

2s

10 Ô¸ ˝ 2 Ô˛

3

1

3

1f

0

3 3 4 4 4 4

1

0

1s

2

14 ¸Ô ˝ 6 ˛Ô

12

7 w 2 9 w 2

20 2 1 2p 3...........……………. Not allowed state as l is negative ...........……………. 1 4 1g 18 ¸ Ô 11 w 2 2 2d 10 ˝ 30 2 Ô 3 0 3s 2˛ 4...........……………. Not allowed state as l is negative ...........…………….

Hence, we find that shells are complete at numbers 2, 8, 20, 40, 70, 112… . This shows that all the magic numbers are not predicted but only the first three lower magic numbers are available. Also if N ≥ 2 we find that different states occur with same energy or states degenerate spin-orbit coupling. The fact that harmonic oscillator potential function predicts only the first three magic numbers but not higher ones, led many scientists to conclude that the shell model was incapable of explaining the magic number. Mayer, Haxel and Jensen suggested a modification, assuming that: 1. There is a strong spin-orbit coupling in the nuclei (i.e., there is large energy dependence on the relative orientation of spin and orbital angular momentum.) 2. The spin orbit splits each energy level into two sub-levels. The sub-level of higher total angular momentum (j = l + 1/2) has less energy than the sublevel corresponding to j = l – 1/2. Thus, the force acting on the nucleons in a nucleus should include a non-central component. This non-central force arises due to interaction between orbital angular momentum (l) and spin angular

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momentum (s). This spin-orbit interaction causes a splitting of energy level into two sub-levels giving them different periodicity. The gap of energy between two sub-levels is proportional to l value. Each energy level will split into (2j + 1) sub-states for given value of j. The energy levels associated with intermediate potential without spin orbital interaction and with spin orbital interaction are shown in Fig. 1.8. Total nucleons S2(2l + 1)

Nucleon per level 2(l + 1)

26

1i (l = 6)

6

3P (l = 1)

Total nucleons Nuclear per level (2j + 1) S(2j + 1) 1 1 12 2 1/2

112 14 22

2f (l = 3)

1h (l = 5)

2

5/2 6 3/2 4 13/2 14

126*

9/2 10 7/2 8

Energy (Not to scale)

11/2 12 3/2 4 2 70 10

3s (l = 0) 2d (l = 2)

18

1g (l = 4)

6

2p (l = 1)

10

1f (l = 3)

40

2

2s (l = 0)

20 10

1d (l = 2)

1/2

2

7/2

8

82*

5/2

6 9/2 10 1/2 2 5/2

6

3/2

4

7/2 3/2 1/2

8 4 2

50*

20* 5/2 1/2 6

2

6 2

1p (l = 1)

1s (l = 0) (a)

8* 3/2

4

1/2

2

2*

(b)

‹‰ǤͳǤͺ౨Single particle energy levels of the nuclear shell model (a) without spin-orbital interaction and (b) with such an interaction. Degeneracy in (a) is broken under (b) except for l = 0 (s-state)

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Achievements 1. The shell model explains the existence of magic numbers. From Fig. 1.8 we find that shell closures occur at nucleon number 2, 8, 20, 50, 82 and 126 which correspond to magic number. 2. The shell model successfully explains the ground state spins and magnetic moments of nuclei. It allows us to predict nuclear angular momentum, e.g., the angular momentum of a closed shell is zero as per Pauli’s exclusion principle. 3. The close shell structures (corresponding to N or Z equal to a magic number) explain the great stability and high binding energy of nuclei. 4. This model also explains the phenomenon of nuclear isomerism.

Failures 1. It fails to explain the spin values of a few nuclei. 2. Shell model is not able to explain quadrupole moment of a few nuclei. 3. Shell model fails to explain the magnetic moment of a few nuclei.

1.11

FERMI GAS MODEL

Fermi gas model is a statistical model which assumes the nucleus as a degenerate gas of neutrons and protons. It is considered that it is like a free electron gas in a metal. We consider a gas as degenerate because in this the particles are crowded into low energy state as satisfied by Pauli’s exclusion principle. As we know neutrons and protons move freely within the nuclear volume and the binding potential is generated by the nucleons. Nucleons are fermions having half spin. Fermi-Dirac statistics determine the nature of neutrons and protons.

Proton potential

Neutron potential

Protons

Neutrons

B¢

p

Ef

n

Ef

‹‰ǤͳǤͻ౨Nuclear potential well

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In the first assumption we consider two separate potential wells for protons and neutrons as they have distinguishable fermions. The nuclear potential well is supposed to be rectangular and it is constant inside the nucleus and stops sharply at the end. The neutron potential well is slightly deeper than the proton potential well because of the missing Coulomb repulsion. This concludes that there are more neutron states present, i.e., N > Z hence, heavier the nuclei become. In gas at 0 K maximum energy level is defined as Fermi energy (Ef) and each level is occupied by particles of opposite spin, i.e., one with spin up and other with spin down. It is supposed that the nuclear temperature is quite low so the nucleons will occupy the lowest available state. According to Fermi–Dirac statistics the number of particle states, 2V n= (2p )3 n=

Pf

Úd P 3

o

VPf3 2 3

3p 

where Pf is Fermi momentum and is defined as nˆ Ê Pf =  Á 3p 2 ˜ Ë V¯

1/3

Ê 3 n ˆ 1/3 = hÁ Ë 8p V ˜¯

where V is the volume of the nucleus and Fermi momentum for proton and neutron is defined as Pf, p =

Ê Z ˆ 1/3 ˜ ÁË 9p 4A¯ r0

Pf , n =

Ê N ˆ 1/3 ˜ ÁË 9p 4A¯ r0

Let us consider N = Z = A/2 therefore Pf , p = Pf , n =

 Ê 9p ˆ Á ˜ r0 Ë 8 ¯

1/3

Assuming  = 197 MeV we get Pf , p = Pf , n = 297/r0 MeV/c Therefore, Fermi energy for r0 = 1.2 fm is given as Ef =

Pf2, p 2m

=

Pf2, n 2m

= 33 MeV.

This corresponds to the kinetic energy of the highest orbit. The average binding energy is 7-8 MeV therefore the depth of nuclear well @ 33 + 8 = 41 MeV. This statistical model helps to explain the properties of nucleus in excited states. This model is quite helpful in explaining the phenomena which are sensitive to high momentum of the nucleon spectrum.

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Numerical Problems Problem 1

Predict the characteristics of the ground states of 8O17 and 17

29Cu

63

.

17

Solution: 1. 8O : According to shell model the terms of the nuclide 8O are given as: (1S1/2)2, (1p3/2)4, (1p1/2)2, (1d5/2)1. The number of particles that can be accommodated in each term is indicated by superscripts. As all the shell/sub-shell except the lost are completely filled, the total angular momentum of the nucleus j = 5/2. For the d state l = 2 Parity. Parity = (–1)l = (–1)2 = +1 Parity is even. Magnetic moment. The magnetic dipole moment of a nucleus is given by the relation 1 μ = (j – 1/2)g l + gs for l = j – (a) 2 Ê j ˆ 1 (b) and μ= Á ÈÎ( j + 3/2) gl - g s ˘˚ for l = j + ˜ 2 Ë j + 1¯ where g l = 1 for proton and g l = 0 for neutron. The magnetic moment of the last odd neutron for which l = 2 and j = 5/2, l = j – 1/2 is given by the Eq. (a). e . Therefore, μ = –1.91 bN , where, bN is nuclear magneton = 2m Electric quadrupole. The electric quadrupole moment of a nucleus is given by 1 3 Ê 2 j + 1ˆ 2 3 Ê 2 j + 1ˆ 3 )2 = Q= - Á R ( R A 0 5 Ë 2 j + 2 ˜¯ 5 ÁË 2 j + 2 ˜¯

R0 = 1.2 ¥ 10–15 m and j = 5/2, A = 17

where

1 3 Ê 4ˆ Therefore, Q = - Á ˜ (1.2 ¥ 10 -15 ¥ 17 3 )2 = –0.0326 ¥ 10–28 m2 5Ë 7¯

Q = 0.0326 barn. 2.

29Cu

63

: For 2

29Cu

63

the terms of the nuclides are given as: 4

(1s1/2) , (1p3/2) , (1p1/2)2, (1d5/2)6, (2s1/2)2, (1d3/2)4, (1f7/2)8, (2p3/2)1 As all the shells/sub-shells except the last one are completely filled the total angular momentum of the nucleus j = 3/2. For p state l = 1. Parity. Parity = (–1)l = (–1)1 = –1, parity is odd Magnetic dipole moment. Since j = 3/2, l = 1, l = j – 1/2 The magnetic dipole moment is also given by Eq. (a). Therefore, for last proton m = (j – 1/2 )g l + gs = (3/2 – 1/2)(1) + 2.79 = 2.79 bN

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Quadrupole moment,

1 3 Ê 2 j - 1ˆ 2 3 Ê 2 j - 1ˆ (R0 A 3 )2 R =- Á Q= - Á ˜ ˜ 5 Ë 2 j + 2¯ 5 Ë 2 j + 2¯

3 2 = - ¥ [1.2 ¥ 10–15 ¥ (63)1/3]2 5 5 = – 0.0546 ¥ 10–28 m2 = –0.0546 barn. Problem 2 Show that the number density of 1H1 is about 1014 times greater than the atomic density. Assuming the atom to have the radius of first Bohr orbit. Solution: Radius of the nucleus = 1.3 ¥ 10–15 Therefore, nuclear density =

mp

=

4 p R3 3

mp 4 3 p (1.3) ¥ 10 - 45 3

R = Atomic radius = Radius of first Bohr orbit = 5.29 ¥ 10–11 m mp mp = Atomic density = (me @ 0) 4 3 4 3 pr p ( 5.29) ¥ 10 -33 3 3 Nuclear density ( 5.29) ¥ 10 -33 1.448 ¥ 1014 = @ 1014 = 3 -45 Atomic density 2 . 197 (1.3) ¥ 10 3

Therefore,

Hence, density of nucleus is 1014 times atomic density. Problem 3 Assuming the nucleons inside a nucleus as forming a Fermi gas, calculate the typical nuclear potential depth? Solution: Let us consider that the nucleons move freely inside a spherical volume V of radius R. In the ground state of the nucleus, all the allowed energy levels up to the Fermi level (Ef) all the levels are empty. The Fermi energy is given by a relation 2

h 2 Ê 3n ˆ 3 Ef = Á ˜ 2m Ë 8 ¯ Here the nucleon density (n) =

n=

A Mass = 4 Volume p r3 3

(

3A

4p R0 A

3 ˆ h2 Ê 3 ¥ Therefore, Ef = Á 2m Ë 8p 4p R03 ˜¯

2/3

)

1/3 3

=

3 4p R03

9 ˆ h2 Ê = Á 2m Ë 32p 2 R03 ˜¯

2/3

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=

h2 Ê 9 ˆ Á ˜ 8m Ë 4p 2 ¯

2/3

.

1

(R0 = 1.2 ¥ 10–15 m)

R02

2

(6.62 ¥ 10 -34 )2 Ê 9 1 ˆ3 Therefore, Ef = . -27 Á 2˜ 8 ¥ (1.66 ¥ 10 ) Ë 4(3.142) ¯ (1.2 ¥ 10 -15 )2

= 65 ¥ 10–12 J = 40 MeV. Also there is an additional factor (the binding energy of each nucleon) whose average value is about 8 MeV. Thus, the nucleons are at the depth = 40 + 8 = 48 MeV. Thus, the depth of the potential well is about 48 MeV. Problem 4 Calculate the atomic number of the most stable nucleus for a given mass number A on the liquid drop model. Hence, explain why out of 2He6, 4Be6, and 3Li6 only 3Li6 is stable? Solution: Stable nucleus: According to the liquid drop model, the binding energy (Eb) is given by

( A - 2Z ) Z2 - aa + ap A-3/4 1/3 A A 2

Eb = avA – asA2/3 – ac

The most stable nucleus for a given mass number A is that which has maximum value of binding energy i.e.,

Eb = max then

dEb =0 dZ

dEb ( A - 2Z ) Z = 0 - 0 - 2ac 1/3 + 4 aa +0=0 dZ A A or

-2ac

(Z ) 1/3

A

+ 4 aa

( A - 2Z )

4 aa

A

( A - 2Z )

4 aa -

A

=0 = 2ac

(Z ) A1/3

8 aa Z (Z ) = 2ac 1/3 A A

Z(4aa + acA2/3) = 2aaA Z=

2 aa A 4 aa + ac A

Since ac = 0.7053 and aa = 23.702 MeV

2 3

=

A ac 23 2+ A 2 aa

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A . For light nuclei 0.015 A2/3 @ 0 (neglected) 2 + 0.015 A2/3

Therefore, Z =

Therefore, Z = A/2 for stable nuclei. The result is confirmed experimentally. Out of 2He6, 4Be6, and 3Li6 it is only 6 3Li for which A = 6 and Z = 3 satisfied the relation Z = A/2. Therefore, it is the only stable nucleus out of given three. Problem 5 Using the semi-empirical binding energy formula. Calculate the binding energy of 20Ca40? Solution: The semi-empirical binding energy formula is

( A - 2Z ) Z2 + ap A-3/4 Eb = avA – asA – ac 1/3 - aa A A where av = 14 MeV, as = 13 MeV, ac = 0.60 MeV, aa = 19 MeV, ap = –34 MeV 2

2/3

As A is even 40 and Z is even = 20 avA = 14 ¥ 40 = 560 MeV

Therefore,

asA2/3 = 13 ¥ (40)2/3 = 13 ¥ 11.696 = 152 MeV ac aa

Z2 20 2 = 0.60 ¥ = 0.60 ¥ 116.9 = 70 MeV A1/3 ( 40)1/3

( A - 2 Z )2 A

= aa(0) = 0

ap A-3/4 = –34 ¥ 40–3/4 = –34 ¥ 0.063 = –2.14 MeV Therefore,

Eb = 560 – (152 + 70 + 2.14) = 335.86 MeV

Binding energy per nucleon of Problem 6

20Ca

40

= 335.86/40 = 8.39 MeV/nucleon

Which nuclei would expect to be more stable 3Li

7

or 3Li8; 4Be9, 4Be10 ?

Solution: For a given mass A, the atomic number (Z) of the most stable nucleus is given by A (Derived in Problem 4) Z= 2 + 0.015 A2/3 For A = 7; Z = A = 8; Z =

7 2/3

=

7 = 3.4 2.055

2/3

=

8 = 3.88 2.060

2 + 0.015 (7 ) 8 2 + 0.015 (8 )

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Out of two values of Z for stability 3.4 and 3.88; 3.4 is nearer to 3 therefore 7 is more stable.

3Li

For A = 9; Z = A = 10; Z =

9 2 + 0.015 (9)

2/3

=

10 2 + 0.015 (10 )

2/3

9 = 4.36 2.065 =

10 = 4.80 2.067

Out of two values of Z for stability 4.36 and 4.8; 4.36 is nearer to 4 therefore 9 is more stable.

4Be

Summary 1. There is a central part of the atom called the nucleus. Rutherford and co-workers have established the existence of nucleus by observing non-uniform scattering of a-particle. 2. The volume of the nucleus is very small as compared to the volume of the atom. 3. Initially, electron-proton hypothesis stated that the atomic nucleus consists of electrons and protons. The radioactive emission of b-particles were the reason behind such a hypothesis. 4. But e-p theory could not explain certain known facts. It was simply rejected. 5. Later it was proposed that nucleus consisted of protons and neutrons (exception: hydrogen). The emission of b-particles were easily explained in term of interconversion of neutrons and protons (n Æ p + e– + n ). The emission of a -particle could be due to the combination of two protons and two neutrons. It may exist as such in the nucleus or it may be instantly formed just before its emission. 6. Size of the nuclei depends upon the atomic mass of the nucleus (R a A1/3). Hence, heavier nuclei are bigger than the lighter ones. But density of nuclear matter is same for all nuclei (as it is independent of atomic mass number, A). 7. The density of nuclear matter is very large ( @ 1017 kg m–3) as compared to atomic density. 8. In fact, density distribution is not uniform throughout a given nucleus, it varies. It is maximum near the centre and its value decreases with distance, r. Near the surface (r = R) the density reduces to half of its value near the centre. 9. As the nucleus has no well defined boundary. Hence, the radius of the nucleus can also be defined as the distance from the centre of the nucleus where the density of the nucleus falls to half its maximum value. 10. The masses of nucleons and the nucleus conveniently expressed in terms of atomic mass unit (amu) 1 amu = 1.66 ¥ 10–27 kg. 11. Protons and neutrons are held together inside the nucleus despite the existence of strong repulsive electrostatic force that operates between two proton by a strong attractive force known as nuclear force.

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12. Nuclear forces are the strongest forces, being extremely short ranged (r = range @ 10–14 m), charge independent, spin dependent, attractive in nature and saturated. 13. Nuclear force is not a central force and does not obey the inverse square law 1 ). Hence, it is not a conservative force. (or FN a rT 14. As per meson theory proposed by Yukawa, the nuclear force is due to the fast exchange of p-meson (p +, p –, p 0). 15. Relative magnitude of nuclear force is about 102 times more than electrostatic 16. 17. 18. 19.

20. 21.

22. 23. 24. 25.

26. 27. 28. 29.

force and 1038 times the gravitational force. A nucleus has also been associated with properties like spin, magnetic moment, electric quadrupole moment, etc. Some other properties (not strictly physical) are also associated with the atomic nucleus. These are statistics, parity, packing fraction and binding energy. The behavior of a nucleus as a quantum mechanical system is subjected to Heisenberg’s uncertainity principle, and the law of probability. Binding energy per nucleon (BE/A) curve explain the stability of nuclides. It also explained why even-even nuclei (even proton and even neutron) are metastable than their immediate. Binding energy per nucleon curve also explains the stability aspect of both nuclear fission or nuclear fusion. There is no theory of the nucleus which can put together the available information about nuclei into a unified knowledge we do not have, on the other hand, a well known quantum mechanical theory (model) of atomic structure. How exactly are the various nucleons arranged within the nucleon structure and with what consequences, is a complicated problem. Understanding the behavior of a nucleus is the central problem in nuclear physics. The nuclear forces and the related interaction potential for the nucleons are believed to be responsible for the existence/behavior of a nucleus. Various nuclear models have been proposed to interpret nuclear properties like stability, spin, magnetic moment, binding energy, shape, radioactivity, parity, etc. Some of the prominent nuclear models include, the alpha particle model, the liquid drop model, the shell model and the collective model. The acceptance of a model is closely linked to its explaining the observed nuclear behaviour. No single model can explain everything. Nuclear models are different from atomic models. N. Bohr’s liquid drop model treats the atomic nucleus similar to a liquid drop for some well-known facts. Liquid drop model explains ‘nuclear stability’, ‘radioactivity’ in certain nuclei, nuclear fission, etc. This model does not explain

EƵĐůĞĂƌ^ƚƌƵĐƚƵƌĞĂŶĚ'ĞŶĞƌĂůWƌŽƉĞƌƟĞƐŽĨEƵĐůĞŝപ39



30.

31. 32.

33.

34.

35.

36. 37.

38. 39.

40. 41.

the existence of magic number, nuclear spin and magnetic moments. It is better approximation for heavy nuclei only. Weizsacker’s semi-empirical mass formula has close resemblance to Bohr’s liquid drop model. It actually incorporates some extra factors vis-à-vis Bohr’s model. Masses as well as binding energy of a large number of stable/unstable nuclei have been successfully predicted by this formula. Binding energy as well proton-neutron ratio in a given nucleus are closely linked to whether a nucleus is stable or not. Out of only 274 stable nuclides, nearly 60% (or 162) are even-even type nuclei, 20% (or 55) are even-odd nuclei, 20% (or 53) are odd-even nuclei and only 4 are odd-odd nuclei. Hence, we can say that a nucleus in which the nucleons of same kind occur in pairs is particularly stable and a nucleus that contains an unpaired proton or unpaired neutron is much less stable. Various interaction potentials for nucleons have been tried while solving the time-independent Schrödinger’s wave equation for motion of a particle in a potential well. These include square well potential, harmonic oscillator well potential and wood-sexon potential, etc. The nuclei orbits are the solutions of this Schrödinger’s wave equation, characterized by quantum numbers n, l, ml. The energy profiles of neutrons and protons (as they fill up energy levels) are similar but not exact, since N π Z, in general. The energy level sequence predicted for magic numbers did not exactly match with the observed one. There was some discrepancy. Mayer and Jensen proposed that it was necessary in shell model to incorporate a spin orbital interaction (L–S or J–J coupling between the interacting nucleons). Such n interaction leads to a splitting of energy levels. The discrepancy was thus eliminated. Existence of magic numbers was explained. This model has some drawbacks too. While light nuclei obey L–S coupling, the heavier nuclei obey J–J coupling. The nuclides with N and/or Z equal to 2, 8, 20, 50, 82, 126 show a marked level of stability. These numbers are called magic numbers. Definite experimental evidences exist for these numbers. Magic numbers represent “closed shells” of protons or neutrons. Nucleons also obey Pauli’s exclusion principle. The nuclear spin-orbital interaction does not arise from magnetic forces associated with the spin and orbital motion. But it arises from the spin dependence of strong nuclear force.

40പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

Short Answer Questions 1.

What is the ratio of nuclear density of two nuclei having mass number in 1 : 4 ratio?

Ans. Nuclear density is independent of atomic mass of the nucleus. Since two nuclei having mass number ratio 1 : 4 has the same nuclear density. So ratio of nuclear density is 1 : 1. 2.

What holds the nucleons together inside the nucleus?

Ans. Since the forces among the nucleons are electrostatic forces of repulsion, gravitational force of attraction, and nuclear force of attraction. As the magnitude of nuclear force is 100 times more than electrostatic force. Hence, it is the nuclear force which holds the nucleons together inside the nucleus. 3.

Why it is said that nuclear force is saturated in nature?

Ans. A nucleon inside the nucleus experiences force only due to its nearest neighbouring nucleons and forms bonds only with them and not with next nucleons. Because of this nuclear force is aid to be saturated in nature. 4.

Why the density of a nucleus is more than that of the atom?

Ans. The size of the nucleus is of the order of 10–14 m whereas the size of the atom is of the order of 10–10 m. The mass of the atom increases only by a small amount (additional mass is the mass of orbital electrons). For this reason, the density of nucleus is very large as compared to the density of atom. 5.

What do you mean by charge-independent nuclear forces?

Ans. For a given inter-nuclear separation, it is found that the force between two protons or two neutrons or between a proton and a neutron is practically of equal magnitude (attractive). Hence, it does not matter whether or not the interacting particles are charged or not. 6.

Why do we get maxima and minima peaks in BE curve?

Ans. The five peaks in the BE curve corresponds to 2He4, 4Be8, 6C12, 8O16, and 10Ne20 nuclei. All these nuclei are more stable than their neighbours. None of these nuclei has any unpaired nucleon in them. 7.

What was the need for various nuclear models?

Ans. In the absence of precise and detailed theory of nuclear structure, attempts have been made to correlate various observations of nuclei. Such logical attempts led to certain models for understanding nuclear behaviour. 8.

Why was the name liquid drop model assigned to the model proposed by Bohr and Kalikar?

Ans. This model takes the atomic nucleus as a drop of a liquid. There are certain compelling analogies between the two (see section 1.8, Liquid Drop Model, p. 22).

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9.

Why is the semi-empirical mass formula so called?

Ans. This formula gives an expression for the nuclear mass (MZ, A) in terms of the masses of protons, neutrons and the net binding energy of the nucleus. The values of the parameters av, as, aL, aa and ap corresponding to the terms due to the volume, surface, Coulomb, asymmetry and pairing effects are obtained by empirical fitting of certain known masses in this formula. 10.

A deuteron has a proton and a neutron bound by a strong attractive force. To separate the two, a g -ray photon of 2.2 MeV is required explain?

Ans. In general, whenever a certain number of nucleons are brought closer within a distance @ 10–14 m, they form a nucleus. In the process, there is a net loss of mass (mass defect) and hence, an equivalent amount of energy is lost by the nucleus (E = Dmc2). Unless an equal amount of energy (the binding energy) is provided to the nucleus, the bound nucleons won’t separate to be free. This energy is 2.2 MeV in the case of deuteron and is different for different nuclei. 11.

What are magic numbers?

Ans. It has been found that nuclei for which Z and (or) N value are equal to any of the value 2, 8, 20, 28, 50, 82, 126 are more stable than others. These numbers are popularly known as the magic numbers. 12.

Why should a heavier nucleus contain more neutrons than protons?

Ans. Positively charged protons inside a nucleus have a strong Coulomb’s force of repulsion between them. Neutrons being an electrically neutral particle, provide a sort of shielding and also experience strong nuclear force. Hence, a nucleus having greater number of neutrons does not allow repulsive force to dominate. As such N > Z lends stability to a nucleus to an extent. 13.

What are main successes of the liquid drop model?

Ans. The liquid drop mainly explains:

14.

1. The nuclear stability 2. Radioactivity shown by certain nuclides 3. Nuclear fission of certain nuclides like 92U235. What are the drawbacks of the liquid drop model?

Ans. It does not explain:

15.

1. The existence of magic numbers 2. Precisely the nuclear spin and magnetic moment 3. The variation in the experimentally observed masses and those predicted by the semi-empirical mass formula particularly for light nuclei. What are the achievements of the shell model?

Ans. The main achievements of the shell model are: 1. Explains the existence of magic numbers

42പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

16.

2. Pediction of ground state nuclear spins and magnetic moments 3. Explain the closed shell structure of nuclei leading to greater stability and high binding energy 4. Understanding of nuclear isomerism. What are the limitations of shell model?

Ans. The shell model do not explain 1. The correct values of total angular momentum of some nuclei. 2. The high stability of the four (odd-odd) nuclei 1H2, 3Li6, 5B10 and 7N14 properly. 3. The electric quadrupole moments of nuclei correctly. 4. Why the energy levels of the first excited states in even-even nuclei are lower than those expected from single particle excitation.

Long Answer Questions 1.

2. 3.

4.

5. 6. 7.

8. 9.

Explain the term: mass defect, binding energy and atomic mass unit. Discuss how BE per nucleon varies with the mass number. What is the significance of BE/A curve. Outline the main features of the average BE per nucleon versus the mass number (A) curve. What do you learn about nuclear fission from it? Explain the terms: 1. Nuclear spin 2. Nuclear magnetic dipole 3. Nuclear density 4. Size of the nucleus What is the physical meaning of binding energy and binding energy per nucleon? How does average BE per nucleon vary with the mass number? How do you account for stability of the nucleus? What is the difference between mass defect and packing fraction as applied to a nucleus? What is nuclear force? In what respect does it differ from electrostatic and gravitational force? What do you mean by the following in relation to the nuclear forces? 1. Charge-independence of the nuclear force 2. Spin-dependence of nuclear force 3. Saturation of nuclear forces. Discuss the wave-mechanical properties of nucleus. What is the relation between binding energy and mass defect? Show how binding energy changes with mass number.

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10. 11. 12. 13. 14. 15. 16. 17.

18.

Discuss nuclear charge and nuclear density. Show that nuclear density is independent of the mass number? Justify with examples that the nuclear forces are charge independent and spin dependent? Explain the term binding energy, mass defect, and packing fraction of atomic nuclei? Discuss the basic assumptions of liquid drop model of the nucleus. Discuss how this model is used to estimate the semi-empirical mass formula? What do you understand by magic numbers? Discuss the evidence for their existence? Give a brief outline of shell model and obtain schematic energy level diagram with the inclusion of spin-orbit splitting? What are the main assumptions of the liquid drop model of the nucleus? Give the significance of the various terms of the semi-empirical mass formula? Mention the main postulates of liquid drop model of the nucleus. Discuss the contribution of various terms to the BE of nucleus. Draw these on a graph where does it fail? Discuss the Shell model of the nucleus? What are its merits and demerits?

Multiple-Choice Questions 1. The nucleus consists of (a) neutrons (b) protons (c) neutrons and protons (d) electrons and neutrons 2. Nucleus is (a) positively charged (b) negatively charged (c) neutral (d) both (b) and (c) 3. Proton has the mass (a) 1637 times of an electron (b) 1737 times of an electron (c) 1837 times of an electron (d) 1937 times of an electron 4. In a neutral atom, the electrons are bound to the nucleus by (a) magnetic force (b) electrostatic force (c) friction force (d) centripetal force 5. Atom has approximate diameter of (b) 10–11 m (a) 10–12 m –10 (d) 10–14 m (c) 10 m 6. In isotope (a) Number of electrons = Number of protons (b) They have different chemical properties (c) They have different atomic numbers (d) They have different mass numbers

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7. The difference in the mass of the resultant nucleus and the sum of the masses of two parent nuclear particles is known as (a) Mass defect (b) Solid defect (c) Weight defect (d) Nucleus defect 8. The atomic number is equivalent to which of the following? (a) The number of neutrons in the atom. (b) The number of protons in the atom. (c) The number of nucleons in the atom. (d) The number of a-particles in the atom. 9. Which of the following particles has the smallest mass? (a) Proton (b) Electron (c) Neutron (d) Nucleus 10. Which of the following statements about the mass of an atom is true? (a) It is evenly divided between the protons and the orbiting electrons. (b) It is evenly divided between the nucleons and the orbiting electrons. (c) It is concentrated in the electron cloud. (d) It is concentrated in the nucleus. 11. When a nucleus is divided into its constituents, energy is (a) Created from nothing. (b) Destroyed into nothing. (c) Transformed into visible light. (d) Absorbed by the nucleus which then breaks it apart. 12. When any nuclei having different mass number while same charge number is called (a) Isobar (b) Similar electron (c) Isotope (d) Similar proton 13. How many neutrons uranium-235 has in the nucleus? (a) 231 (b) 100 (c) 143 (d) 243 14. The density of atom is uniform, who proposed this law? (a) Rutherfords model (b) Bohrs model (c) J.J.Thomson model (d) None of these 15. Which of the following is correct during fusion of hydrogen into helium: (a) Mass is increased (b) Mass is reduced (c) Energy is absorbed (d) Energy is released 16. How do we define the mass number? (a) Number of protons in a nucleus (b) The complement of the atomic number

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17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

(c) The number of protons plus the number of neutrons in the nucleus (d) Number of neutrons in a nucleus Number of protons which is inside of the nucleus represents (a) Atomic mass (b) Atomic number (c) Atomic count (d) Radioactivity level Positive charge in an atom was intense in a small region called (a) Atomic mass (b) Electron (c) Proton (d) Nucleus Which of the following is correct in the case of one atomic mass unit (AMU) is equal to (b) 1.66 ¥ 10–24 g (a) 1.66 ¥ 10–20 g (d) 1.66 ¥ 10–26 g (c) 1.66 ¥ 10–22 g What is the mass on neutrons? (a) 1839 times of an electron (b) 1739 times of an electron (c) 1639 times of an electron (d) 1939 times of an electron Quadrupole moment for prolate spheroid (ellipsoid) nucleus is (a) Positive (b) Negative (c) Zero (d) Yet unknown The nucleus radius is proportional to (a) A (b) A1/3 (c) A2/3 (d) A1/2 The existence of tiny, massive and positively charged nucleus as confirmed from the experiment on (a) The passage of X-ray through matter (b) The estimation of electron-proton distance in hydrogen atom (c) Exceptionally non-uniform and large angle of scattering of a-particle by matter (d) b -disintegration Nuclear forces are (a) Short range forces (b) Long range forces (c) Short as well as long range forces (d) Yet not known at all Magnitude of nuclear forces between p – p (fpp), n – p (fnp π fpn) and n – n (fnn) have the following relation (b) fnn > fpp > fnp (a) fnn = fnp = 0 and fpp > 0 (c) fnn = fpp = fnp (d) fnn < fpp < fnp The nucleus of an atom of heavy elements contains (a) electrons and mesons (b) protons and neutrons

46പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

27.

28.

29.

30.

31.

32.

33.

34.

35.

(c) electrons and neutrons (d) protons, neutrons and photons The BE group shows a cyclic recurrence of peak of stability at mass number interval of (a) 3 (b) 4 (c) 5 (d) 6 The pair of nuclides 20Ca40 and 19K39 are (a) Isotopes (b) Isobars (c) Isotones (d) Isomers Mirror nuclei are those nuclei for which (a) There are same number of protons (b) There are same number of neutrons (c) The number of protons is equal to the number of neutrons (d) The number of protons in one equals the number of neutrons in the other Parity is not conserved in (a) a-decay (b) b-decay (c) g -decay (d) All of the above Nuclear fission can be best explained by (a) The shell model (b) The liquid drop model (c) The Thomson model (d) None of the nuclear model Semi-empirical mass formula as given by (a) Einstein (b) Bohr (c) Weizsacker (d) Jenson Nuclei with magic numbers are (a) Unstable (b) Stable (c) Metastable (d) Stable and unstable depending upon conditions Magic numbers are certain numbers which are (a) Equal to the Z-value of a nucleus (b) Equal to the N-value of a nucleus (c) Equal to the Z and/or N-value of a nucleus (d) Pure magic in relation to some nuclei In the liquid drop model, it is assumed that (a) Nucleons do not interact with each other (b) Nucleons exert an influence on one another only at close range (c) Nucleons exert an influence on one another only at large range (d) None of the above is a valid reason

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36. The shell model was proposed by (a) N. Bohr (b) Weizsacker (c) Mayer (d) A. Bohr 37. The example of nuclear fusion is (a) The formation of water from hydrogen and oxygen (b) Formation of barium and krypton from uranium (c) The formation of helium from hydrogen (d) None of the above 38. If m is the mass of electron, the value of Bohr’s magneton (in SI) is given by (a)

eh 4p m

(b)

eh 4p mc

(c)

e 4p m

(d)

e2 h 4p mc

39. In an atom of size ª 10–10 m, the electron revolves around the nucleus of size ª 10–15 m. The space between nucleus and electron contains (a) Air at low pressure (b) Air at normal pressure (c) Ether (d) Vacuum 40. Neutron has (a) Positive magnetic moment (b) Negative magnetic moment (c) Sometime positive and sometime negative (d) No magnetic moment

Answers to Multiple-Choice Questions 1. 8. 15. 22. 29. 36.

(c) (b) (d) (b) (d) (c)

2. 9. 16. 23. 30. 37.

(a) (b) (c) (c) (b) (c)

3. 10. 17. 24. 31. 38.

(c) (d) (b) (a) (b) (a)

4. 11. 18. 25. 32. 39.

(b) (d) (d) (c) (c) (d)

5. 12. 19. 26. 33. 40.

(c) (c) (b) (b) (b) (b)

6. 13. 20. 27. 34.

(d) (c) (a) (b) (c)

7. 14. 21. 28. 35.

(a) (c) (a) (c) (b)

2 Nuclear Reactions

2.1

INTRODUCTION

The knowledge about the structure of the nuclei comes from the experiments in which a particular nuclide is bombarded with a beam of high energy particles or gamma rays. During this process of bombardment, the projectile or incident particle interacts with the nuclei of the target in a number of ways. New nuclei and particles may be formed; the analysis of which gives the information regarding the structure of atomic nuclei. The first artificial nuclear reaction was performed by Rutherford in 1919. He bombarded nitrogen nuclei with a-particle (2He4) from a radioactive source and showed that protons were given out according to the reaction 7N

14

+ 2He4 Æ 8O17 + 1H1

In this reaction, 7N14 nucleons called target nucleus with an a-particle referred to as incident particle or projectile; a proton (p) is ejected out and an oxygen (8O17) nucleus known as recoil or product nucleus is formed. Symbolically, the above reaction can also be written as 7N14 (a, p) 8O17. Hence, a reaction in which a certain species after certain suitable bombardment undergoes a change of character is known as nuclear reaction. There are many types of nuclear reactions depending upon the nature of the particle involved such as a, b, proton, deuteron, a-particles, neutrons, electrons, etc.

2.2 2.2.1

TYPES OF NUCLEAR REACTION Nuclear Particle Reaction and Scattering

A particle in the incident beam may come close enough to one of the target nuclei so that the reaction causes expulsion of the same kind of particle from the target nucleus. The process is called scattering. Scattering, elastic or inelastic depends upon whether the kinetic energy is conserved or not. Alternatively, the incident particle may penetrate the nucleus and interaction may therefore cause emission of a different kind of particle or more than one particle. This process is called reaction of transmutator, since target nucleus has been transmuted into different nuclides.

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A familiar example of elastic scattering is that of a-particles from the gold foil (Rutherford’s a-ray scattering experiment). It is expressed in the form 2He

4

+

79Au

197

Æ

79Au

197

+ 2He4

The target nucleus remains unaffected. In general, if a particle a incident on a target X, then elastic scattering process is written as: a+X Æ X+a The outgoing particles remain same. In elastic scattering, the kinetic energy is not conserved but a part of the energy of the incident particle is taken up by the target nucleus which goes to the higher energy quantum state later decay to ground state radiating the excess energy in the form of g -rays (photon). Example: 1H1 + 3Li7 Æ 3Li7 *(excited) + 1H1 3Li

7 *Æ

3Li *

7

+g

or in general a + X Æ X + a. X* Æ X + g The outgoing particle remains same. The target nucleus is left in an excited state which ultimately emits its energy as g -ray.

2.2.2

Radioactive Capture

A low energy proton or neutron may be captured by the target which gets transferred into a residue nucleus in the high energy excited state. The residual nucleus may decay to the ground state with the emission of one or more g -rays. This is called radioactive capture. Example: 1H1 + 1H

2.2.3

1

12Mg

26

Æ

13Al

27

+g

+ 6C12 Æ 7N13 + g

Photo Disintegration

In this reaction, the target nucleus is bombarded by electromagnetic radiation. If the quantum energy of the radiation is high enough one or more particles may be liberated. 1H

2

+ g Æ 1H 1 + 0n 1

This reaction requires photon of energy 2.25 MeV. If the energy is not sufficient even to remove a neutron, the nucleus may enter one of its excited states and later emit excess energy as radiation. Example:

36K

83

+g Æ

36K

83*

(excited state) Æ

36K

83

+g

Radioactive capture and photo disintegration are inverse processes.

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2.2.4

Induced Fission

In fission reaction the target nucleus after interacting with the projectile undergo a more violent transformation. The compound nucleus splits mainly two lighter nuclei of comparable mass and at the same time it emits a few neutrons. A typical example is: 92U

235

+ 0n 1 Æ

56Ba

141

+

36Kr

92

+ 3 0n1 + 200 MeV

The product neutrons can be slowed down, and in turn, they induce new fission events and a chain reaction is set up. When the projectile has sufficient large energy, the target nucleus slits into 20 or 30 nucleons and some a-particles leaving behind a number of nuclei of significant smaller mass. This is known as spallation.

2.2.5

Compound Nucleus Reaction

The concept of formation of compound nucleus was given by Bohr and Wheeler. According to them when a bombarding particle has sufficiently low energy (known as thermal particle), it will be absorbed by the target nucleus and a compound nucleus is formed. The kinetic energy of the incident particle will represent excitation energy of the compound nucleus. Because of the collisions between the incoming nucleon and the nucleons of the target nucleus, this energy would be shared between a number of particles. Hence, the excitation energy of each nucleon will increase. Due to excitation, the compound nucleus may split, e.g., 1D

2.2.6

2

+

20K

39

Æ [21Ca41] Æ

20K

40

+ 1H 1

Heavy Ion and High Energy Reaction

When the incident particle is heavier than a-particle, the reaction is known as heavy ion reaction. This type of reaction is known as heavy ion reaction. This type of reaction was observed at Barkley in 1950 with accelerated carbon ion and aluminum target. 27 12 Æ 1 38 13Al + 6C 0n + 19Kr This type of reaction has the properties of both compound nucleus and of stripping and pickup mechanism. In high energy reaction above 150 MeV, a new kind of particles like p +, p –, p0 and strange particles are ejected along with nucleons.

2.2.7

Spontaneous Decay

The a and b decay processes of a radioactive nuclei are the nuclear reactions of this type.

2.2.8

Disintegration

This is the most general type of reaction in which the incident particle is absorbed and a different particle is ejected. Thus, it is the case of nuclear rearrangement.

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Example: 2He4 + 3Li7 Æ 5B10 + 0n1 2He

4

+ 7N14 Æ 8O17 + 1H1

If the striking particles are g -ray photons of high energy, then the corresponding reaction is called photo disintegration.

2.3

CONSERVATION LAWS DURING NUCLEAR REACTION

Just like chemical reactions obey certain rules or laws, a nuclear reaction also obeys some fundamental laws or without even any one of these laws, a nuclear reaction is not allowed. These laws have been experimentally verified. The following are the various conservation laws that appear to be valid in all nuclear reactions.

2.3.1

Law of Conservation of Charge

Total charge is always conserved in all types of nuclear reactions, i.e., total charge of the products must be equal to the total charge of initial particles. Example: 7N14 + 2He4 Æ 8O17 + 1H1+ Energy Total charge on the reactant side = 7e + 2e = 9e Total charge on the product side = 8e + 1e = 9e Thus, total charge is conserved or SZe = constant

2.3.2

Law of Conservation of Mass Number

Total mass number or the total number of nucleons before and after the reaction remains same. This is because nucleons can neither be created nor be destroyed. In the above example: Total number of nucleons before reaction = 14 + 4 = 18 Total number of nucleons after reaction = 17 + 1 = 18 Hence, the total number of nucleons or the mass number is conserved, i.e., SA = constant or SDA = 0.

2.3.3

Conservation of Mass-Energy

If nuclear reaction neither kinetic energy nor the rest mass is conserved by itself. However, the total sum of mass and energy in a nuclear reaction remain unchanged. For example, a fast moving particle having rest mass ma and kinetic energy Ea is incident on a target nucleus X then it is supposed to be at rest with rest mass MX and a new particle b is given out with rest mass mb, kinetic energy Eb and the product nucleus Y also has kinetic energy EY then

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Total sum of mass-energy before reaction = MXc2 + (mac2 + Ea) Total sum of mass-energy after reaction = MYc2 + EX + (mbc2 + Eb) According to the law of mass energy conservation: MXc2 + (mac2 + Ea) = (MYc2 + EY) + (mbc2 + Eb)

2.3.4

Conservation of Linear Momentum

The total linear momentum of the particles taking part in a nuclear reaction must be the same before and after the reaction, e.g., in the above reaction O + mava = MYvY + mbvb

2.3.5

(As target nucleus X is at rest)

Conservation of Angular Momentum

In a nuclear reaction, the vector sum of the total angular momentum of the reactant must be same as that of the product. The total angular momentum is composed of      intrinsic spin angular momentum ( s ) and orbital angular momentum ( l ) { J = ( l ) + ( s )} .

2.3.6

Conservation of Spin and Statistics

Depending upon the spin, a particle is either a boson (having zero or integral spin and obey B-E statistics) or a fermion (having half integral spin and obey F-D statistics). A nuclear reaction is subject to the condition that the nature of the statistics will not change for the reactants as well as the products. The spin character of an isolated system also remains constant.

2.3.7

Conservation of Parity

The total parity of a system is the product of the intrinsic parities of the target nucleus and the bombarding particle. The net parity before and after the reaction must be equal. No violation of parity has been observed in a nuclear reaction for strong nuclear forces. However, parity does not appear to be conserved in weak interaction.

2.3.7.1

Physical Quantities not Conserved

Quantities such as magnetic moment and electric quadrupole moments of the reacting nuclei are not conserved. These moments depend upon the internal distribution of mass, charge and current within the nuclei involved and thus are not subjected to conservation laws.

2.4

KINEMATICS OF NUCLEAR REACTION: Q-VALUE OF A REACTION

A nuclear reaction can be analyzed quantitatively in terms of the masses and energies of the particles and nuclei involved. This analysis is one of the main sources of information about nuclear properties is called kinematics of the nuclear reaction. The analysis is smaller to that for chemical reaction except that the relativistic relation between the mass and energy must be taken into consideration. Q value of nuclear

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reaction is defined as the total energy released or absorbed in the reaction. Consider a fast moving particle a having rest mass ma and kinetic energy Ea incident on a target nucleus X supposed to be at rest with rest mass MX. A new particle b is given out with kinetic energy Eb and the product nucleus Y also has a kinetic energy EY. This reaction is represented by the equation a+X Æ Y+b or briefly X(a, b)Y and the Q-value of the reaction is given by the relation Q = (EY + Eb) – Ea According to mass-energy conservation, total mass energy before reaction must be equal after reaction, i.e., (MXc2 + mac2 + Ea) = (MYc2 + EY) + (mbc2 + Eb) Therefore, Q = EY + Eb – Ea = MXc2 + mac2 – MYc2 – mbc2 = {(MX + ma) – (MY + mb)} c2 = Dmc2

(2.1)

where Dm is also defined as the mass defect. Hence, Q value of a nuclear reaction may be defined as the difference between the kinetic energy of the product and that of the incident particle. It is also equal to the mass-energy equivalence of the mass defect. The nuclear reaction X(a, b)Y is shown in Fig. 2.1 y Y

a

 Va

Projectile (ma)

 Vy

Product nuclear (MY) f q X Target nuclear (MX)

x b

 vb rated (m particle b)

Libe

‹‰ǤʹǤͳ౨Kinematics of nuclear reaction

Using the law of conservation of linear momentum along X-axis and Y-axis, we have mava = MY vY cos f + mbvb cos q 0 = MY vY sin f – mbvb sin q

(2.2) (2.3)

54പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

From Eqs. (2.2) and (2.3) mava – mbvb cos q = MY vY cos f mbvb sin q = MY vY sin f

and

(2.4) (2.5)

Squaring and adding Eqs (2.2) and (2.5) we get ma2 va2 + mb2 vb2 – 2 mamb va vb cos q = MY2 vY2

(2.6)

Since kinetic energy of the particles a, b and Y are Ea = ½ mava2, Eb = ½ mbvb2 and EY = ½ MY vY2 (Kinetic of target nucleus X is zero as it is at rest.) Therefore, va2 =

2Ea 2Eb 2EY ; v b2 = and vY2 = put in Eq. (2.6) ma mb mY 1

1

Ê 2E ˆ 2 Ê 2E ˆ 2 2maEa + 2mbEb – 2mamb Á a ˜ Á b ˜ cos q = 2MYEY Ëm ¯ Ëm ¯ a

maEa + mbEb – 2(mamb Ea Eb)

b

1/2

cos q = MYEY

1 Êm ˆ Êm ˆ 2 (ma mb Ea Eb ) 2 cos q EY = Á a ˜ Ea + Á b ˜ Eb MY Ë MY ¯ Ë MY ¯

(2.7)

Put Eq. (2.7) in Eq. (2.1), we get the Q value or the reaction energy as 1 Ê mb + MY ˆ Ê m - MY ˆ 2 2 cos q m m E E E + ( ) Q = Ea Á a b a b a b ÁË M ˜¯ M Ë MY ˜¯ Y Y

(2.8)

If the masses are not known accurately we can still find the Q-value to a good approximation, using the mass number. Equation (2.8) gives the Q-value in terms of measured parameters Ea, Eb and q. If the product particle ejected at an angle q = 90° to a collimated beam of incident particle a then cos q = cos p/2. Therefore,

Ê m + MY ˆ Ê m - MY ˆ + Eb Á b Q = Ea Á a ˜ Ë MY ˜¯ Ë MY ¯ Ê Ê m ˆ m ˆ Q = Eb Á 1 + b ˜ - Ea Á 1 - a ˜ MY ¯ MY ¯ Ë Ë

It is possible to measure Ea, Eb, ma, mb and MY quite accurately as also the value of Q and hence the above relation can be verified.

2.5

PHYSICAL SIGNIFICANCE OF THE Q-VALUE OF A REACTION

The Q-value of a nuclear reaction is given by Q = (EY + Eb) – Ea

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ55



where Ea, Eb, EY are the kinetic energies of the projectile particle (a) incidenting on the target nucleus X, new light particle given out, and (b) product nucleus Y. If Q is positive, i.e., the process is accompanied by the libration of energy, it is known as an exothermic or exoergic reaction. In this case the kinetic energy of the products of the reaction is greater than kinetic energy of reactants, i.e., (EY + Eb) > Ea. Also in this reaction mass defect (Dm) is positive. If Q value is negative, the process is accompanied by the absorption of energy, it is known as endothermic or endoergic reaction.

Numerical Problems Problem 1 Show by mass-energy calculation whether reactions (i) 14N (a, p) O17, (ii) Li7(p, a) He4 are exothermic or endothermic. Given atomic masses 7N14, 2He4, 8O16 and 1H1 (p) are 14.00753, 4.00260, 17.00450 and 1.00814 amu, respectively. Solution: (i) The nuclear reaction

14

N (a, p) O17 is represented as

2He

4

+ 7N14 Æ 8O17 + 1H1

Mass defect, Dm = m(7N14 + 2He4) – m(8O17 + 1H1) = {(14.00753 + 4.00260) – (17.00450 + 1.00814)} = – 0.00251 amu Q = Dmc2 = Negative As Q-value of the reaction is negative. Hence, it is endothermic. (ii) The nuclear reaction Li7(p, a) He4 is represented as 3Li

7

+ 1H1 Æ 2He4 + 2He4

Mass defect of the reaction Dm = m (3Li7 + 1H1) – m(2He4 + 2He4) = {(7.01822 + 1.00814) – (4.00260 + 4.00260)} = 0.0216 amu Q-value of the reaction, Dmc2 = 0.0216 ¥ c2 = Positive As Q-value of the reaction is positive. Hence, it is exothermic. Problem 2 Calculate the Q-value for the formation of Pb30 in the ground state in the reaction Si29(d, n) P30 from the following cycles of nuclear reactions. P31 + g Æ P30 + n – 12.37 MeV P31 + p Æ Si28 + 2He4 + 1.909 MeV Si28 + d Æ Si29 + p + 6.246 MeV 2d Æ 2He4 + 23.834 MeV

(a) (b) (c) (d)

56പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

Solution: Relation (a) can also be written as: P30 + n Æ P31 + g + 12.37 MeV

(e)

Formation of P30 from Si29 is given by the reaction Si29 + d Æ P30 + n + Q

(f)

Now adding Eqs. (b), (c), (d), (e) and (f), we get P31 + p + Si28 + d + 2d + P30 + n + Si29 + d Æ Si28 + 2He4 + 1.909 + Si29 + p + 6.246 + 2He4 + 23.834 + P31 + g + 12.37 + P30 + n + Q or 4d = 2 2He4 + Q + 44.359 MeV From Eq. (d) 2 2He4 + 47.668 MeV = 2 2He4 + Q + 44.359 MeV Therefore, Q = 47.668 – 44.359 = 3.309 MeV. Problem 3 Calculate the energy required to remove the least tightly bound neutron from Ca40. Give mass of Ca40 = 39.962584 amu, mass of Ca39= 38.970691 amu, and m(n) = 1.008665 amu. Solution: The removable least tightly bound neutron from Ca40 can be represented as 20Ca

40

Æ

20Ca

39

+ 0n 1

Mass defect = m (Ca40) – m (Ca39) – m (0n1) = 39.962584 – 38.970691 – 1.008665 = – 0.016767 amu As the mass defect is negative. Hence, energy required is Q = Dm(amu) ¥ 931 = 15.6 MeV Problem 4 Find the threshold energy required to start the reaction P31 (n, p) Si31. Given mp = 1.00814, mn = 1.00898, MP = 30.98356 and MSi = 30.98515. Solution: The nuclear reaction P31 (d, n) Si31 is represented as P31 + 0n1 Æ Si31 + 1H1 + Q Mass defect, Dm = Mp + mn – MSi – mH = (30.98356 + 1.00898 – 30.98356 – 1.00814) = – 0.00075 amu Q-value = Dm ¥ 931 = – 0.00075 ¥ 931 = – 0.698 MeV

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ57



Ê M + MN ˆ Threshold energy = – Q Á P ˜¯ MP Ë Ê 30.98356 + 1.00898 ˆ = 0.698 Á ˜¯ = 0.719 MeV. Ë 30.98356 Problem 5 When a nucleus of Li7 is bombarded with a proton, two a-particles are formed. Calculate the kinetic energy of the a-particle assuming the K.E. of a-particle is due to the Q-value of the reaction. Solution: The nuclear reaction is represented as Li7 + 1H1 Æ 2 2He4 Let m(Li)= 7.016004 amu, m(1H1) = 1.007825 and m (2He4) = 4.002603 Mass defect = m(Li) + m (1H1) – 2m (2He4) = 7.016004 + 1.007825 – 2 ¥ 4.002603 = 0.018628 amu Q-value of the reaction = Dm ¥ 931 = 0.018623 ¥ 931 = 17.34 MeV As the energy is being carried by a-particles as their KE. Therefore, KE of each a-particle is = 17.34/2 = 8.67 MeV. Problem 6 Calculate the Q-value of the reaction Be9(d, n) B10 given that M(Be9) = 9.012182, B10= 10.012939, M(d) = 2.014102, M(n) = 1.008665. Solution: The reaction can be represented as Be9 + 1H2 Æ B10 + 0n1 Mass defect of the reaction = M (Be9 + 1H2) – M (B10 + 0n1) = (9.012182 + 2.014102) – (10.012939 + 1.008665) = 0.004681 amu = Dm ¥ 931 = 4.36 MeV.

2.6

NUCLEAR CROSS SECTION

Nuclear cross section, usually represented by s, is a measure of the probability that a bombarding (incident) particle would interact in a certain way with the target nucleus. In order to understand this, we have to visualize each target nucleus as presenting a certain area, called nuclear cross section to the incident particles such that if it strikes within this area lead to the nuclear reaction in question, while the ones hitting outside this area do not induce nuclear reaction. Hence, greater the cross section, the greater the likelihood of the interaction or reaction. The nuclear cross section, in general, does not represent the geometrical cross section of the target nucleus. In fact, it may be greater than, equal to or less than the geometrical cross section (= pR2, R = radius of the nucleus) of the target nucleus.

58പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

Thus, nuclear cross section may be defined as either; 1. the probability that an event may occur when a single nucleus is exposed to a beam of particles of total flux one particle per unit area; or 2. the probability that an event may occur when a single particle is shot perpendicularly at the target consisting of one particle per unit area.

2.6.1

Determination of Cross Section

Suppose we have a slab whose face has area A and thickness dx. Then volume of the slab = Adx. Let n be the number of atoms per unit area of the material of the slab. Total number of nuclei in the slab = nAdx If the cross section of each nucleus is s for a particular interaction. The aggregate cross section for all the nuclei in the slab = snAdx Let N be the number of incident particles in a bombarding beam and dN the number of particles that interact with the nuclei of the slab, then

Number of interacting particles Ag ggregate cross section = Number of incidenting particles Target area dN s nAdx = = s ndx N A dN /N s= ndx

(2.9)

Here s gives the cross section per nucleus or microscopic cross section. Here dN is the number of particles removed by the slab thickness, dx. A n - atoms/m3 (N-dN) particles emerge from slab Flux of N-incidenting particles dx Cross section (s) per atom

‹‰ǤʹǤʹ౨Relationship between cross section and incident beam intensity

Now let us consider the same beam of particles incident on a slab of finite thickness x. If each particle can interact only once we may think that dN is the number of particles removed from the beam while passing through the first small thickness dx. Obviously, the number of particles will further decrease as the beam passes through

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ59



subsequent such thickness. Hence, the quantity dN/N is negative and Eq. (2.9) is slightly modified as dN = -s ndx (2.10) N Therefore, the total effect of the finite slab thickness on the beam intensity is given by integrating Eq. (2.10) N

Ú

N0

Therefore,

x

dN = -s n Údx N 0

Ê Nˆ = -s nx log Á Ë N0 ˜¯ N = N0e–snx

(2.11)

is the number of incidenting particles surviving which are coming out of the slab of thickness x. The number of particles decreases exponentially with the increase in the thickness of the slab. The quantity sn = m is called attenuation coefficient. It is dimension (length)–1 or m–1. Using Eq. (2.11) the intensity of the beam coming out of the slab is given by I = I0e–snx = I0e–mx

(2.12)

Unit of s. s has the dimension of area. The unit normally used for cross section is barn. 1 barn = 1 b = 10–28 m2 = 100 fermi2

2.6.2

Differential Cross Section

In many nuclear reactions the product particles are not produced in an isotropic manner. Detector dN

dW

q

Flux of N-particles

Target slab

Fig. 2.3 Differential scattering cross section s (q, Q)

In such case if dN is the number of product nuclei emitted per unit time in a small solid angle dW at some angle q with the direction of incident beam, then

60പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

dN = ndxds N where ds is the small cross section corresponding to the small solid angle d:. ds 1 dN = ndx N dW dW dN ds 1 dW = (2.13) dW N ndx ds is called differentially cross section of the nucleus reaction and represent the dW probability. Therefore,

2.7

THE CONCEPT OF COMPOUND NUCLEUS

Niel Bohr in 1936 proposed the theory of compound nucleus for interpreting nuclear reaction. According to this theory the nuclear reaction takes place in two ways. 1. Formation of compound nucleus 2. Decay of the compound nucleus giving rise to product nucleus and emitted particles.

2.7.1

Formation of Compound Nucleus

The incident particle strikes a target nucleus and the two combine to form a new nucleus, called compound nucleus, whose atomic number and atomic mass are the sum of atomic number of the incident particle and target nucleus, and the sum of their mass number, respectively, i.e., Incident particle + Target nucleus Æ Compound nucleus The compound nucleus has no memory how it was formed. Since its nuclei are mixed together regardless of their origin. If X is the target nucleus and a the incident particle then interaction between X and a leads to the formation of a compound nucleus C as a + X Æ C* The incident particle captured by the target nucleus makes available to compound nucleus, an energy equal to its kinetic energy plus its binding energy in compound nucleus. This energy is called excitation energy and is given by Ê MX ˆ + EB E = Ea Á Ë MX + Ma ˜¯

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ61



where Ea is the kinetic energy of the incident particle, MX and Ma are the masses of target nucleus and incident particle, respectively, and EB is the binding energy of the incident particle in the compound nucleus. The compound nucleus has a lifetime relatively long compared with the time taken by a nucleon to travel across the nucleus. This is because of the random way in which the excitation energy is distributed in the compound nucleus. The time taken by a particle to travel across the nucleus is known as the natural nuclear time. The nuclear time ( @ 10–17 to 10–22s) is relatively much smaller than the lifetime of compound nucleus which is @ 10–15 to 10–18s.

2.7.2

Decay Scheme of Compound Nucleus

During the lifetime (10–16 s) of nucleus there can always be statistical fluctuations in the energy distribution among nucleons. The energy of the incoming particle is distributed among all the nucleons in a random manner. At any instant, the excitation energy may be shared among several nucleons. If the excitation is large enough the nucleon or a group of nucleons may escape and the compound nucleus may disintegrate into the product nucleus and an outgoing particle, i.e., Compound nucleus Æ Product nucleus + Emitted particle The emitted particle may be a proton, a neutron, a-particle or a g -ray. The excitation energy must be concentrated on the particle or group of particles in order to separate it from the compound nucleus is called the separation or dissociation energy and is usually of the order of 8 MeV. Bohr along with Wheeler explains the formation and decay of a compound nucleus on the basis of “liquid drop model” of the nucleus. According to this model the target nucleus having spherical shape due to the nuclear force similar to the liquid drop having spherical shape due to surface tension. When a incident particle strikes the target nucleus, it is completely absorbed by the target nucleus forming a compound nucleus. The energy of the incident particle is shared by the nucleons of the target nucleus thereby increasing the excitation energy of the nucleus. The excitation energy tries to deform the spherical shape of the nucleus whereas nuclear force tries to keep its spherical shape and the final break gives rise to product nucleus and the outgoing particle. 13Al

2.8

27

+ 1H1 Æ [14Si28] Æ Æ

11Na

Æ

14Si

24

28

14Si

27

+ 0n 1

+ 3 1H 1 + 0n 1

+g

ENERGY LEVELS OF COMPOUND AND PRODUCT NUCLEUS

During the relatively long lifetime of a compound nucleus (as compared to natural nuclear time) it forgets how it was formed and the disintegration. Therefore, it is independent of the mode of formation of compound nucleus. This hypothesis is known as the independent hypothesis and it has been experimentally verified by

62പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

Ghosal. So it is said to exist in a ‘quasi-stationary state’, which suggests that although the compound nucleus exists for relatively longer time, it can still disintegrate by emitting one nucleon or combination of nucleons. These quasi-stationary states are generally known as the ‘virtual levels’ in contrast to the bound states or bound levels which can decay only by emitting gamma radiations. The most stable energy state of the nucleus is the lowest energy level or ground state (zero potential energy). Corresponding to this state the energy of the nucleus is the binding energy calculated from mass defect. For the excitation energy less than the binding energy the corresponding energy levels are bound energy levels. In the normal state of nucleus the lowest bound levels are all occupied. If the excitation energy is less than about 8 MeV, the nucleon will be in one of the normally unoccupied bound states. The regions of the virtual levels contain those energy levels that can be occupied during a nuclear reaction while the nucleus exists in an intermediate state as a compound nucleus. These levels are closely related to the phenomenon of resonance. Virtual Levels Zero put energy Bound levels (unoccupied) 8 MeV

Bound levels (occupied)

‹‰ǤʹǤͶ౨Energy levels of nuclei

If the energy of the incident particle is such that the total energy of the systemincident particle plus target nucleus is equal to the excitation energy of one of the virtual levels of the compound nucleus, the probability that the compound nucleus will be formed is much greater than if the energy falls in the region between two levels. The phenomenon is similar to the resonance turning of radio-circuit. Deexcitation of bound levels can take place only by the emission of g -radiations whereas de-excitation from a virtual energy level can take place in a variety of ways either by particle emission or g -emission. The presence of discrete energy levels in the nuclei has been confirmed by experimental results.

2.8.1

Level Width

The excitation energy of a compound nucleus, whether bound or virtual has a definite mean lifetime. It is defined as the average time for which a nucleus remains in an excited state before de-excitation itself by emission of g -rays or a particle. If l is the

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ63



disintegration or decay constant, i.e., the probability per unit time of the emission of a particle or g -rays then 1 l In the discussion of energy states excited by nuclear reaction we use a quantity l proportional to the disintegration constant. This quantity is called level width. Ta =

Mean lifetime,

 Ta Level width has the units of energy. According to uncertainty principle,

Therefore, level width, = l =

DE ◊ Dt =

h = 2p

where DE is the uncertainty in energy and DT is the uncertainty in time. The mean lifetime Ta of the excited nuclear state is identified with uncertainty DT. If we put DT = Ta in the above relation we get DE =

 = t (level width) Ta

Hence, level width of an excited state is the spread of energy of this state. Hence, from the relation we find that a large value of mean lifetime Ta gives a narrow spread of energy level, i.e., the energy level is fine and sharp (low level width). On the other hand a small value of mean lifetime (Ta) gives large spread of energy level, i.e., energy level is broad and diffuse. Now,

t= =

2.9

1 6.62 ¥ 10 -34 h ¥ = Joules 2p Ta 2p Ta

6.62 ¥ 10 -34 2p ¥ 1.6 ¥ 10 -1 Ta

eV =

6.6 ¥ 10 -16 eV Ta

NUCLEAR TRANSMUTATION

When high energy incident particles having energy of a few MeV are made to strike certain target nuclei, many distinct possible reactions can occur. This all depends upon the overall conditions under which such an interaction takes place. “If the basic character of the target nucleus gets changed, it is said to have undergone a transmutation. A large number of reactions have so far been studied and documented in scientific literature. We thus have reactions like: (1) Transmutation by protons: (p, a), (p, n), (p, g), (p, d), (p, 3n), (p, 4n), (p, 2d), etc. (2) Transmutation by deuterons: (d, a), (d, p), (d, n), (d, t), (d, 2n), (d, 3n), (d, 2p), etc. (3) Transmutation by neutrons: (n, a), (n, p), (n, g ), (n, d), (n, 2n), etc.

64പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

(4) Transmutation by a-particle: (a, p), (a, n), (a, g ), (a, 2n), (a, 4n), (a, 3p), (a, 4p), etc. (5) Transmutation by photon (g -radiations): (g, n), (g, p), (g, d), (g, He3), etc. (6) Transmutation by tritium (1H3): (t, d), (t, n), (t, He3), etc. (7) Transmutation by heavy ions: (C, n), (N, n), etc.

2.9.1

Transmutation by Protons

The first nuclear transmutation using protons as projectile particle was carried out by Cockcroft and Wilton (1932) using lithium. 1. The (p, a) reaction: The reaction can be represented by 1H

1

+ 3Li7 Æ [4Be8]* Æ 2He4 + 2He4

The outgoing a-particles so produced was confirmed by photographing their tracks in a cloud chamber. This reaction also established Einstein mass-energy equivalence (DE = Dmc2). Precisely measuring the mass of reactants and products, the Q-value of the reaction comes out to 0.0186 amu ¥ 931 = 17.31 MeV. Also the experimental Q-value of the reaction is 17.33 MeV. The agreement points to the success of the relation, DE = Dmc2. A (p, a) reaction may be represented as (standard notation) ZX

+ 1H1 Æ [Z+1DA+1]* Æ

A

A–3 Z–1Y

+ 2He4

A few familiar examples are: 11 5B 23

+ 1H1 Æ [6C12]* Æ 4Be8 + 2He4

1 11Na + 1H 19 1 9F + 1H 27 1 13Al + 1H

Æ [12Mg24]* Æ

20 4 10Ne + 2He Æ [10Ne20]* Æ 8O16 + 2He4 Æ [14Si28]* Æ 12Mg24 + 2He4

2. The (p, n) reaction: A general (p, n) reaction may be represented as ZX

A

+ 1H1 Æ [Z+1DA+1]* Æ

A Z+1Y

+ 0n 1

Typical examples of this reaction are: 5B 11Na

11

+ 1H1 Æ [6C12]* Æ 6C11 + 0n1

23

+ 1H1 Æ [12Mg24]* Æ

12Mg

58

+ 1H1 Æ [29Cu59]* Æ

29Cu

+ 1H1 Æ [Z+1DA+1]* Æ

Z+1D

28Ni

23

58

+ 0n 1 + 0n 1

3. The (p, g ) reaction: ZX

A

A+1

A few examples are: 3Li

7

+ 1H1 Æ [4Be8]* Æ 4Be8 + g

+g

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ65

 14

+ 1H1 Æ [8O15]* Æ 8O15 + g

27

+ 1H1 Æ [14Si28]* Æ

7N 13Al

14Si

28

+g

4. The (p, d) reaction: The end product may contain deuteron (1H2) when some target nucleus is bombarded with high energy protons. We will represent such a reaction, in general, as ZX

A

+ 1H1 Æ [Z+1DA+1]* Æ ZYA–1 + 1H2

A compound nucleus may or may not be formed. Though there is no change in the z-value of the nuclide, its A-value is reduced by one unit. Such reactions are more often encountered with heavier targets and protons of very high energy. + 3Li7 Æ [4Be8]* Æ 3Li6+ 1H2 9 1 10 * 8 2 4Be + 1H Æ [5Be ] Æ 4Be + 1H 1H

1

5. The (p, 3n), (p, 4n), (p, 2d) reaction: In such reactions, there are two or more (emitted) particles in the product side. It is obvious because the incident protons have kinetic energy more than 20 MeV. The compound nucleus is more energetic and hence, more unstable. Due to sufficient excitation energy, a compound nucleus may allow the release (expulsion) of two or more particles. The examples are: 88 Æ [39Y89]* Æ 39Y86+3 0n1 38Sr 63 Æ [30Zn64]* Æ 30Zn60+4 0n1 1H + 29Cu 1 65 Æ [30Zn66]* Æ 28Ni62+ 2 1H2 1H + 29Cu 1H 1

2.9.2

1

+

Transmutation by Deuteron

There are many reactions initiated by deuteron (1H2) as an incident particle. For this purpose, a deuteron of several MeV’s can be obtained with a cyclotron or electrostatic generator. 1. (d, a) reaction: A general (d, a) reaction may be represented as ZX

A

+ 1H2 Æ [Z+1DA+2]* Æ

A–2 Z–1Y

+ 2He4

A few examples are: 8O

16

+ 1H2 Æ [9F18]* Æ 7N14 + 2He4

20 2 22 * 18 4 10Ne + 1H Æ [11Na ] Æ 9F + 2He 27 2 29 * 25 4 13Al + 1H Æ [14Si ] Æ 12Mg + 2He 6 2 8 * 4 4 3Li + 1H Æ [4Be ] Æ 2He + 2He

2. The (d, p) reaction: It is not always that reaction induced by deuterons yields only a-particles. Occurrence of deuteron-proton reaction is often observed. A general notation for such reaction can be

66പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ ZX

A

+ 1H2 Æ [Z+1DA+2]* Æ ZYA+1 + 1H1

Hence, atomic mass number increases by one unit whereas atomic number (Z) remains same. This type of reaction is normally exoergic with positive mass change. The product nucleus is the only isotope of the target nucleus. Some of the reactions of this type are: 7 3Li 31 15P 12 6C

+ 1H2 Æ [4Be9]* Æ 3Li8+ 1H1 + 1H2 Æ [16S33]* Æ + 1H2 Æ [7N14]* Æ

32 1 15P + 1H 13 1 6C + 1H

3. The (d, n) reaction: Neutrons can also be produced as a result of deuteron bombardment. A general reaction of this type is: ZX

A

+ 1H2 Æ [Z+1DA+2]* Æ

A+1 Z+1Y

+ 0n 1

The product nucleus is the isotope of compound nucleus. Few examples of (d, n) reaction are: 7 3Li 12 6C 209

+ 1H2 Æ [4Be9]* Æ 4Be8+ 0n1

+ 1H2 Æ [7N14]* Æ 7N13+ 0n1 + 1H2 Æ [84Po211]* Æ 84Po210+ 0n1

83Bi

4. The (d, t) reaction: When the emitted particle is tritium (1H3) on using deuterons as projectiles hitting against certain targets, the reaction is known as (d, t) reaction. This type of reaction has low probability (d, t) reaction, in general, represented as: ZX

A

+ 1H2 Æ [Z+1DA+2]* Æ ZYA–1 + 1H3

As the product nucleus has some value of Z as that of target nucleus. So it is an isotope of target nucleus. For example: 3Li

7

+ 1H2 Æ [4Be9]* Æ 3Li6+ 1H3

9 2 4Be + 1H 31 2 15P + 1H

Æ [5B11]* Æ 4Be8+ 1H3 Æ [16S33]* Æ

15P

30

+ 1H 3

5. The (d, 2n), (d, 3n) and (d, 2p) reaction: In this kind of reaction, Q is usually negative so that the incident particle must be having very high energy (> 200 MeV say).

2.9.3

Transmutation by Neutron

Due to the neutral character, neutrons are a class apart from the charged particles say, a-particles or protons. They have proved more effective in inducing nuclear transmutation. It is precisely due to the fact that they do not face repulsive Coulomb forces at or around the nucleus of the target atoms. It makes the neutron to have more penetration power. Thus, even low energy neutrons can be more effective for this purpose. Hence, we have more reactions induced with neutrons than with any other particle.

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ67



The interaction of neutrons with nucleus may capture the neutrons to form temporarily a compound nucleus. This compound nucleus will have the same atomic number but its mass number will be increased by one. The compound nucleus is generally unstable and emits energy in the form of a g -ray photon, a-particle, a proton, a single neutron or pair of neutrons. These reactions are respectively known as (n, g), (n, a), (n, p), (n, 2n), etc. 1. The (n, g ) reaction: It is also called radioactive capture of the incident particle (neutrons). It can be represented by the reaction ZX

A

+ 0n1 Æ [ZDA+1]* Æ ZYA+1 + g

The compound nucleus emits one or more g -rays so that the final nucleus is only an isotope of the target nucleus. The (n, g ) reaction has been observed in almost all the elements. However, the Q-value are always positive, and the excess energy is carried (exo-ergic reaction) away by gamma rays. Few typical examples of (n, g ) reactions are 1H

1

+ 0n 1 Æ [ 1H 2] * Æ 1H 2+ g

1H 27

2

+ 0n 1 Æ [ 1H 3] * Æ 1H 3+ g

13Al 238 92U

+ 0n1 Æ [13Al28]* Æ + 0n1 Æ [92U239]* Æ

28 13Al + 239 + 92U

g g

2. The (n, a) reaction: When a nucleus is bombarded with a neutron and the product nucleus decays with the emission of a-particle, the general reaction is represented as ZX

A

+ 0n1 Æ [ZDA+1]* Æ

A–3 Z–2Y

+ 2He4

The final product is a new element whose atomic number is 2 less than the target nucleus and atomic mass is 3 less than the target nucleus. Few examples of this reaction are: 10 5B 27

+ 0n1 Æ [5B11]* Æ 3Li7+ 2He4

+ 0n1 Æ [13Al28]* Æ

24 4 11Na + 2He 14 1 15 * 11 4 7N + 0n Æ [7N ] Æ 5B + 2He 14 1 15 * 7 4 7N + 0n Æ [7N ] Æ 3Li + 2 2He

13Al

3. The (n, p) reaction: In some cases, it is possible that compound nucleus formed by the capture of neutron emit proton. The general reaction (n, p) represented as A 1 A+1 * Æ A 1 ] ZX + 0n Æ [ ZD Z–1Y + 1H In fact, the reaction involves the replacement of a proton by a neutron. As such, the atomic number of product nucleus reduces by one but atomic mass remains same as that of target nucleus. Hence, product nucleus is an isobar of target nucleus. For example,

68പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ 12Mg

24

+ 0n1 Æ [12Mg25]* Æ

11Na

24

+ 1H 1

The product in all cases of (n, p) reaction is radioactive and decay with the emission of b-particle. For example, 11N

24

Æ

12Mg

24

+

–1e

0

The final product is same as the target nucleus: 13Al

27

+ 0n1 Æ [13Al28]* Æ

65

+ 0n1 Æ [29Cu66]* Æ

29Cu

12Mg

27

28Ni

+ 1H 1

65

+ 1H 1

4. The (n, n) reaction: Sometimes, the bombarding neutron produces the target nucleus in an excited state which passes on to the ground state with the emission of a neutron accompanied by a g -ray. Example: 13Al

27

+ 0n1 Æ [13Al28]* Æ

13Al

27

+ 0n 1

The product nucleus 13Al27 is in a metastable state and decay with the emission of g -ray. 5. The (n, 2n) reaction: When the kinetic energy of the bombarding neutron is about 10 MeV, then the incident neutron is captured by the target nucleus followed by the emission of two neutrons. With the emission of two neutrons the atomic mass (A) is reduced by one unit while atomic number (Z) remains same. Hence, the product nucleus is an isotope of target nucleus. The general (n, 2n) reaction is ZX

A

+ 0n1 Æ [ZDA+1]* Æ ZYA–1 + 2 0n1

Examples: 39

+ 0n1 Æ [19K40]* Æ

38 1 19K + 2 0n 12 1 13 * 11 1 6C + 0n Æ [ 6C ] Æ 6C + 2 0n 197 + 0n1 Æ [80Hg198]* Æ 80Hg196+ 2 0n1 80Hg 19K

6. The (n, 3n) reaction: If the energy of the incident neutron approaches 30 MeV, then sufficient energy is available to overcome the potential barrier, three neutrons or even two neutrons and a proton are ejected from the compound nucleus. Bombardment of 90 MeV neutron may give rise to (n, d) and (n, t) reactions.

2.9.4

Transmutation by a-particles

The fact that spontaneous disintegration does take place in certain atoms gave a valuable hint, that is, the possibility of causing a similar disintegration in ordinary (non-radioactive) nuclides. Alpha particles seemed most likely to behave as effective

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ69



projectile for this purpose. It is because they possessed quite large energy and momentum. The Coulomb’s force of repulsion could possibly decrease the probability of disintegration of the target nuclei. But if lighter targets were used, the scattering could be greatly reduced hence, disintegration probability increases. 1. The (a, p) reaction. The first artificial disintegration was carried by Rutherford in 1919. Nitrogen atoms on being bombarded by a-particles emit protons of high energy. Rutherford and Chadwick extended this work to other elements. They found that all elements from boron to potassium, except carbon and oxygen, gave same result. They also found that in some cases the energy of emitted protons was even higher than that of a-particles. Two explanations were offered: one the nuclei of the bombarded atom simply loses a proton as a result of collision with the fast a-particle. Second: A a-particle is first captured by the nucleus of the atom it hits, the unstable compound nucleus so formed then emits a proton. In the first case, an a-particle will exist after collision whereas it should disappear in second case. Blackett actually confirmed in 1925 from the study of tracks produced by a-particle passing through cloud chamber, that the second option is correct. The above nuclear reaction may be represented as 7N

14

(target) + 2He4(projectile) Æ [9F18]*(compound nucleus) Æ 8O17+ 1H1(proton) Some other reactions are: 10 5B 23

+ 2He4 Æ [7N14]* Æ 6C13+ 1H1

+ 2He4 Æ [13Al27]* Æ

11Na

28 4 32 * 14Si + 2He Æ [16S ] 39 4 43 * 19K + 2He Æ [21Se ]

Æ Æ

26 1 12Mg + 1H 31 1 15P + 1H 42 1 20Ca + 1H

In general (a, p) reaction can be represented as ZX

A

+ 2He4 Æ [Z+2DA+4]* Æ

A+3 Z+1Y

+ 1H 1

Hence, in each case, atomic number of product nucleus increased by one unit and atomic mass by three units. In some cases, the product nucleus may be radioactive, which disintegrates with the emission of b-particles. For example, 5B

11

+ 2He4 Æ [7N15]* Æ 6C14*+ 1H1 14* Æ 14 0 6C 7N + –1e

2. The (a, n) reaction: The general (a, n) reaction can be written as ZX

A

+ 2He4 Æ [Z+2DA+4]* Æ

A+3 Z+2Y

+ 0n 1

Hence, the atomic number of the product nuclei increased by two units and atomic mass by three units. Some well known (a, n) reactions are: 3Li

7

+ 2He4 Æ [5B11]* Æ 5B10+ 0n1

70പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ 9 4 13 * 12 1 4Be + 2He Æ [6C ] Æ 6C + 0n 14 4 18 * 17 1 7N + 2He Æ [9F ] Æ 9F + 0n 23 4 27 * 26 1 11Na + 2He Æ [13Al ] Æ 13Al + 0n 27 4 31 * 30 1 13Al + 2He Æ [15P ] Æ 15P + 0n 40 4 44 * 43 1 18Ar + 2He Æ [20Ca ] Æ 20Ca + 0n

3. The (a, g ) reaction-radioactive capture: In certain cases, it is observed that a compound nucleus formed after the capture of incident a-particle acquires stability without emitting a particle. In such a case, the compound nucleus emits g -ray photon. Example:

3Li

7

+ 2He4 Æ [5B11]* Æ 5B11+ g

4. The (a, 2n), (a, 4n), (a, 3p) and (a, 4p) reactions: If the incident a-particle possesses a large energy the products may contain two or more nucleons (protons, neutrons). In other words, these reactions are essentially of the (a, p) and (a, n) type.

2.9.5

Transmutation by Photon (g -radiation)

The reaction brought out by high energy photons (g -rays) are classified as ‘photo disintegration’. These reactions are endoergic and have threshold energies of the order of 10 MeV. As a photon has zero rest mass, it can supply only kinetic energy to the nuclear reaction. 1. The (g, n) reaction: Examples: 2 1H 9

4Be 31 15P

+ g Æ [ 1H 2] * Æ 1H 1+ 0n 1 + g Æ [4Be9]* Æ 4Be8+ 0n1 + g Æ [15P31]* Æ

15P

30

+ 0n 1

2. The (g, p) reaction: Examples: 4Be 25

12Mg

9

+ g Æ [4Be9]* Æ 3Li8+ 1H1

+ g Æ [12Mg25]* Æ

11Na

24

+ 1H 1

3. The (g, d), (g, t), (g, a) reaction: Sometimes, compound nucleus may emit a deuteron (1H2), a tritium (1H3) and an a -particle (2He4). Examples: 9 4B 11 5B 12 6C

+ g Æ [5B10]* Æ 4Be8+ 1H2 + g Æ [5B11]* Æ 4Be8+ 1H3 + g Æ [6C12]* Æ 4Be8+ 2He4

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ71



4. With g -rays of energy more than 10 MeV reaction of the type (g, n, p), (g, 2n), (g, n, 2p) may occur. Example: 16 8O 27

13Al

2.9.6

+ g Æ [8O16]* Æ 6C11+ 30n1 +2 1H1 + g Æ [13Al27]* Æ

11Na

24

+ 0n1 +2 1H1

Transmutation by Tritium

The reaction of this category is (t, d), (t, n), (t, He3). Some of the reactions are 6

+ 1H3 Æ [4Be9]* Æ 3Li7+ 1H2

¸ 59 3 62 * 60 2 Ô 27Co + 1H Æ [28Ni ] Æ 27Co + 1H ˝ (t, d) reaction 63 3 66 * 64 2Ô 24Cu + 1H Æ [30Zn ] Æ 29Cu + 1H ˛ 3Li

1H 16S 13Al

2.9.7

2

32

27

+ 1H3 Æ [2He5]* Æ 2He4+ 0n1 + 1H Æ [17Cl ] Æ 3

35 *

+ 1H3 Æ [14Sr30]* Æ

17Cl

34

12Mg

+ 0n

27

1

+ 2He3

Ô¸ (t, n) reaction ˝ ˛Ô

(t, He3) reaction

Transmutation by Heavy Ions

Reaction with particles heavier than the a-particle as projectile has also been observed. 12 14 16 accelerated to more than 90 MeV can result in transmutations. 6C , 7N and 8O Here, either we get a transfer reaction (involving exchange or transfer of one or more nucleons) or a reaction in which relatively a large number of particles are emitted. 7N

14

+

14 10 Æ 11 13 7N + 5B 5B + 7N 27 Æ 25 16 13Al 13Al + 7N (nucleon Æ 12Mg27+ 8O14

transfer)

These direct reactions do not involve the formation of a compound nucleus. 197 79Au 12 238 6C + 92U 14 238 7N + 92U

6C

2.10

12

+

Æ [85At209]* Æ Æ [98Ci250]* Æ Æ [99Es252]* Æ

205 + 4 0n 1 85At 244 + 6 0n 1 98Ci 246 + 6 0n 1 99Es

DISCOVERY OF NEUTRON (A USEFUL PROBE FOR NUCLEAR REACTION)

How sharp guess work and logical speculation often uncover marvels of nature is beautifully demonstrated in the discovery of neutron. Initially, it was assumed that the atomic nucleus was composed of proton and electrons. This hypothesis failed and paved the way for another hypothesis. According to this, an atomic nucleus contained protons and neutrons only. This hypothesis was possible only after the discovery

72പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

of neutron (1932). There are various reasons to support this picture of nucleus. A neutron has got a prominent position mainly for two reasons: first, it has helped us to understand the structure of an atom finally, and second, it has revolutionalized the field of nuclear reaction (induced). It serves as an excellent probe for such reactions. In 1930, when two German physicists W. Bothe and H. Becker bombarded a beryllium (4Be9) target with a-particle from a sample of polonium (84Po210) and found that highly penetrating electrically uncharged (neutral) radiation was emitted. These radiations could pass through a few centimeters thick lead (Pb) sheet (Fig. 2.5a). They assumed that these radiations are high energy gamma rays (em waves of very small wavelength @ 10–14 m). The possible reaction therefore was thought to be 2He

4

+ 4Be9 Æ [6C13]* Æ 6C13 + g (photon) ?

Polonium (Source)

a a

?

a a Lead as absorber

Beryllium target

Fig. 2.5(a)

Many physicists performed different experiments to find the properties of these unknown radiations. In one such experiment, Irene Curie and her husband Fredericks Joliot observed in 1932 something more interesting. When the same radiations struck a paraffin slab, protons were knocked out of the slab (Fig. 2.5b) Polonium source of a-particles

a a

Protons of energy 5.7 MeV

a a

? Beryllium target

Paraffin slab

‹‰ǤʹǤͷȋ„Ȍ౨Protons of up to 5.7 MeV are escaped when aǦ’ƒ”–‹…Ž‡••–”‹‡•’ƒ”ƒˆϐŽ‹‡•Žƒ„

They confirmed that the energy of these protons was anything up to 5.7 MeV. Suspecting this to be a case of Compton-like interaction between those very short wavelength photons and the protons, it was calculated that these photons must have at least energy equal to 55 MeV only then could the ejected protons have energy up to 5.7 MeV. Obviously, the assumption that the new radiations consisted of high energy g -rays was totally false. It had to be rejected. The problem of the identity of the bombardment products from 4Be9 target hit by a -particles was solved by Chadwick. In a series of experiments on the recoil of nuclei which were struck by the rays

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ73



coming from the bombarded 4Be9 he made up his mind about neutrons. Chadwick, Rutherford’s co-worker, proposed in 1932 that unknown radiation consisted of neutral particles whose mass is comparable to the proton mass. Because they were uncharged, these particles were called neutrons. The neutron mass accounted nicely for the observed proton energies (£ 5.7 MeV ): two comparable masses colliding head on with each other just exchange their energies. Thus, a moving neutron can transfer almost whole of its kinetic energy to a proton at rest. This also implies that the neutron energy was also anything up to 5.7 MeV, not the 55 MeV required if g -rays were to cause same effect. In the light of Chadwick assertion about neutron, a new correct equation can be written as 4He

2.10.1

4

+ 4Be9 Æ [6C13]* Æ 6C12 + 0n1

Detection of Neutrons

Since neutrons are uncharged particles and produce only a very low ionization in passing through the matter. These observations lead us to the fact they cannot be detected directly by any instrument say, a GM counter or a cloud chamber which depends for its action on the ionization caused by the particle entering it. It becomes, therefore an indirect case of detection for such particles. The detection of neutrons depends on the secondary effects which are caused by their interaction with nuclei. Some notable secondary effects are: 1. The absorption of neutron by a nucleus with an immediate emission of a fast charged particle. 2. The absorption of a neutron forming an unstable compound nucleus that undergoes fission. 3. The absorption of a neutron forming a radioactive nuclide whose activity can be measured. 4. The incident neutron may even be scattered by a light (low Z) nucleus such as a proton. The recoil light nuclei produce ionization that can be measured. One of the most frequently used detectors is based on the reaction 5B

10

(n, a) 3Li7

with

Q = +2.78 MeV

The most commonly used method for detecting fast neutrons is based on the ionization produced by the recoiling protons in the elastic scattering of neutrons by hydrogenous materials. Such hydrogen containing materials can be kept inside a cloud chamber or a nuclear emulsion.

2.10.2

Estimation of Mass of Neutron

The mass of the neutron was estimated by Chadwick using the method of head on collision and applying the principle of conservation of kinetic energy and linear momentum. Suppose a neutron of mass mn moving with velocity vn ejected from a 4Be9

74പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

target makes a head on elastic collision with hydrogen nucleus (proton) in paraffin which is initially at rest. Then both move in the same direction as the direction of motion of neutron before collision. If mp is the mass of hydrogen nucleus and vp is the velocity after collision. Let vn¢ be the velocity of neutron after collision then Neutron

Proton  vp

v n¢

vn mn

v=0 mp

mp

Neutron

Collision

Before collision

After collision

Fig. 2.6

Then according to law of conservation of kinetic energy and linear momentum we have 1 1 1 mn vn2 + 0 = mn vn2¢ + mp vp2 (2.14) 2 2 2 mn vn + 0 = mn vn¢ + mp vp

and From Eq. (2.15) vn¢ =

mn vn - mp vp mn

(2.15)

. Put in Eq. (2.14)

Ê mn vn - mp vp ˆ 1 1 1 mn vn2 + 0 = mp vp2 + mn Á ˜¯ mn 2 2 2 Ë mn vn2

=

mp vp2

+

2

(mn vn - mp vp )2 mn

mn2 vn2 = mn mp vp2 + mn2 vn2 + mp2 vp2 - 2mn mp vn vp

mn mp vp2 + mp2 vp2 - 2mn mp vn vp = 0 mp vp [mn vp + mp vp - 2mn vn ] = 0 As mp vp π 0 we get mn vp + mp vp - 2mn vn = 0 vp =

2mn vn mn + mp

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ75



If in place of hydrogen nuclei, we have a nitrogen nucleus then the above equation is equivalently put as

Therefore,

vN =

2mn vn mn + mN

vp

mn + mN mn + 14 = mn + mP mn + 1

vN

=

(∵ masses of nitrogen nucleus and hydrogen nucleus is 14 amu and 1 amu respectively.) If we know the recoil velocities vp (the hydrogen nucleus) and vN (nitrogen atom), the mass neutron (mn) was found. The value so obtained gave mn = mp a very good approximation. We take mn = 1.675 ¥ 10–27 kg.

2.10.3

Neutron Properties

The most well known properties of a neutron are: 1. It is a neutral fermion (spin = 1/2 ) and is not deflected by electric and magnetic fields. 2. It is a fundamental building block of matter. 3. It has no independent (free) existence, i.e., it is not stable outside the nuclei. A free neutron immediately decays according to the equation n Æ p + e– + n e with a mean life of 15.5 min. 4. Being uncharged, it can penetrate deeper into a few centimeters thick lead sheet. 5. On being proton captured, it forms deuteron 0n

1

+ 1H1 (proton) Æ 1H2 (deuteron) + g (2.23 MeV)

6. Because of zero charge on it, a neutron has a lower ionization power. 7. Its electrical neutrality is of greater value; without any obstruction (attraction and repulsion from negatively and positively charges present in the atom) a neutron can reach right up to an atomic nucleus. It is thus, an excellent probe for inducing nuclear transmutation. We have many types of reactions possible with neutrons such as (n, p), (n, n), (n, a), and (n, g ) reactions. 8. They can exchange their energies when they collide against lighter nuclei (almost at rest). As such faster neutrons get slowed down. This fact is of great value in choosing a moderator. H2O, D2O (heavy water), carbon (graphite), beryllium, etc., are most well known moderators. 9. They are used for producing a variety of radioisotopes that find applications in various fields. 10. The nuclear fission reaction {(n, n) reaction} is the most prized gift to mankind for producing nuclear power.

76പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

2.10.4

Importance of Neutrons

1. Neutrons have helped in understanding the nuclear structure and hence all matter in relation to how it is arranged. 2. Hundreds of nuclear reactions with neutrons as probes have been used for various purposes. 3. Nuclear fission owes a great deal for its success to neutron (slow) probes. This reaction is key to generating nuclear power. The share of nuclear power over the conventional energy sources is quite high in some developed countries. 4. The radioisotopes produced using neutron as projectile particle on various targets find tremendous use in medicine, agriculture, industry and pure research.

2.11 2.11.1

RUTHERFORD SCATTERING (COULOMB’S SCATTERING) Scattering of a-particle

Fundamental discoveries concerning our knowledge of atomic structure have been developed from Rutherford investigation of scattering of a-particles by matter. If a sharply defined pencil beam of a -particles is allowed to fall upon a photographic plate in vacuum, the shadow image formed has clean and sharp edges. When an air or some other gas is introduced in the path of a -particle, or a screen having thin foil of metal or mica is placed in the path, the image becomes diffused. This spreading out of a stream of a -particles in passing through the thin layers of matter is called scattering. Geiger and Marsden made a detailed study of scattering of a-particles by thin film of gold, platinum and other metallic elements of high atomic weight. It was observed that a majority of the a -particles were scattered through small angles, but a few of them deviated through 90° and very few (1 out of 1000) retraced their path, i.e., deviated through about 180°. This large angle scattering is known as anomalous scattering and could not be explained according to Thomson’s concept that the atom consisted of a positive sphere of electrification embedded with negative electrons. To explain the large scattering angle of a-particles, Rutherford proposed that in the atom the whole positive charge and almost the whole mass of the atom is concentrated in a very small central region (later on called the nucleus), about which the electrons revolve in circular orbits. He assumed that the force of repulsion between the nucleus of an atom of the scattering foil and the incident a-particle obeyed Coulomb’s inverse square law. Assuming further that for elements of large atomic number like gold, platinum, etc., the nuclei of the atoms concerned were so massive compared with the a-particle that they remained almost at rest. Rutherford was able to calculate correctly, the large angle of scattering of a-particles. Suppose an a-particle at any instant is at point P is travelling initially with constant velocity v in a straight line along PO where point P is at very large distance w.r.t. the

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ77



nucleus at N as compared with the diameter of the nucleus (atom). As the a-particle approaches the nucleus, its velocity goes on decreasing due to Coulomb’s force of repulsion. On reaching the point A it is deflected and when it is again outside the field of the nucleus, it travels along OP¢. The path PAP’ is a hyperbola with the nucleus N at its outer focus. P V

X

N

a

O q

p

f

q

A v

M

P¢

‹‰ǤʹǤ͹౨a-ray scattering diagram

2.11.2

Angle of Scattering (f)

The angle of scattering (f) is the angle between the asymptotic direction of approach of the a-particle PO produced and asymptotic direction in which it move back OP’.

2.11.3

Impact Factor

Draw NM perpendicular to PO produced then = p represents the impact parameter. The impact parameter is the minimum distance to which the a -particle would approach the nucleus if there were no force between them.

2.11.4

Distance of Closest Approach

Let initially the a -particle be at a very large distance from the nucleus possessing kinetic energy E. If moved towards the nucleus along the line joining the centre of positively charged nucleus, its K.E. goes on decreasing as work has to be done against the electrostatic force of repulsion and potential energy goes on increasing.

a-particle

A + d

‹‰ǤʹǤͺ౨a-ray scattering

At a particular point A known as turning point the whole kinetic energy will convert into potential energy. Momentarily, a-particle comes to rest and retrace its path.

78പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

The distance (d) between the point (A) (known as turning point) and the centre of the nucleus is called the distance of closest approach. If Ze is the charge of the nucleus, the electrostatic potential energy of the system of two charged particles at a distance d, 1 q1q2 1 ( Ze )( 2e ) = d 4pŒ0 d 4pŒ0 Since KE of the a-particle at infinite distance is equal to the potential energy at a distance d. U=

Therefore,

E=

1 2Ze 2 4pŒ0 d

or distance of closest approach d=

1 2Ze 2 4p Œ0 E

Hence, distance of the closest approach depends upon the initial kinetic energy of a-particle.

2.11.5

Relation among f, p and d

Let initially a-particle be at P travelling along a straight line PO with constant velocity V. As it approaches the nucleus its velocity goes on decreasing up to the point A, where it is deflected and changes direction. Let v be the velocity of a-particle be at A then Kinetic energy of a-particle at P = 1/2 mV2 Kinetic energy of a-particle at A = 1/2 mv2 x

Potential energy of a-particle at A =

1

Ú 4pŒ0

(Ze )(2e ) dr

0

U =

r2

2Ze 2 4pŒ0 x

According to law of conservation of energy 1/2 mV2 = 1/2 mv2 +

2Ze 2 4pŒ0 x

V2 = v2 +

4Ze 2 4pŒ0 mx

v2 = V2 -

˘ È 4Ze 2 4 Ze 2 = V2 Í1 2 ˙ 4p Œ0 mx Î 4p Œ0 mxV ˚

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ79



È b˘ v2 = V2 Í1 - ˙ Î x˚ where b =

(2.16)

4Ze 2 4pŒ0 mV 2

The angular momentum of a -particle at A = mvx The angular momentum of a-particle at P = (mV)(p) Therefore,

mVp = mvx v= V 2 p2 x

2

Vp put in Eq. (2.16) x

È b˘ = V 2 Í1 - ˙ Î x˚

È b˘ p2 = x2 Í1 - ˙ = x [x – b] Î x˚ Since from the coordinate geometry of hyperbola Eccentricity = sec q and Focal distance, ON = eccentricity ¥ OA = (sec q) (a) ON = a sec q Also, ON = p cosec q (from Fig. 2.8) p cosec q = a sec q

Therefore, Now,

AN = ON + OA = a sec q + a = a (1 + sec q) x = p cot q (1 + sec q) = p x= =

p (cos q + 1) sin q p qˆ Ê 2 cos 2 ˜ Á Ë q q 2¯ 2 sin cos 2 2 q 2 = p cot q q 2 sin 2

p cos =

cos q Ê 1 ˆ 1+ Á sin q Ë cos q ˜¯

(2.17)

80പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

Put the value of x in Eq. (2.17) qÈ q ˘ p2 = p cot Í p cot - b ˙ 2Î 2 ˚ q q qÈ q ˘ p = cot Í p cot - b ˙ = p cot 2 - b cot 2 2 2Î 2 ˚ b cot

q q ˆ Ê = p Á cot 2 - 1˜ Ë 2 2 ¯ q ˆ Ê p Á cot 2 - 1˜ Ë q 1 ˘ 2 ¯ È = p Ícot b= q q˙ 2 cot cot ˙ Í Î 2 2˚ q q˘ È È 2q 2q˘ Í cos 2 - sin 2 ˙ Í cos 2 sin 2 ˙ = pÍ b = pÍ q q˙ q q ˙ Í sin cos ˙ Í sin cos ˙ Î Î 2 2˚ 2 2 ˚ Ê cos q ˆ b = 2p Á = 2p cot q Ë sin q ˜¯

tan q =

2p b

The angle of diffraction of a-particle f = p – 2q f p = -q 2 2 cot

2p f Êp ˆ = cot Á - q ˜ = tan q = Ë2 ¯ b 2

p=

f 1 Ê 4 Ze 2 ˆ f b cot = Á cot 2 2 2 Ë 4p Œ0 mV 2 ˜¯ 2

Ê Ze 2 ˆ f p = impact parameter = Á ˜ cot 2 4 p Œ E Ë 0 ¯ where E =

(2.18)

1 mV2 = initial KE of the a-particle. 2

Equation (2.18) gives the relation between the impact parameter (p) and scattering angle, f.

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ81



Now the distance of closest approach, d =

2Ze 2 4p Œ0 E

Ze 2 = d/2 . Put in Eq. (2.18) 4pŒ0 E p=

d f cot 2 2

(2.19)

is the relation between the impact parameter (p), distance of closest approach (d) and scattering angle (f).

2.11.6

Rutherford Scattering Formula

All those incidenting a-particles aimed so as to strike the circumference of the circle drawn about the nucleus and having a radius p will be deflected at an angle f. All a-particles which strike within the shaded area (pp2) will be deflected through an angle greater than f. The area pp2 is called cross section and denoted by s. s = p p2

\

a-particle

f p pp2

‹‰ǤʹǤͻ౨Rutherford a-particle scattering

To calculate the total target area presented by all the nuclei within a foil of area A and thickness t, we assume that single scattering takes place, i.e., the foil is so thin that in passing through it an a -particle is scattered by any one of the nuclei. If N0 is the Avogadro number, W is the atomic weight of the scattering material and r is its density then the number of atoms per unit volume of the scatterer, n=

N0 r W

Volume of the foil = At Therefore, number of nuclei in the foil = nAt Hence, the target area for scattering by at least an angle f = s (Ant) As the incident beam cannot be aimed to strike anyone of the nucleus in the foil, because its total area is very large as compared to the cross section s, the probability that any one incident a-particle will be scattered at an angle greater than f is given by

82പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

N s s ( Ant ) = = s nt Ni A

where Ni is the number of a-particles incident on the foil and Ns is the number scattered particles at an angle greater than f. Therefore,

Ns = s nt = pp2nt Ni

Put

Ns Ze 2 f = f and p = cot , we have Ni 4pŒ0 E 2 2

È Ze 2 ˘ 2f f = pnt Í ˙ cot 4 p Œ 2 E 0 ˚ Î

(2.20)

This is Rutherford scattering formula. In an actual experiment, the detector measures the a-particles, scattered between the angle f and (f + df). The fraction of incident a-particles so scattered is found by differentiating Eq. (2.20), i.e., 2

È Ze 2 ˘ f 2 f df = –pnt Í ˙ cot cosec df 2 2 Î 4p Œ0 E ˚ The negative sign indicates that as f increases f decreases. If we place a fluorescent screen at a distance r from the foil and the scattered a-particles are detected by the scintillation they produce, then a-particles scattered between f and (f + df) will reach to a zone of a sphere of radius r and width rdf. The radius of the zone is r sin f. Hence, area of the screen struck by the a-particles, f f ds = (2pr sin f)(rdf) = 2pr2 sin f df = 4pr2 sin cos df 2 2 Total number of a-particles incident on the foil = Ni Therefore, number of a-particles scattered per unit area striking the screen = Nf N |df| = i ds 2 È Ze 2 ˘ f 2f Ni pnt Í ˙ cot cosec df 2 2 Î 4p Œ0 E ˚ = f f 4pr 2 sin cos df 2 2 Nf =

Ni nt Z 2 e 4

(8p Œ0 ) 2r 2E2sin 4 f2

(2.21)

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ83

 rdf r sin f df

r

f

Foil

Fig. 2.10

The quantitative verification of Rutherford’s Model was given by Geiger and Marden.

2.12

NUCLEAR FISSION

The phenomenon of disintegration of a heavy nucleus into two or more light nuclei more or less comparable mass with the release of a large amount of energy is known as nuclear fission. For example, if a thermal neutron (neutron with energy 0.03 eV) is bombarded against uranium atom (92U235), it will disintegrate into two nuclei. Barium (56Br141) and krypton (36Kr92) with the release of three neutrons and about 200 MeV energy. The nuclear reaction of uranium is given as 0n

1

+

92U

235

Æ

56Ba

141

+

36Kr

92

+ 30n1 + 200 MeV

The energy so released is due to the mass defect. The 92U235 nuclei split up into 141 and 36Kr92. They may divide into nuclei of several pairs of elements lying in 56Ba the central region of periodic table with slightly unequal nuclear masses. These are known as fission fragments. Hence, other made of 92U235 fission is 0n

1

+

92U

235

Æ

92U

236

Æ

40Zr

98

+

52Te

135

+ 30n1 + Energy

If we plot the percentage yield as the function of mass number (A) we get the curve as shown in Fig. 2.11.

% age yield

60

80

100

120 140 Mass numbers

160

‹‰ǤʹǤͳͳ౨—…Ž‡ƒ”ϐ‹••‹‘’”‘†—…–•

180

84പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

The yield is maximum at A = 95 and A = 139 and the distribution is called asymmetric. The yield is minimum at A = 118 which corresponds to symmetric splitting of 92U235 nucleus. Nuclear fission may occur spontaneously or it may be included. Most of the heavy nuclei with Z > 82 show spontaneous fission. For such nuclei the number of protons is very large and hence, exert large Coulomb’s force of repulsion among them which even exceeds the nuclear binding force. Induced nuclear fission takes place when a bombarding particle shares its energy with the nucleons of heavy nucleus and makes it unstable. If the projectile particle is positively charged, it requires sufficient energy to overcome the repulsive force exerted by the nucleus. But for neutral particle like neutron, it is neither affected by the presence of electric field produced by orbital electron nor positive charge on the nucleus.

2.13

BOHR AND WHEELER THEORY OF NUCLEAR FISSION

Bohr and Wheeler explained the phenomenon of nuclear fission on the liquid drop model of the nucleus. A nucleus is in equilibrium having spherical shape due to nuclear force similar to liquid drop which is having spherical shape due to surface tension. When a projectile particle such as neutron is bombarded on the nucleus it is completely absorbed by the nucleus and a compound nucleus is formed. The energy of the incoming particle is shared by all the nucleons of the two compound nucleus due to which their excitation energy increases, and oscillations are set up in the compound nucleus. The excitation energy or oscillations tries to deform the shape of the nucleus. Whereas nuclear force (surface tension) tries to keep its spherical shape. Hence, there is a competition between them. The competition will destroy the shape of the nucleus from spherical to elliptical and then to dumbbell. Finally, Coulomb’s force of repulsion between the two halves of this dumbbell exceeds the nuclear forces holding the nucleons, the nucleus breaks up into two segments and fission is said to take place. The different steps of nuclear fission of

92U

235

are shown in Fig. 2.12.

n

n

n

+ Energy U235

U236

Neutron

Unstable nucleus

Ba

Fig. 2.12

For spherical charge drop, the surface energy is given as Es = 4pR2s = 4pR02A2/3 s = asA2/3

Kr

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and the Coulomb’s energy, Ec =

3 Z(Z - 1)e 2 3 Z 2 e 2 @ 5 R 5 R0 A1/3

= ac

Z2 A1/3

As explained above to begin the fission process the shape of the nucleus should be deformed. This means R must become large resulting in an increase of surface energy and decrease of Coulomb’s energy. The net change in energy DE a (2 Es – Ec) (for two spherical balls Ba and Kr) Hence, DE = 0 if 2 Es = Ec or

ac

Z2 = 2 ¥ asA2/3 A1/3 2as Z2 = . Since as = 18 MeV, ac = 0.8 MeV ac A Z2 2 ¥ 18 @ 45 = A 0.8

Therefore,

Thus, according to Bohr-Wheeler theory, spontaneous fission occurs in nuclides Z2 Z2 values greater than 45. If is less than 45 fission is not expected to occur with A A unless some particle is captured by the nucleus which supplies the activation energy required.

2.14

SOURCE OF ENERGY IN NUCLEAR FISSION

Since nuclear fission of

92U

235

1

92U

0n

+

by a thermal neutron is represented as 235

Æ

56Ba

141

+

36Kr

92

+ 3 0n 1 + Q

The total mass of the particle before reaction is more than the total sum of masses of the particles after reaction. The decrease in mass results in the corresponding release of energy, i.e., Q = [{mu + mn} – {mBa + mKr + 3mn}] c2 Since we know that Mass of 92U235 = 235.1175 amu Mass of 0n1 = 1.00898 amu

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Total mass before nuclear reaction = 235.1175 + 1.00898 = 236.12648 amu Mass of

50Ba

141

= 140.9577 amu

Mass of 36Kr92 = 91.9264 amu Mass of three neutrons = 3 ¥ 1.00898 = 3.02694 amu Total mass after nuclear reaction = 140.9577 + 91.9264 + 3.02694 = 235.91104 amu Mass defect, Dm = 236.12648 – 235.91104 = 0.21544 amu Energy release = Dm ¥ 931 = 0.21544 ¥ 931 = 200.5 MeV. Out of this energy, 170 MeV is carried by the fission fragments as their kinetic energy 5 MeV by fission neutrons and rest of the energy in the form of g -rays.

2.15

CHAIN REACTION

The nuclear fission of

92U

0n

1

235

+

by thermal neutron is represented as

92U

235

Æ

50Ba

141

+

36Kr

92

+ 3 0n 1 + Q

Hence, the disintegration of uranium nucleus by thermal neutron produces three secondary neutrons. The neutrons so produced then cause the fission of three other 235 nuclei producing 9 neutrons, which in turn, can bring about the fission of 9 92U 235 nuclei and so on. Due to the multiplication of neutrons in this manner, the 92U whole of the uranium material will disintegrate in a short time producing a large amount of energy. Such a fission reaction is called uncontrolled chain reaction. It is so called because once the fission reaction gets initiated, then it proceeds at such a fast pace that it cannot be stopped and energy so produced can bring disastrous effects. An atom bomb is an example of an uncontrolled chain reaction. The uncontrolled chain reaction is shown in Fig. 2.13. Following points must be considered for continuous chain reaction: 1. Leakage of neutrons from the system. Some of the secondary neutrons produced may escape out of the system and will not take part in further fission. This leakage may be reduced by designing the system appropriately. 2. Absorption of neutrons by impurities. The secondary neutrons produced in the fission may be absorbed by impurities which are not fissionable. This loss can be reduced by using fissionable material free from impurities.

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ87

 Ba n n n U235

Ba n

U235

n

Neutron

U

235

Kr Ba n n n

n

Kr

Kr U235

Ba n n n Kr

‹‰ǤʹǤͳ͵౨Uncontrolled chain reaction

3. Absorption of neutrons by uranium-238. The natural uranium consists of U233, U235, U238 having relative abundance of 0.006 %, 0.71% and 99.28%, respectively. It is found that U238 is fissionable with fast moving neutrons and absorbs slow or thermal neutrons whereas U235 is fissionable with slow neutrons. As the relative abundance of U235 is less than U238. Hence, there is more possibility of collision of neutrons with U238 and are absorbed by them. Hence, in uncontrolled chain reaction we never use natural uranium. 4. Critical size or mass. In order to have sustained chain reactions in a sample of U235, it is required that the number of neutrons lost due to leakage or absorption should be much smaller than the number of neutrons produced in the fission process. This is possible if the size of uranium block has a size greater than a certain critical value. It is called critical size or mass.

2.15.1

Controlled Chain Reaction

To achieve a controlled chain reaction an arrangement is made for the particle absorption of neutrons available from each U235 fission as shown in Fig. 2.14. It shows that for controlled chain reaction one neutron is absorbed out of three neutrons produced in the fission of 92U235. The neutron is absorbed by cadmium rods which act as controlling rods. The system in which the nuclear chain reaction may proceed in a controlled manner is called nuclear reactor. Neutron

U235 Neutron

Neutron absorber

U235 Neutron

Neutron

Neutron absorber

‹‰ǤʹǤͳͶ౨Controlled chain reaction

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2.15.2

Neutron Reproduction Factor (K)

Neutron reproduction factor may be defined as the ratio of the rate of production of neutrons to the rate of loss of neutrons due to leakage and absorption, i.e., Reproduction factor, K =

Rate of production of neutrons Rate of loss of neutrons

1. If K = 1, the chain reaction will be steady or sustained. The size of the fissionable material used is said to be the critical size and its mass is the critical mass. This is required for steady power generation. 2. If K > 1, the chain reaction accelerates resulting in an explosion. This stage is said to be supercritical. This is how an atom bomb works. 3. If K < 1, the chain reaction gradually comes to halt. This stage is said to be subcritical.

2.16

NUCLEAR FUSION

The process in which two or more light nuclei fuse together to form a heavy nucleus with the liberation of large amount of energy is called nuclear fusion. For example, two deuterons can fuse together to form a helium nucleus with the release of 24 MeV energy. The reaction is given as 1H

2

+ 1H2 Æ 2He4 + 24 MeV

The above nuclear fusion reaction is energetically possible only if the mass of 2He4 nucleus is less than the sum of the masses of the two deuteron nuclei. The energy release is due to mass defect. Due to nuclear fusion of light nuclei into a heavy nucleus leads to increase in binding energy per nucleon. To carry out the fusion of two nuclei, they must be brought so much close to each other that they overcome the electrostatic force of repulsion and come within the attractive range of nuclear force. This is possible only when they approach each other with kinetic energy of the order of 0.1 MeV or more. This is possible only by raising the temperature of the two nuclei to about 107K. At this temperature, the thermal motion of the atom is with kinetic energy of the order of 0.1 MeV. For this reason, the nuclear fusion reaction is also known as thermonuclear reaction. The temperature of the order of 107K can be obtained first by the nuclear fission. Once the nuclear fusion begins the energy liberated is about seven times more than nuclear fission of some mass. The energy released in the case of sun or stars is due to nuclear fusion.

2.17

THERMONUCLEAR REACTIONS AND THE STELLAR ENERGY

The sun radiates energy at a tremendous rate of 4 ¥ 1016 Js–1. The stars also radiate a huge amount of energy. Further, they have been radiating energy for several billions of years. It has been estimated that chemical processes, like burning or nuclear fission cannot account for such a large amount of radiations for several years. Whereas no

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ89



known chemical process can be a source of such large amount of energy. Possibility of nuclear fission as the cause of this energy is also ruled out on the basis of the fact that the sun does not have the required abundance of heavy nuclei of fissionable matter. On the other hand about 98% of the mass of the sun consists of hydrogen and helium gases. Thus, we conclude that thermonuclear reactions are the main source of energy of the sun as well as of the stars. Two types of cyclic processes are proposed to be occurring in the sun and stars as explained below.

2.17.1

Carbon-Nitrogen Cycle

This process was proposed by Bethe. A proton reacts with Carbon-12 producing Nitrogen-13 and releasing energy. The Nitrogen-13 is unstable and emits a positron (e+) producing Carbon-13 which further reacts with proton, producing Nitrogen-14 and releasing energy. The Nitrogen-14 further reacts with proton producing Oxygen-15 and energy. The Oxygen-15 is unstable which emits a positron and some energy converting itself into Nitrogen-15. The Nitrogen-15 reacts with proton producing Carbon-12 and Helium. Hence, Carbon acts as a nuclear catalytic agent. The whole process can be represented as 1 12 Æ 13 1H + 6C 7N +Q1 7N 1H

Æ 6C13 + 1e0

1

+ 6C13 Æ 7N14 +Q2

1

+ 7N14 Æ 8O15 +Q3

1H

1H

13

1

+

15 8O 15 7N

Æ 7N15 + 1e0 +Q4 Æ 6C12 + 2He4

All the above nuclear reactions, when summed up gives the following net nuclear reaction 4 1H1 Æ 2He4 + 21e0 + Q i.e., four protons combine together to form a helium atom and two positrons along with the release of energy equal to 24.7 MeV. The carbon atoms with which the reaction starts returns at the end of the complete cycle and hence acts as a catalytic agent.

2.17.2

Proton-Proton Cycle

Another nuclear fusion cyclic process, which can account for the release of same amount of energy, is as below 1H

1

+ 1H 1 Æ 1H 2 + 1e 0 + Q 1

1

+ 1H2 Æ 2He3 + Q2

1H 1

1H

+ 2He3 Æ 2He4 + 1e0 + Q3

90പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

Hence, the reaction can be summed up as: 41H1 Æ 2He4 + 2 1e0 + Q The net energy released in the reaction is also 24.7 MeV. It is found that carbonnitrogen cycle takes place at rather late stage in the life of a star. The proton-proton cycle takes place at comparatively low temperature and carbon-nitrogen cycle occurs at a high temperature. However, in the interior of our sun, the temperature is of the order of 2 ¥ 106 K at which both of the above mentioned processes are equally probable. Stars having temperature more than that of the sun obtain their energy from carbon-nitrogen and those at lower temperature from proton-proton cycle.

Numerical Problems Problem 1 Complete the nuclear reaction (a) 35Br70 + 1H2 Æ ? + 0n1 (b) 17Cl35 + ? Æ 16S32 + 2He4 (c) 4Be9 + 2He4 Æ 3 2He4 + ? (d) 3Li6 + ? Æ 4Be7 + 0n1 Solution: Using the law of conservation of atomic number and atomic mass for nuclear reaction we have: (a) 35Br70 + 1H2 Æ 36Kr70 + 0n1 (b) 17Cl35 + 1H1 Æ 16S32 + 2He4 (c) 4Be9 + 2He4 Æ 3 2He4 + 0n1 (d) 3Li6 + 1H2 Æ 4Be7 + 0n1 Problem 2 In the relation 1H2 + 1H2 Æ 2H3+ 0n1 + Q, the energy released is 3.26 MeV. If the masses of 1H2 and 2He4 are 2.014102 amu and 3.016049 amu, respectively, find the mass of the neutron? Solution: Since Q = (2m(1H2) – m(2H3) – m(0n1)) ¥ 931 MeV If the masses are measured in amu Therefore, 2 ¥ 2.014102 – 3.016049 – m(0n1) =

3.26 931

m(0n1) = 1.008656 amu Problem 3 Calculate the energy released in kWh when 0.1 kg of 3Li7 is converted into 2He4 by proton bombardment. Given that mLi = 7.0183 amu, mHe = 4.004 amu and mH = 1.0081 amu. Solution: The nuclear reaction is represented as 3Li

6

+ 1H1 Æ 22He4+ Q

The Q-value of the reaction = (mLi + mH – 2 mHe) ¥ 931 = (7.0183 + 1.0081 – 2 ¥ 4.004) amu ¥ 931 = 17.13 ¥ 1.6 ¥ 10–13 J

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ91



=

17.13 ¥ 1.6 ¥ 10 -13 kWh , 1 J = 3600 kWh 3600

Hence, energy released by 0.1 kg of Li7 =

17.13 ¥ 1.6 ¥ 10 -13 0.1 = 6.53 ¥ 109 kWh. ¥ 3600 7.0183 ¥ 1.66 ¥ 10 -27

Summary 1. The reaction in which a certain species after suitable bombardment undergoes a change of character is known as ‘nuclear reaction’. Rutherford carried out the first induced nuclear reaction, i.e., N14(a, p)O17. 2. We study the nuclear reactions to know: (a) The quantitative aspects of mass and energy balance. (b) Inner aspects of ‘nuclear structure’ and nuclear spectroscopy. (c) How some ideas about nuclear forces can be tested. (d) More about ‘excited nuclear state’ and decay schemes from such energy levels. 3. Nuclear reaction involves only the inner nuclear structures of species and very large ( @ MeV) Q-values. 4. Nuclear reactions are similar to chemical reactions, but the basic difference between the two is that, a chemical reaction involves “change in the outer” electronic structure and very small value of energy (Q = ± , a few KeV) whereas nuclear reaction involves changes in the inner nuclear structure of a species and very large values of energy (Q = ± MeV). 5. Before the discovery of the neutron in 1932, only about ten nuclear reactions were known. These reactions were (a, p) type. But after the discovery of neutron and other particles of suitable energies were available from particle accelerators, many more reactions were studied experimentally. These reactions are divided into various categories. 6. Nuclear reactions have also been classified on the basis of energy of the probes (projectile particle) and A-value of the target nuclei. 7. Just as a chemical reaction obeys certain rules or laws, a nuclear reaction also obeys some fundamental laws. In other words, without even any one of these laws a nuclear reaction cannot be allowed. 8. James Chadwik was successful with the discovery of neutrons in 1932. He had performed a series of experiments to confirm the identity of bombardment products from BE target hit by a-particle. 9. Being electrically neutral, neutrons are a special class of probes. In fact, we have a number of reactions induced with neutrons than with any other particles. Neutrons possess some typical properties.

92പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

10. The prized reaction, induced with slow (thermal) neutrons, is the fission of U-235 that forms the basis of generation of nuclear power. 11. The mathematical analysis of a nuclear reaction X (a, b) Y using the conservation of mass energy and momentum is known as kinematics of the given nuclear reaction. 12. Disintegration energy or the reaction energy of a nuclear reaction is known as its Q-value. It is in MeV. 13. If Q is positive, i.e., the reaction is accompained with the liberation of energy. It is known as an exothermic or exoergic reaction. This reaction occurs if the mass defect (Dm) is positive. 14. If Q is negative, i.e., the reaction is accompanied by the absorption of energy. It is known as endothermic or endoergic reaction. In such a reaction mass defect is negative. 15. Nuclear cross section is an important parameter to indicate how much effective area the target atom or nucleus offers to the incident beam of particles. A probe striking within this area interacts with the target nucleus. Higher cross section value mean the reaction is more likely to occur and vice versa. 16. Nuclear cross section is different from geometrical cross section. The geometrical cross section of 48Cd113 is extremely small ( @ 1.06 barns), while its nuclear cross section is extremely large ( @ 20,000 barns) for neutrons. For this reason, cadmium rods are excellent control rods (absorbs fast moving neutron) in a nuclear reactor. Only few mm thick rods of cadmium are needed for this problem. 17. Bohr has given the theory of compound nucleus: A compound nucleus is formed when the incident particle is captured by the target nucleus. The energy of incident particle is quickly distributed among all the nucleons of the compound nucleus. Hence, compound nucleus is in the excited nucleus. 18. The compound nucleus has a lifetime of the order of 10–16s (i.e., 10–14s to 10–18s). The lifetime of the compound nucleus is sufficiently long as compared to the natural nuclear time, i.e., the time taken by a nuclear particle to pass through the nucleus ( @ 10–27 s). 19. Theory of compound nucleus is more convincing for high Z nuclear targets, provided the energy of incident particles is not very high. 20. With high energy particles striking certain target nuclei, so many distinct possible reactions with or without the formation of a compound nucleus, can occur. Such basic character changing reaction for the target nuclei is called “nuclear transmutation”. 21. A large number of reactions have been studied with probe energy varying from few MeV to about 100 MeV. Such reactions are denoted like (a, p), (p, n), (n, g), (d, a), (g, n), etc.

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ93



Short Answer Questions 1. What do you mean by nuclear reaction? 2. How will you identify a nuclear reaction? 3. Why chemical reactions are brought about by energies ~ keV, and nuclear reactions cannot be? 4. Can any nuclear reaction occur if it is energetically possible? 5. Which conservation laws are obeyed during a nuclear reaction? 6. What is the Q value of the nuclear reaction? What is its significance? 7. What are two specific types of reactions based on their Q-value? 8. Neutrons are considered better bombarding projectiles for nuclear reaction, why? 9. What do you mean by nuclear cross section? What is its physical significance? 10. How is the differential scattering cross section different from nuclear crosssection?

Long Answer Questions 1. What are different types of nuclear reactions? Discuss with examples? What is nuclear cross section? 2. What is the role of various conservation laws in a nuclear reaction? Mention the laws? 3. What is nuclear reaction? Mention the conservation laws in nuclear reaction? 4. How nuclear reactions are classified? What conservation laws are obeyed? 5. Describe the kinematics of nuclear reaction and obtain an expression for its Q-value? 6. Define Q-value of a nuclear reaction? Derive an expression for the Q-value of a nuclear reaction X (a, b) Y in terms of kinetic energies of incident and product particles and nuclei; assume the target nucleus to be at rest? 7. What are endoergic and exoergic reactions? Define and calculate the threshold energy of an endoergic nuclear reaction that proceeds through the formation of a compound nucleus? 8. What do you mean by Q-value of a reaction? What is its significance? How can it be determined experimentally? 9. Why are neutrons useful as bombarding agents in nuclear reactions? 10. How do you define nuclear cross section? What physical meaning you assign to it? Derive the formula. 11. Explain the concept of compound nucleus with assumptions? What are the possible reactions that may occur when a high energy particle approaches a nucleus? 12. Describe the compound nucleus model of nuclear reaction?

94പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

13. Complete the following reaction. Name these reactions also: (a) 3Li6 + ? Æ 4Be7 + 0n1 (b) 2He4 + 4Be9 Æ ? + 0n1 (c) 0n1 + 13Al27 Æ ? + 2He4 (d) 1H1 + 9C19 Æ 8O16 + ? (e) 15P31 Æ 14Si30 + ? [Ans. (a) 1H2 (b) 6C12 (c) 11Na24 (d) 2He4 (e) 1H1] 14. Calculate the energy in the reaction 0n

1

+ 3Li6 Æ 1H3+ 2He4

Given: mass of Li = 4.00260 amu mass of 1H3 = 3.016049 amu [Ans. 4.8 MeV] mass of 0n1= 1.008665 amu 7 15. A 0.01 mm tick 3Li target is bombarded with a beam intensity of 1013 protons/s. As a result 108 neutrons/s are produced. Calculate the cross section for this reaction. Given density of Li = 500 kg/m3. 16. Find the Q-value of the reaction: (a) 2He4 + 7N14 Æ 8O17+ 1H1 + Q (b) 2He4 + 4Be9 Æ 6C12+ 0n1 + Q Given mass of 2He4 = 4.002603 amu, 7N14 = 14.003074 amu, 8O17 = 16.999133 amu, 4Be9 = 9.012186 amu, 6C12 = 12.000 amu, 1H1 = 1.007825 amu, 0n1 = 1.008665 amu. [Ans. (a) – 1.193 MeV; (b) 5.7 MeV]

Multiple-Choice Questions 1. Nuclear reaction X (a, b)Y requires that KE of the probe must be at least of the order of (a) A few eV (b) A few keV (c) A few MeV (d) A few BeV 2. 1 barn represents (b) 10–30 m2 (a) 10–26 m2 –28 2 (c) 10 m (d) 10–32 m2 3. In a transformation reaction, the target end the product nuclei are (a) Totally unrelated to each other (b) Close neighbours in the periodic table (c) Temporarily stable (d) Mirror image of each other



EƵĐůĞĂƌZĞĂĐƟŽŶƐപ95

4. A typical nuclear reaction is acceptable only if it ensures the law of conservation (a) Charges (b) Charge and mass number (c) Mass number and momentum (d) Charge, mass number, momentum, mass energy, spin, parity and statistics 5. The first ever nuclear reaction was produced by (a) Rutherford (b) Bohr (c) Curie (d) Enrico Fermi 6. The compound nucleus theory does not hold good for the nuclei which are (a) Light (b) Intermediate (c) Heavy (d) Very heavy 7. Nuclear cross sections have the dimensions of (a) Area (b) Length (c) Volume (d) Density 8. Differential cross section is measured in (a) m2 (b) m3 (c) steradian (d) m2/steradian 9. If Bd, and Ba are the binding energies of deuteron and a-particle then (b) Bd > Ba (a) Bd < Ba (c) Bd = Ba (d) None 10. If n is the number of atoms per m3 in a material, s is the cross section of the material for some incident particles, then mean free path (l) of these particles in the material is equal to n 1 s (a) (b) ns (c) (d) n s ns 11. The theory of compound nucleus was given by (a) S.M. Bose (b) N. Bohr (c) Abdul Salam (d) Dirac 12. The level width T of an excited state of a nucleus is related to its mean lifetime ( t ) according to the relation h  1 (a) T = (b) T =  t (c) T = (d) T = t t t 13. Neutron was discovered by (a) Bothe and Becker (b) I. Curie and F. Joliot (c) Rutherford (d) J. Chadwick 14. Cadmium is an excellent choice for making ‘central rod’. (a) It is a silvery white soft material (b) It can slow down fast neutrons (c) It can greatly absorb the neutrons after they have been slowed down in a moderator (d) It is not costly

96പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

15. Neutrons have greater penetration power than allowed to pass through matter. It is because (a) They are light particles (b) They are bulky particles (c) They have no free existence (d) They are electrically neutral 16. An excited energy level that corresponds to an excitation energy greater than binding energy of the particle in compound nucleus is called a (a) Virtual level (b) Bound level (c) Ground level (d) All of the above 17. Stripping reactions are (a) Direct reaction (b) Compound nucleus reaction (c) Indirect reaction (d) None of the above 18. Which one is not a direct reaction? (a) Stripping reaction (b) Oppenheimer-Phillips Reaction (c) Knock-out reaction (d) Inelastic scattering of protons 19. Theory of compound nucleus is applied to (a) Spallation (b) Stripping reaction (c) Knock-out reaction (d) Radioactive capture of a-particle 20. When a nuclear reaction takes place, which of the following is true about the reaction I. The energy is conserved II. The electric charge is conserved III. Mass is conserved IV. The number of nucleons is conserved (a) I and II only (b) I, II, III only (c) III only (d) I, II, IV only 21. The name of the following nuclear reaction is

22. 23. 24. 25.

1

+

92U

235

Æ

+ 3 0n 1 (a) Fission (b) Fusion (c) a-deacy (d) b-decay What is the missing element from the equation: 4Be9+? Æ 6C12 + 0n1 (a) –1n0 (b) +1e0 (c) 1H1 (d) 2He4 14 4 What is the missing element from the equation: 7N +2He Æ8O17 + ? (b) 1H1 (c) 1H2 (d) 2He4 (a) –1e0 Which of the following is the correct product of a-decay 88Ra226 Æ 2He4 + ? (b) 85Pa232 (c) 87Fr232 (d) 86Rn222 (a) 86Th232 In a nuclear reactor, the role of a moderator is to (a) Absorb neutrons moving around (b) Considerably reduce the KE of the fast neutrons 0n

56Ba

141

+

36Kr

92

EƵĐůĞĂƌZĞĂĐƟŽŶƐപ97



26.

27.

28.

29.

30. 31. 32.

33. 34.

(c) Scattering fast moving neutrons out of the reactor zone (d) Keep the proper ratio of neutrons Which two statements are correct for nuclear reaction? (a) The Q-value of the nuclear reaction is variable. (b) The disintegration energy of a nucleus is the Q-value of the reaction involving that nucleus. (c) The Q-value of a reaction is nothing but the energy balance of that nucleus. (d) The Q-value of the reaction tells whether the reaction is reversible or irreversible. The nuclear reaction cross section of a nucleus (a) Different from geometrical cross section (b) Identically the same as its geometrical area (c) Always greater than geometrical cross section (d) Always less than geometrical cross section The cross section of a particular nucleus is (a) Always fixed values for all probes (incident particles) (b) Always different for different probes (c) Actual value that depends upon the nature of interaction (d) Not well defined Which of the following is correct for the number of neutrons in the nucleus? (a) N = A – Z (b) N = Z – A (c) N = Z + A (d) N = Z (e) N = A How many electrons are in the 6C12 atom? (a) 12 (b) 6 (c) 18 (d) 3 23 How many neutrons are in the 11Na atom? (a) 12 (b) 11 (c) 18 (d) 24 20 How many nucleons are in the 10Ne atom? (a) 12 (b) 30 (c) 18 (d) 10 (e) 20 How many protons are in the 7N14 atom? (a) 14 (b) 6 (c) 7 (d) 10 What law did Ernest Rutherford use to estimate the size of the nucleus? (a) Conservation of nucleon number (b) Conservation of angular momentum (c) Conservation of linear momentum (d) Conservation of energy

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35. Why are nuclear energy levels more complex than electron energy levels? (a) Nuclear energy levels depend only on attractive forces. (b) Nuclear energy levels depend on attractive and repulsive forces. (c) Nuclear energy levels are an order of one hundred times as great as electron energy levels. (d) Electron energy levels depend on the interaction between neutrons and electrons. 36. Isotopes of an element (a) have the same number of protons and electrons, but a different number of neutrons. (b) have the same number of protons and neutrons, but a different number of electrons. (c) have different number of protons. (d) have different number of electrons. 37. Binding energy is (a) the amount of energy required to break a nucleus apart into protons and neutrons. (b) the amount of energy required to break a nucleus apart into protons and electrons. (c) the amount of energy required to break a nucleus apart into electrons and neutrons. 38. If mH is the atomic mass of Hydrogen, mn is the mass of a neutron, and M is the atomic mass of the atom, which of the following is the mass defect formula? (b) Dm = Z ◊ mH + N ◊ mn + M (a) Dm = Z ◊ mH + N ◊ mn – M (d) Dm = Z ◊ mH – N ◊ mn + M (c) Dm = Z ◊ mH – N ◊ mn – M 39. When nucleons form a stable nucleus, binding energy is (a) created from nothing. (b) destroyed into nothing. (c) transformed into visible light. (d) released as high energy photons or particles. 40. What is the missing element from the following equation 88Ra226 Æ ? + 2He4 (b) 86Rn220 (c) 86Rn228 (d) 86Rn222 (a) 86Rn230

Answers to Multiple-Choice Questions 1. 8. 15. 22. 29. 36.

(c) (d) (d) (d) (a) (a)

2. 9. 16. 23. 30. 37.

(c) (a) (a) (b) (b) (a)

3. 10. 17. 24. 31. 38.

(d) (c) (a) (d) (a) (a)

4. 11. 18. 25. 32. 39.

(d) (b) (d) (b) (e) (d)

5. 12. 19. 26. 33. 40.

(a) (c) (d) (b, c) (c) (d)

6. 13. 20. 27. 34.

(a) (d) (d) (a) (d)

7. 14. 21. 28. 35.

(a) (c) (a) (c) (b)

3 Radioactivity

3.1

INTRODUCTION

There are about 2450 known isotopes of 100 odd elements in the Periodic Table. There are a few unstable nuclei which are above as well as below the nuclear stability curve. These unstable nuclei in order to achieve more stability break into smaller fragments and this phenomenon is known as nuclear fission and the whole process is called radioactivity. The phenomenon of spontaneous disintegration of nuclei with the emission of electromagnetic radiations or other particles like a-particle, b-particle, and g -rays.

3.2

RADIOACTIVITY

The phenomenon of spontaneous disintegration of nuclei of certain heavy elements of atomic number (A) greater than about 206 with the emission of highly penetrating radiations such as a-particles, b-particles and g -rays is called natural radioactivity. The elements which exhibit this property are called radioactive elements. The phenomenon of natural radioactivity was discovered by a French Physicist Henry Becquerel in 1896. He observed that uranium salts possessed a peculiar property of affecting a photographic plate even when the plate was in a light-proof package. Then he thought it must be due to certain active radiations emitted by uranium salts. These radiations were called Becquerel rays. The experimental evidence reveals that the phenomenon of radioactivity is not at all affected by the imposed conditions of temperature, pressure, chemical combination, etc. Therefore, electrons orbiting around the nucleus were not responsible for radioactivity. Hence, the radioactivity must be the property of heavy nucleus. The emission of active radiations indicates that the parent nucleus is unstable. We find that from the binding energy curve that BE/A decreases with the increase in the value of A for nuclei A > 140. This indicates that heavy nuclei are unstable. The heavy nuclei have large value of Z (atomic number) and hence, contain large number of protons. The mutual repulsion of protons reduces the binding effect of nuclear forces. This is the main cause of relative instability of heavy nuclei. Radioactivity results from this instability.

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3.2.1

Types of Nuclear Radiations

Experimental investigation has proved that radiations emitted by radioactive element consists of three kinds radiation depending upon their penetration in matter. One type of radiation which has small penetration power and can be easily stopped by an aluminium foil of thickness 0.02 mm is known as a-ray. a-rays possess two unit of positive charge and four units of atomic mass. Because of the large mass and charge, velocity of a-particles lies between 1.4 ¥ 107 ms–1 and 1.7 ¥ 107 ms–1. a-particles have large ionizing power. Each a-particle produces a large number of ions before being absorbed. a-particles are deflected by electric field and magnetic field. a-particles also produce fluorescence in certain substances like zinc-sulphide. They also affect photographic plate slightly. a-particles on being stopped, produce heating effect. a-particle is equivalent to helium nucleus (2He4). Other type of radiation with a comparatively large penetration power and which can easily pass through a few mm thick aluminium foil is known as b-particle. b-particles carry one unit negative charge (i.e., –1.6 ¥ 10–19 C) and mass 9.1 ¥ 10–31 kg which is same as the charge and mass of electrons. Hence, b-particles are nothing 1 th velocity of but a stream of fast moving electrons. They move with velocity 10 light. They are also deflected by electric field and magnetic field. b-particles can also produce fluorescence and can affect the photographic plate. b-particles have large penetration power but small ionization power. The last kind of radiation with maximum penetration power but small ionization power known as g -ray. g -rays do not carry charge. Hence, they are not deflected by electric field and magnetic field. As the rest mass of g -rays is zero. Hence, they can move with velocity equal to velocity of light (v = c).

3.3

MODES OF RADIOACTIVE DECAY

The unstable nuclei having sufficient nucleonic energy decay spontaneously, releasing one or more types of radiations. i.e.,

Æ

X (Parent)

Y (Daughter)

+

R (Radiation)

The decay of radioactive nuclei (sample) is governed by the law of probability. Such processes are totally random throughout in the sense that we cannot predict which particular nucleus (atom) will decay first and when. Disintegration is purely by chance. Different ways or modes of radioactive decay are classified as under. 1. a-decay ZX

A

Æ

A–4 Z – 2Y

+ 2He4 (a-particle)

a-decay is possible only if the mass of the parent nucleus is greater than the mass of daughter nucleus.

ZĂĚŝŽĂĐƟǀŝƚLJപ101



2. b-decay is of three types: (a) Electron emission: ZXA Æ

A Z+1Y

+–1e0 + n (anti-neutrino)

Æ

A Z –1Y

+ 1e0 + n (neutrino)

(b) Positron emission:

ZX

A

(c) Electron capture: In this decay mode, the nucleus captures an electron from the atomic shell resulting in another nucleus and a neutrino. ZX

A

+

–1e

0

Æ

A Z – 1Y

+n

3. g-decay: It involves two sub-models. (a) Gamma emission: In this decay, a nucleus is in excited state it gets rid of excess excitation energy by emitting g -ray photon. *

ZX

A

Æ ZXA + g (photon)

(b) Internal conversion: Here a nucleus is in excited state, hands over its excess energy of excitation to one of the orbital electrons so that the latter is ejected. – – – – – –

+

g

+ a

b

+ +

Radioactive sample

‹‰Ǥ͵Ǥͳ౨Disintegration of radioactive nuclei

3.4

NATURE AND PROPERTIES OF NUCLEAR RADIATION

The radiations emitted by radioactive substances are of three types: 1. a -rays 2. b-rays 3. g -rays

Nature 1. Alpha rays consist of helium nucleus ( ++2 He 4 ) and possess two units of positive charge and four units of atomic mass.

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1 th the 10 velocity of light (v @ 3 ¥ 107 ms–1). They have a negative charge equal to 1.6 ¥ 10–19 C and mass equal to 9.1 ¥ 10–31 kg. 3. g -rays are electromagnetic waves of very short wavelength ranging from 0.005 to 0.5 Å. They move with velocity equal to the velocity of light in vacuum (3 ¥ 108 ms–1). 2. Beta particles are the stream of fast moving electrons moving with

Properties of a-particles 1. The a-particles shoot out from the radioactive material with very large velocities ranging from 1.4 ¥ 107 ms–1 to 1.7 ¥ 107 ms–1. The velocity of a-particle depends upon the radioactive substance from which they are ejected and for a given substance it is always the same. 2. They produce intense ionization in the gas through which they pass. The ionization power of a-particle is about 100 times more than that of b-particles and about 10,000 times more than that of g -rays. 3. They affect a photographic plate. 4. They produce fluorescence in substances like zinc sulphide or barium platinocyanide. 5. Range of a-particles: The distance that an a-particle can travel in air at atmospheric pressure is called its range in air. The range of an a-particles depends upon the radioactive substances from which the rays are given out. It also depends upon the nature of the medium and the velocity with which it comes out. The range is proportional to v3. 6. The a-particles are scattered when they pass through a thin sheet of mica, gold foil, etc. The divergence of a-particles from its straight line path is 2 to 3 degrees. 7. The rays are deflected by electric and magnetic fields showing that they are charged particles. The a-particles are deflected towards the negative plate. 8. They produce a heating effect. The evolution of heat is due to the stopping of a-particles by the substance. 9. When exposed to a-particles the body suffers incurable burns.

Properties of b-particles 1. b-particles shoot out from radioactive substance with very high velocity about 1 th the velocity of light. The velocity of all b-particles given out by a substance 10 is not the same. 2. They produce ionization in air but the number of ions produced is hardly 1 th of those produced by a-particles. The small ionization power of 100

ZĂĚŝŽĂĐƟǀŝƚLJപ103



3. 4. 5. 6. 7.

b -particles is due to the fact that they have small mass and move with very high velocity. They affect a photographic plate and their effect is greater than that of a-particles. They produce fluorescence in barium platinocynaide. They can penetrate through large thickness of matter, i.e., they can easily pass through 1 cm thickness of aluminium sheet. They are more readily scattered when they pass through matter, because their mass is very small as compared to the mass of atomic nuclei. They are deflected by electric field and magnetic field. Their direction of deflection indicates that they are negatively charged particles. The e/m values has been found same as that of cathode rays. This indicates that b-particles are the streams of fast moving electrons.

Properties of g-rays 1. They move with same velocity as the velocity of light (3 ¥ 108 ms–1) in vacuum. 2. They ionize the gas through which they pass but the ionization is very small. 3. They affect the photographic plate and their effect is greater than that of b-particles. 4. They produce fluorescence in barium cyanide, etc. 5. They are more penetrating than even b-particles and can easily pass through 300 cm thickness of iron. 6. They are not deflected by electric field and magnetic field. This property shows that they are electromagnetic waves of very small wavelengths. 7. They are deflected by crystals in the same way as X-rays. 8. They can cause the photoelectric emission if allowed to fall on metal surface. 9. The photon of g -rays have no rest mass. They represent a small mass only because of momentum (m = p/c) or energy (m =

3.5

E ). c2

LAW OF RADIOACTIVE DISINTEGRATION

First Law Radioactivity is a spontaneous disintegration of nuclei of certain radioactive elements and it does not depend upon external factors like temperature, pressure, etc., i.e., a radioactive element is in the state of disintegration depends upon the law of chance.

Second Law During disintegration of nucleus, either an a-particle or a b-particle is emitted. It is never that both the particles are emitted simultaneously. Also nucleus never emits

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more than one a-particle or b-particle at a time. g -rays always come after the emission of a-particle or b-particle.

Third Law With the emission of a-particle, a new nucleus is formed whose mass number reduces by four units and atomic number by two units i.e.,

ZX

A

Æ

A–4 Z – 2Y

+ 2He4

With the emission of b-particle, a new nucleus is formed whose mass number remains same but atomic number increases by one unit ZX

A

Æ

Z + 1Y

A

+ –1e0

If a nucleus emits g-rays, there is no change in the atomic mass and atomic number *

ZX

A

Æ ZX A + g

(For g -rays emission, the parent nucleus must be in the excited state which goes to the ground state with the emission of one or more g -rays.)

Fourth Law Rate of disintegration of a radioactive atom at any instant depends upon the number of radioactive atoms present in the sample. This is also known as radioactive decay law. Radioactive disintegration,

dN aN dt dN = -l N dt

(3.1)

where N is the number of atoms left at time t and l is constant called disintegration constant. (Negative sign indicates that with time number of atoms left go on decreasing.) From Eq. (3.1) dN = -l dt N Integrating the equation on both sides within the limits When t = 0, N = N0 (original number of atoms) t = t, N = N N

Ú

N0

t

dN = -l Údt N 0

(3.2)

ZĂĚŝŽĂĐƟǀŝƚLJപ105



|log N|N t t0 N0 = -l || log

N = -l t N0 N = e -lt N0 N = N 0 e - lt

Therefore,

(3.3)

is the number of atoms left after time t. Equation (3.3) shows that radioactive decay obeys the exponential law as shown in Fig 3.2. N0

N = N0e–lt

N

O

t

‹‰Ǥ͵Ǥʹ౨Exponential decay curve of radioactive nuclei

When and when

3.6

t = 0, N = N0 e0 = N0 t = f, N = N0 e–f = 0

DISINTEGRATION CONSTANT (l)

Since radioactive decay law is N = N 0 e - lt Put t = 1/l; N = N0e–1 = N0/e Thus, the disintegration constant of a radioactive element is the reciprocal of time at the end of which, the number of atoms decreases to 1/e times the original number of atoms (N0) of the sample.

3.7

HALF-LIFE (T )

Half-life of a radioactive element is the time during which number of radioactive atoms left is equal to half of its original number (t = 0). It is also defined as the time during which half of the radioactive atoms disintegrate.

106പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

N So, t = T(half-life); N = 0 2 Since N = N0 e - l t Therefore, when t = T, N =

N0 2

N0 = N0 e -lt 2 e lT = 2 lT = loge2 = 2.303 log10(2) = 2.303 ¥ 0.3010 = 0.693 T(half-life) =

0.693 . l

Hence, half-life varies inversely the decay constant of the radioactive material. Since decay constant (l) is characteristic of the radioactive sample taken, its half-life (T) too is characteristic property of the sample. Also the value of l or T for a given radioactive sample cannot be changed by any method. Remarks: 1. As discussed above for t = T, N = N0/2 2

In another half-life (i.e., after t = 2T); N =

1 Ê N0 ˆ 1 Ê 1ˆ ÁË ˜¯ = N0 = ÁË ˜¯ N0 2 2 4 2 3

1Ê N ˆ 1 Ê 1ˆ After the end of the third half-life (t = 3T); N = Á 0 ˜ = N0 = Á ˜ N0 Ë 2¯ 2Ë 4 ¯ 8 and so on. Hence, after n half-lives n

Ê 1ˆ t = nT; N = Á ˜ N0 Ë 2¯ or

t = nT,

N Ê 1ˆ =Á ˜ N0 Ë 2 ¯

n

2. The rate of disintegration or count rate of radioactive sample is also called activity of the sample which is directly proportional to the number of atoms left in the sample. A N Ê 1ˆ = =Á ˜ A0 N0 Ë 2 ¯ 3. As A0 = N0l = Initial activity at t = 0 A = Nl = Activity at time t

t/T

ZĂĚŝŽĂĐƟǀŝƚLJപ107



N = N 0 e - lt

Since Therefore,

A = A0 e - lt

i.e., both N and A decrease exponentially with time. 4. Number of atoms in the excited state, N x = N0 e - (Ex - E0 )/kT where E0 is the energy in ground state.

3.8

ACTIVITY AND ITS UNITS

Activity of the radioactive sample is defined as the rate at which radioactive atom disintegrates. Therefore,

Activity, A =

dN = lN dt

N = N 0 e - lt

As

- lt - lt Therefore, A = l N0 e = A0 e (where A0 = lN0 initial activity when t = 0) Hence, activity of radioactive sample also obeys exponential law.

Unit of A (a) Curie (Ci) The activity of a radioactive sample is said to be one curie if it undergoes 3.7 ¥ 1010 disintegrations per second. 1 Ci = 3.7 ¥ 1010 disintegration/s 1 mCi = 3.7 ¥ 107 disintegration/s 1 μCi = 3.7 ¥ 104 disintegration/s (b) Becquerel (Bq) It is the SI unit of activity. One Becquerel is the activity of radioactive sample if it undergoes one disintegration per second. 1 Bq = 1 disintegration/s (c) Rutherford (rd) The activity of a radioactive sample is said to be one Rutherford if it undergoes 106 disintigrations/second. 1 rd = 106 disintegration/s 1 m rd = 103 disintegration/s 1 μ rd = 1 disintegration/s

3.9

AVERAGE LIFE OR MEAN LIFE

The phenomenon of radioactivity is random and statistical. It is statistical because the number of particles (atoms) is very large, and random because we just cannot predict

108പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

which of the atom in a given sample will decay first and when. It is a process totally dependent on chance; as such the theory of probability is applicable. Average life of a radioactive sample is defined as the ratio of total life of all N0 atoms to the total number of atoms. Tav = - lt Since N = N0 e , we have

Total life of all N0 atoms Total number of atoms (N0 )

dN = - l N 0 e - lt dt - dN = l N0 e - lt dt

(3.4)

where dN are the radioactive atoms which disintegrate in small time dt second after the time t or the number of atoms disintegrating in the time interval t and (t + dt). Hence, each dN lives a life of t second. Total life of all dN atoms = tdN = tlN0e–lt (only magnitude) Total life of all N0 atoms 

=



- lt Ú tdN = Ú tl N0 e dt 0

0



- lt = l N0 Ú te dt 0

e - lt e - lt - 2 = l N0 t -l l



0

1 ˘ N È = l N 0 Í0 + 2 ˙ = 0 Î l ˚ l Therefore, average life, Tav =

Total life of all N0 atoms N0 /l = N0 N0

Therefore,

1 l

Tav =

The mean life or average life of a radioactive sample is inversely proportional to its decay constant. It is of great practical importance to see a relation between half-life and mean life. Since half-life, T = 0.693/l Mean life,

Tav =

1 l

ZĂĚŝŽĂĐƟǀŝƚLJപ109



Therefore,

T = 0.693 Tav

or

Tav =

3.10

T = 1.443 T 0.693

LAW OF SUCCESSIVE DISINTEGRATION (RADIOACTIVE GROWTH AND DECAY)

It was observed experimentally that radioactive nuclei naturally occur in three distinct groups. These groups are called three radioactive series. Each series represents a succession of radioactive stages and is said to represent a chain disintegration: A parent nucleus decays into a daughter nucleus, which further decays into another daughter nucleus and so on. The chain stops only when the end product is stable. A Æ B Æ C Æ …… Æ X (Stable) Obviously, during the intermediate stage we shall have only a mixture of various nuclei (A, B, C, …… etc.), also note that any two adjacent nuclides are parent and daughter. It is represented in Fig. 3.3 giving the decay and recovery curves for various species.

Number of atom

B

C

A

Time

‹‰Ǥ͵Ǥ͵౨Decay and recovery curve for A, B, C, etc

It should be noted from the graph that the specimen B (daughter A) starts building up in strength. When t = 0, the parent A is at peak. Likewise, C starts building up in strength when its parent B is at its peak and so on. Each nuclide is thus a parent of the very next nuclide that succeeds it. If at t = 0, the number of atoms A = N0 and those of B (yet to be formed) = 0, then after some time, their respective numbers will change. Let their numbers be now N1 and N2. If l1 and l2 are the disintegration constants of A and B, then Rate of decay of A =

dN1 = l1 N1 (Rate of formation of B = Rate of decay of A) dt

110പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

Rate of decay of B =

dN 2 = l2 N2 (Rate of formation of C = Rate of decay of B) dt

Therefore, net increase in the number of atoms of B is equal to rate of decay of B into C. So,

l1N1 = l2N2 = gain of B + loss of B =

dN 2 dt

(3.5)

-l t But N1 = N0 e 1 for A, so the above equation becomes

dN 2 = l1N0 e - l1t - l2 N 2 dt dN 2 + l2 N 2 = l1N0 e - l1t dt l t Multiplying by e 2 , the above relation becomes

e l2t

dN 2 + l2 N 2 e l2t = l1N0 e - l1t e l2t dt d (N 2 e l2t ) = l1N0 e( l2 - l1 )t dt

(3.6)

On integration, this results in N 2 e l2t = l1

N0 e( l2 - l1 )t + C ¢ (Constant) (l2 - l1 )

(3.7)

If t = 0, N2 = 0 put in Eq. (3.7), we get Ê l1 ˆ C¢ = - Á N0 put in Eq. (3.7) Ë l2 - l1 ˜¯ N 2 e l2t = l1

=

N2 =

N0 e( l2 - l1 )t Ê l1 ˆ N0 (l2 - l1 ) ÁË l2 - l1 ˜¯

l1N0 ( l2 - l1 )t - 1] [e l2 - l1 l1N0 - l1t l2t (e -e ) l2 - l1

(3.8)

Equation (3.8) gives the number of atoms B at any instant. To begin with, we had only the atoms of A; with the passage of time we start having atoms of B, then C, and so on. Obviously, while the number of parent atoms decreases, that of the daughter

ZĂĚŝŽĂĐƟǀŝƚLJപ111



atoms increases. Soon a situation is reached when the rate of decay (or the lN values) of the various atoms A, B, C …… are equal dN1 dN 2 dN 3 = = = …… = 0 dt dt dt (As N1, N2, N3 …… do not change with time) From equation (3.5) l1N1 – l2N2 = 0 (Therefore, l2N2 – l3N3 = 0 (Therefore,

dN 2 =0) dt

dN 3 = 0 ) and so on. dt

Thus, for the equilibrium state, we must have l1N1 = l2N2 = l3N3 = …… etc. We have two cases of radioactive equilibrium.

3.10.1

Permanent or Secular Equilibrium

If the half-life T1, T2 of the species A and B are such that T1 >>> T2, then their decay constant obeys l1 rn (the effective nuclear radius) the nuclear force becomes very weak due to increase of distance and balance the Coulomb repulsion force. The Coulomb’s force of repulsion experienced by a-particle due to daughter nucleus of charge (Z – 2)e, F=

1 (2e )(Z - 2)e 4pe0 x2

The potential energy of the a-particle is equal to the amount of work done in bringing a-particle from infinity to a distance x against the electrostatic force of repulsion. Therefore, potential energy at a distance x, x

Vx =

x

(Z - 2)e ¥ 2e dx 4pe0x2 0

Ú - Fdx = – Ú



=

2(Z - 2)e 2 4pe0 x

The maximum value of potential energy occurs where x = r0 (= 10–14 m) Vmax = For

92U

238

2(Z - 2)e 2 9 ¥ 109 ¥ 2(Z - 2) ¥ (1.6 ¥ 10 -19 )2 = 4pe 0 r0 r0

the maximum value of potential energy,

Vmax

9 ¥ 109 ¥ 2(Z - 2) ¥ (1.6 ¥ 10 -19 )2 = 41.47 ¥ 10–13 J = 25.9 MeV = -14 10

This gives the height of potential barrier or depth of potential well for a-particle. At a distance less than r0, i.e., within the nucleus, powerful nuclear attractive force comes into play which holds the nucleus together. As the a-particle is prevented from escaping, it has a negative potential energy within the nucleus. This potential energy is assumed to be constant and equal to –V0. Hence, the shape of the potential well is represented by constant potential energy region up to x = r0 given by Vx = –V0 and Coulomb’s repulsive potential for x > r0 is given by Vx =

1 2(Z - 2)e 2 1 2Zd e 2 . = x x 4 p e0 4 p e0

where Zd = (Z – 2) atomic number of daughter nucleus or the height of the potential barrier at a distance x from the centre of the nucleus is

ZĂĚŝŽĂĐƟǀŝƚLJപ123



Vx =

the maximum height:

3.17

Vmax =

1 2Ze 2 4 p e0 x 1 2Ze 2 (x = r0) 4 p e 0 r0

THEORY OF a-DECAY

The nuclei with mass number more than 210 are unstable. It is because the volume of such nuclei is very large (V μ A) and short range nuclear force could not balance the long range electrostatic force of repulsion. To attain the stable configuration, a nucleus emits a -particle. With the emission of a-particle, mass number decreases by 4 units. This process continues until the volume decreases to such an extent that nucleons come within the attractive range. Now the nuclear force dominates the electrostatic force of repulsion and the nucleus attains a stable configuration. a-decay of radioactive nuclei ZXA is given by ZX

A

Æ Z – 2YA – 4 + 2He4 + Q

The amount of energy release is due to the mass defect. The kinetic energy with which a -particle comes out of the nucleus can be calculated by using the law of conservation of linear momentum and law of conservation of energy. We assume that decay of parent nucleus occur at rest. According to the law of conservation of linear momentum 0 = mYvY + mHevHe Therefore,

vY =

-mHe vHe mY

(3.9)

According to the law of conservation of energy 1 1 2 mY vY2 + mHe vHe = Q from Eq. (3.9) 2 2 2

1 Ê - mHe vHe ˆ 1 2 mY Á + mHe vHe = Q ˜ 2 2 Ë mY ¯ 2 2 vHe 1 mHe 1 2 + mHe vHe = Q 2 mY 2

Ê mHe ˆ 1 2 mHe vHe + 1˜ = Q Á 2 Ë mY ¯ 1 mY Ê ˆ 2 mHe vHe = Kinetic energy of a -particle = Á ˜ Q 2 + m m Ë He Y¯

124പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

1 Ê A - 4ˆ 2 mHe vHe Q = Á Ë A ˜¯ 2

(3.10)

Knowing the value of A and Q, energy with which a -particle comes out of a radioactive nucleus can be calculated. But from different a-particle emitters it has been found that kinetic energy of a-particle varies from 4 MeV to 8 MeV. But the potential barrier for a U238 nucleus is about 26 MeV. According to classical theory a-particles cannot come out of the nucleus as E < V0. In that case kinetic energy of the a-particle is negative and momentum is imaginary. Thus, it is very difficult to understand how a-particle lying within the nucleus can jump over a potential barrier which has more height than the total energy of the a-particle. However, the escape of a-particle from a radioactive nucleus can be explained on the basis of quantum theory by the application of wave mechanics. Let us consider one-dimensional potential barrier of constant height and width. Consider a particle of mass m incidenting on a potential barrier of height V0 and width b. Let the total energy E of the particle be less than the height of the potential barrier, i.e., E < V0. The potential barrier is represented as V(x) = 0 for x < 0 (Region I) = V0 0 £ x £ b (Region II) = 0 for x > b (Region III) V(x)

II

V0

I

III

E < V0 x=0

x=b

x

According to quantum mechanics the motion of the particle is governed by Schrödinger’s wave equation. One-dimensional Schrödinger’s wave equation is ∂2j 2m + (E - V )j = 0 ∂x 2  2 For Region I: V(x) = 0 ∂2j I ∂x Let

2

+

2m (E)j I = 0 2 2m (E) = a 12 2

ZĂĚŝŽĂĐƟǀŝƚLJപ125



∂2j I ∂x

2

+ a 12j I = 0

(3.11)

For Region II: V = V0 ∂2j II ∂x

2

-

2m (V0 - E)j II = 0 2 2m (V0 - E)j II = a 22 2 

Let

∂2j II ∂x

2

- a 22j II = 0

(3.12)

For Region III: V = 0 ∂2j III ∂x 2

+

2m (E)j III = 0 2

∂2j III ∂x 2

+ a 12j III = 0

(3.13)

Solutions for Eqs. (3.11), (3.12) and (3.13) are j I = Ae ia1x + B e - ia1x j II = Ce

-a 2 x

j III = Ee

ia1x

+ De + Fe

a2x

- ia1x

(3.14) (3.15) (3.16)

Since the particle cannot be reflected back in the third region therefore F = 0 Eq. (3.16) becomes j III = Ee ia1x

(3.17)

where A, B, C, D, E are constants and their values can be calculated from boundary conditions The wave function must be continuous at x = 0 and x = b Therefore, j I (0) = j II (0) from Eqs. (3.14) and (3.15) A+B=C+D

(3.18)

Also, j II (b) = j III (b) from Eqs. (3.15) and (3.17) Ce The first derivative

-a 2 b

+ De

a 2b

= Ee

ia1b

∂j must be continuous at x = 0 and x = b ∂x

(3.19)

126പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

∂j I ∂x

i.e.,

= x=0

∂j II ∂x

x=0

i a1A – i a1B = –a2C + a2D i a1(A – B) = –a2(C – D) ∂j II ∂x

and

= x=b

–a2Ce -a 2b + a2D e –C e

-a 2 b

+ De

a 2b

∂j III ∂x

(3.20)

x=b

= ia1E e ia1b ia 1 ia1b Ee a2

(3.21)

Ê ia 1 ˆ ia1b ÁË 1 - a ˜¯ e

(3.22)

a 2b

=

From Eqs. (3.19) and (3.21), we get C=

D=

Eea 2 b 2

Ee -a 2 b 2

2

Ê ia 1 ˆ ia1b ÁË 1 + a ˜¯ e

(3.23)

2

From Eqs. (3.18) and (3.20), we get ˆ a ia b Ê A + B = Ee 2 Á cosh a 2 b - i 1 sinh a 2 b˜ a2 Ë ¯

(3.24)

ˆ a ia b Ê A – B = iEe 2 Á cosh a 2 b - i 1 sinh a 2 b˜ a2 Ë ¯

(3.25)

From Eqs. (3.24) and (3.25), we get ˘ È a ˆ i Êa A = Ee ia1b Ícosh a 2 b + Á 2 - 1 ˜ sinh a 2 b ˙ 2 Ë a1 a 2 ¯ ˚ Î B=

iEe ia1b 2

ÈÊ a 2 a 1 ˆ ˘ + ˜ sinh a 2 b ˙ ÍÁ ÎË a 1 a 2 ¯ ˚

(3.26)

(3.27)

ZĂĚŝŽĂĐƟǀŝƚLJപ127



Transmission Coefficient (T)

T= T=

v3 E * E v1 A* A E* E A* A

T= –

=

but v3 = v1

E* E È ˘ È ˘ Êa a ˆ a ˆ i Êa E* e - ia1b Ícosh a 2 b - i Á 2 - 1 ˜ sinh a 2 b ˙ Ee ia1b Ícosh a 2 b + Á 2 - 1 ˜ sinh a 2 b ˙ 2 Ë a1 a 2 ¯ Ë a1 a 2 ¯ Î ˚ Î ˚ 1 2

a ˆ 1Êa cosh a 2 b + Á 2 - 1 ˜ sinh2 a 2 b 4 Ë a1 a 2 ¯ 2

T=

T=

1 2 2 È 4a a + a 24 + a 14 - 2a 12a 22 ˘ 1 + sinh 2 a 2 b Í 1 2 ˙ 4a 12a 22 Î ˚ 1 2

Ê a 2 + a 22 ˆ 2 1+ Á 1 ˜ sinh a 2 b 2 a a Ë 1 2 ¯

We know that

sinh(a2b) =

ea 2b - e -a 2b e -a 2b 2a 2b = - 1) (e 2 2

2a b 2a b When a 2b >>>> 1, then e 2 - 1 @ e 2

Then

sinh(a 2b) =

e 2a 2b 4

sinh2(a 2b) =

and

e -a 2b 2a 2b ea 2 b (e )= 2 2

Put the value of sinh2(a 2b) in the expression of T and neglecting 1 as compared to sinh2(a 2b) term, we get

T=

2

1 2

Ê a 2 + a 22 ˆ e 2a 2b 1+ Á 1 ˜ 4 Ë 2a 1a 2 ¯

Ê 2a a ˆ = 4 Á 2 1 22 ˜ e -2a 2b Ë a1 + a 2 ¯

128പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ 2

Ê 4a a ˆ = Á 2 1 22 ˜ e -2a 2b Ë a1 + a 2 ¯ The variation in the coefficient of the exponential in the above equation with E and V0 is negligible as compared with the variation in the exponential term itself. Further coefficient is never far from unity. Hence, we find that transmission coefficent is not zero, or there is a finite probability of the particle to penetrate through the barrier and reach region III. This theory was proposed by Gamow in collaboration with Gurney and Condon and is known as Gamow’s theory of a-decay. V0 E j II

jI

j III x=b

x=0

‹‰Ǥ͵Ǥ͹౨Gamow’s tunneling effect

According to the Gamow theory: An a -particle in constant motion is contained in the nucleus by the surrounding potential barrier. It bounces back and forth from the barrier walls. In such collision with the wall there is definite probability that the particle will leak through the barrier.

3.18

THEORY OF b-DECAY: THE NEUTRINO THEORY

b -particles are a stream of fast moving electrons having one unit negative charge and mass equal to 9.1 ¥ 10–31 kg. When they come out of the nucleus, they have velocity 1 th the velocity of light. The radioactive nuclei (in which ratio of neutrons to 10 protons (n/p) is greater than that of stable nuclei) emit b -particles. Since we know that electrons do not exist inside the nucleus. It was proposed that neutron inside the nucleus is unstable and decays into proton and electron. An electron so produced by the spontaneous conversion of neutron into proton comes out of the nucleus in the form of b -particle. Thus, in b-decay, the mass number (A) remains the same but atomic number increases by one, i.e., from Z to (Z + 1), this process continues. The number of neutrons decreases and of protons increases until the stage when ratio n/p is close to a stable nuclei. The b -decay of a neutron is given as n Æ p + e–

(3.28)

ZĂĚŝŽĂĐƟǀŝƚLJപ129



and b-decay of radioactive nucleus is given as

No. of b+-particles

ZX

A

Æ

Z+1Y

A

5

+

–1e

64

4

0

+Q

(3.29)

Cu b +

3 End point energy

2 1 0.2

0.4

0.6

0

No. of b–-particles

Energy of b+-particles (MeV)

4 3

64

Cu b –

2 1 End point energy 0

0.2

0.4

0.6

Energy of b –-particles (MeV)

‹‰Ǥ͵Ǥͺ౨Continuous energy spectrum of e– from the b-decay

The amount of energy released is due to the mass defect, i.e., Q = (mX – mY – me)c2

(3.30)

Since parent and daughter nuclei are very heavy and hence remain at rest before and after disintegration. Hence, the energy so released is taken up by the b -particle. Emitted energy possessing maximum value of energy is defined as end point energy. However, the experimental graph shows that energy spectrum of b-particle is continuous and can have energy ranging from zero to a certain well defined maximum limit known as end point energy (Fig. 3.8). Also Eq. (3.28) or Eq. (3.29) does not obey the law of conservation of linear momentum and law of conservation of angular momentum (spin). Since p, n, e– are 1 fermions having spin angular momentum . Hence, total spin angular momentum 2 before disintegration is not same as after disintegration.

130പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

3.18.1

Neutrino Hypothesis

All these difficulties have been overcome by Pauli by predicting theoretically the existence of another particle called neutrino and its antiparticle anti-neutrino. The neutrino as well its antiparticle have zero rest mass (mass less) carry no charge 1 and have spin angular momentum ± . They travel with the speed of light. In 2 accordance with this theory the b-decay of neutron and radioactive nuclei is given by n Æ p + e– + v ZX

A

Æ

Z+ 1Y

A

+

–1e

0

+ v +Q

(3.31) (3.32)

Now the energy so released is carried by the electron and anti-neutrino. If the electron carries the maximum energy (end point energy) then the energy of antineutrino is zero. Hence, b-particle can have any value of energy from zero to maximum energy. The energy may be shared by the two particles (e– , v ) in any proportion. Hence, it explains the continuous b-ray spectrum. Equations (3.31) and (3.32) also obey the law of conservation of spin angular momentum.

3.19

THEORY OF GAMMA DECAY

A nucleus can exist in one of the quantized energy states, similar to an atom. A stable nucleus is normally found in its lowest energy state (ground state). An excited nucleus can fall back (or return) to its ground state with the emission of energy of range MeV in the form of photon. The energy range lies in the gamma region. Hence, the nucleus returns from the excited to ground state with the emission of one or more gamma photons. An alternative to gamma-decay, an excited nucleus in some cases may return to the ground state by giving up its excitation energy to one of the orbital electrons around it. The process can be thought of as a sort of photoelectric effect from within. It is better to regard this internal conversion as a direct transfer of energy from a nucleus to an electron (in a given atom). The emitted electron will have kinetic energy equal to the lost excitation nuclear energy minus the binding energy of the electron in the atom. Most excited nuclei have very short half-lives against g -rays, but a few remain excited for several hours. Such intermediate unstable states of the nucleus are known as its metastable states. A metastable state has appreciable lifetime. A long lived excited nucleus is called isomer of the same nucleus in its ground state. In other words, a nucleus and its isomer differ only in their energy states obviously, an isomer is a nucleus but not in its ground state but in the metastable state. From the gamma ray spectra, it has been found that inside the nucleus, nucleons (p and n) are arranged in different energy levels just like electrons outside the nucleus. With the emission of a-particle or b-particle, generally daughter nucleus remains

ZĂĚŝŽĂĐƟǀŝƚLJപ131



in the excited state which returns to the ground state with the emission of one or more gamma rays. But only those nuclear transitions are allowed for which the law of conservation of energy, momentum, angular momentum, and parity hold good. These rules governing the transitions (similar to atomic transitions) are known as selection rules. More than one gamma ray only are emitted if the daughter nucleus is not stable; being unstable, it will undergo gamma-emission and so on. The process will stop only when the end product is stable. 27Co

60

b-particle * Ni 60

(Unstable)

28

g (1.17 MeV)

When

g (1.33 MeV) 28Ni

60

(Stable)

Fig. 3.9

3.19.1

Properties of Gamma Rays

1. The g -rays are electromagnetic radiations, i.e., they can move with velocity equal to the velocity of light. 2. The g -rays have no charge or they cannot be deflected by the electric and magnetic fields. 3. The energy of the g -quanta is 10 keV to 5 MeV. 4. The penetration power of g -rays is very high. 5. The wavelength of g -rays is 1 Å to 2 ¥ 10–13 m. 6. The spin of g -rays is I = 1. 7. The lifetime of g -rays is in the range 10–7 to 10–11 s. 8. Gamma ray energies can be measured by the following methods: Wavelength measurement, absorption measurement, secondary electron methods (photoelectric method, Compton spectrometer), energy measurement by photodisintegration, and scintillation spectrometer method.

3.19.2

Internal Conversion

When a nucleus passes from a higher excited state to a lower excited state or ground state the difference in energy of the two states is emitted as a g -ray. An excited nucleus may also transfer its surplus energy directly to one of its orbital electrons, so that it ejects from the atom with kinetic energy E is given by

132പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

E = Ee – W where Ee is the available excitation energy, i.e., energy given out by the excited nucleus in passing from higher energy level to a lower energy level and W is the binding energy of the ejected electron in its shell of origin (i.e., energy required to eject electron from its orbit). The process is known as internal conversion to distinguish it from the ejection of electron from an atom by a high energy photon incidenting from outside which is known as external conversion. The emitted electron is called conversion electron. Thus, internal conversion and emission of g -rays from the nucleus are two alternative ways of accomplishing the same nuclear transition. The internal conversion is not a two-step process in which g -ray photon is first emitted and then it knocks out an orbit electron. It is a one-step process in which the excited nucleus interacts directly with the orbital electron, transferring its excitation energy to it. The energy of the ejected electron (b-particle) has discrete values. Since Ee and W have only discrete values of energy. The corresponding b-particle energy spectrum is therefore, a line spectrum having discrete energies. We can put the relation E = Ee – W = (E2 – E1) – W as the available excitation energy Ee is equal to the difference in energies of excited state (E2) and ground state (E1). The line with the lowest energy corresponds to conversion of K-electron and its energy is given by Ek = (E2 – E1) – Wk The next line corresponding to the conversion of L-electron and its energy is given by EL = (E2 – E1) – WL

3.19.3

Difference between Internal Conversion and Photoelectric Effect

In internal conversion the excited nucleus transfers its surplus energy directly to one of its orbital electrons so that it gets ejected from the atom with kinetic energy, E = Ee – W In photoelectric effect, when a high energy photon strikes a metal from outside, it ejects a free electron from the metal known as photoelectrons. If n is the frequency of the incident photon and v is the velocity with which the photo-electrons is ejected then hn = 1/2 mv2 + W where W is the work function of photo metal, i.e., energy required to eject the electron from the metal surface with zero velocity.

ZĂĚŝŽĂĐƟǀŝƚLJപ133



Thus, in internal conversion energy is supplied from within the nucleus and the ejected electrons are orbital electrons of the atom. In photoelectric effect energy is supplied from outside in the form of high energy photon and the electron ejected is a free electron in the metal.

3.19.4

Internal Conversion Coefficient

The total transition probability l from a nuclear state a to a nuclear state b is given by l = l g + lc where l g and l c are the partial decay constants for g -emission and internal conversion respectively. Internal conversion coefficient a is defined as the ratio of internal decay constant l c to the g -emission decay constant lr. a=

lc lg

If NC is the total number of conversion electrons emitted in a given time and Ng the number of g -ray photons emitted in the same transition during the same time, then Conversion coefficient, a =

Nc N l . Hence, a = c = c Ng lg Ng

If a g-ray has a definite value, then due to internal conversion it should emit group of electrons. Thus, for each g-ray there should be several conversion lines corresponding to the ejection of electrons from K, L, M… shell. If lK, lL, lM… are the transition probability for transitions from K, L, M… etc. shell, then partial internal conversion constant aK, aL, aM… are defined as aK = Hence,

lK l l , aL = L , aK = M lg lg lg

a = aK + aL + aM + … a=

lK + lL + l M + º N + NL + N M + º = K lg Ng

where NK, NL, NM, … is the number of conversion electrons emitted for the transition from K, L, M… shells in a given time and Ng is the number of g-ray photons emitted in the same transitions in the same time.

134പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

Numerical Problems Problem 1

Obtain the amount of

of 8 nCi strength. Half-life of Solution: Since

27Co

27Co 60

60

necessary to provide a radioactive source

is 5.3 years.

dN = 8 mCi = 8 ¥ 3.7 ¥ 107 disintegration/s dt

Half-life, T = 5.3 year = 5.3 ¥ 365 ¥ 24 ¥ 60 ¥ 60 = 1.67 ¥ 108 s l=

As

0.693 0.693 = = 4.14 ¥ 10-9 s–1 T 1.67 ¥ 108

dN dN = lN fi N = dt l dt

As

N= Mass of 27Co60, m = = Mass of

27Co

60

8 ¥ 3.7 ¥ 107 = 7.15 ¥ 1016 4.14 ¥ 109

M ¥ N (M = Molecular mass, NA = Avagadro number) NA 60 ¥ 7.15 ¥ 10 -6 g 6.023 ¥ 10 23

= 7.12 ¥ 10–6 g

Problem 2 The half-life of 38Sr90 is 28 years. What is the rate of disintegration of 15 mg of the isotope. Solution: T = Half-life = 28 years = 28 ¥ 365 ¥ 24 ¥ 3600 = 28 ¥ 3.154 ¥ 107 s mN A 15 ¥ 10 -3 ¥ 6.023 ¥ 10 23 = 90 Number of atoms in 15 mg of Sr = M = 1.0038 ¥ 10 20 Rate of disintegration,

dN 0.693 ¥N = lN = dt T =

Problem 3 half-life.

0.693 ¥ 1.0038 ¥ 10 20 = 7.877 ¥ 1010 Bq. 28 ¥ 3.154 ¥ 107

One gm of Ra226 has an activity of one curie. Calculate the mean life and

Solution: Rate of disintegration of Ra =

dN = 1 curie = 3.7 ¥ 1010 disintegration/s dt

ZĂĚŝŽĂĐƟǀŝƚLJപ135



Number of atoms in one gm of Ra226 =

mN A 1 ¥ 6.023 ¥ 10 23 = 226 M

dN dN = lN fi l = dt N dt

Since

l=

3.7 ¥ 1010 ¥ 226 = 1.38 ¥ 10 -11 s 23 6.023 ¥ 10

Mean life, Tav =

1 1 = 7.25 ¥ 1010 s (2298 years). = -11 l 1.38 ¥ 10

Half-life,

0.693 0.693 = = 5 ¥ 1010 s (1585 years). l 1.38 ¥ 10 -11

T=

Problem 4 Half-life of radon is 3.8 days. After how many days will 1/10th of a radon sample remain behind? Solution: Since half-life, T = 3.8 days Therefore, l =

0.693 0.693 = = 0.1824 day–1 T 3.8

Let t be the time in which 1/10th of the random sample remain behind then N 1 = = e -lt N0 10 elt = 10 fi lt = loge(10) = 2.303 ¥ 1 t= Problem 5 Curie?

2.303 2.303 = = 12.62 days l 0.1824

Calculate the activity of 1 gm of Bi209. Half-life of Bi is 2.7 ¥ 107 years in

Solution: T = 2.7 ¥ 107 = 2.7 ¥ 107 ¥ 365 ¥ 24 ¥ 3600 = 8.5 ¥ 1014 s–1 Disintegration constant, l =

0.693 0.693 = = 8.15 ¥ 10–16 s–1 T 8.5 ¥ 1014

Number of atoms in one gram, N =

mN A 1 ¥ 6.023 ¥ 10 23 = 209 M

N = 2.88 ¥ 1021 Since A =

dN = lN = 8.15 ¥ 10–16 ¥ 2.88 ¥ 1021 = 23.472 ¥ 105 disintegration/s dt

136പEƵĐůĞĂƌĂŶĚWĂƌƟĐůĞWŚLJƐŝĐƐ

A=

2.3472 ¥ 106 = 63.6 ¥ 10–6 curie = 63.6 μCi. 3.7 ¥ 1010

Problem 6 Half-life of a certain radioactive sample is 15 hours. How long will it take for 93.75% of its material to disintegrate? Solution: Amount of material to disintegrate in time t sec = 93.75% Amount of material left after time t sec = (100 – 93.75)% = 6.25% = 6.25/100 = 1/16 of its original amount Since

N Ê 1ˆ =Á ˜ N0 Ë 2 ¯

n

after n half-times n

Therefore,

n

1 Ê 1ˆ Ê 1ˆ Ê 1ˆ = Á ˜ fi Á ˜ =Á ˜ ¯ Ë 2¯ Ë 2¯ Ë 16 2

4

Therefore, n = 4 t = n fi t = nT = 4 ¥ 15 = 60 hr T Problem 7 A certain radioactive substance has half-life of 30 days. What is the disintegration constant (l)? Solution: Since t =

0.693 0.693 fil= l t

Disintegration constant, l =

0.693 = 0.0231 day–1. 30

Problem 8 Consider 1 gm of pure radium emitting a-particle. Calculate the electric current represented by this emission. Solution: Disintegration reaction of radium is 88Ra

226

Æ

86Rn

222

+ 2He4

Number of Ra atoms in 1 gm of radium, N=

mN A 1 ¥ 6.023 ¥ 10 23 = 226 M

Number of atoms disintegrate in half of the time period, N¢ = N/2 =

1 ¥ 6.023 ¥ 10 23 2 ¥ 226

Number of a-particles released = Number of radium atoms disintegrate =

1 ¥ 6.023 ¥ 10 23 2 ¥ 226

ZĂĚŝŽĂĐƟǀŝƚLJപ137



Total charge, q = N¢(2e) =

6.023 ¥ 10 23 ¥ 2 ¥ 1.6 ¥ 10–19 = 4.264 ¥ 102 C 2 ¥ 226

Current due to emission of a-particle, 4.264 ¥ 10 2 7 I = q/T = 7 (1 Y = 3.2 ¥ 10 s) 1590 ¥ 3.2 ¥ 10 = 0.84 ¥ 10–8 s = 0.0084 μA. Problem 9 A certain radioactive element disintegrates for an interval equal to its mean life. (a) What fraction of it remain? (b) What fraction has disintegrated? Solution: Since t = Tav =

1 and N = N0 e–lt l

Number of atoms left after t = Tav =

1 l

N = N0 e–lt = (a)

N0 e

N = fraction of material (atoms) left = 1/e = 37% N0

(b) Fraction of material disintegrate = (1 – 1/e) = 63%

Unsolved Numerical Problems 1. 2. 3. 4. 5.

6.

The radius of oxygen nucleus (8O16) is 2.8 ¥ 10–15 m. Calculate the radius of [Ans. 6.55 ¥ 10–15 m] the lead nucleus (82Pb205). Compare the radii of two nuclei with mass number 1 and 27 respectively. Also find ratio of their density. [Ans. 1 : 3, 1 : 1] Calculate the binding energy of alpha-particle. Given that mp = 1.0073 amu, mn = 1.0087 amu, mass of a-particle = 4.0015 amu. [Ans. 28.4 MeV] 209 Calculate binding energy per nucleon of 83Bi , given that m (83Bi209) = 208.980388 amu, mn = 1.008665 amu, mp = 1.007825 amu. [Ans. 1639.37 MeV] Calculate (i) mass defect, (ii) binding energy, and (iii) binding energy per nucleon for a 6C12 nucleus. Given that: mass of 6C12 nucleus = 12.0000 amu, mp = 1.007825 amu, mn = 1.008665 amu. [Ans. (i) 0.09894 amu (ii) 92.1 amu (iii) 7.67 MeV/N] The neutron separation energy is defined as the amount of energy required to remove a neutron from a nucleus. Obtain the neutron separation energies of the 20Ca41 and 13Al27 from the following data. mn = 1.008665 amu, m(20Ca40) = 39.962591 amu,

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m(20Ca41) = 40.962278 amu, m(13Al26) = 25.986895 amu, m(13Al27) = 26.981541 amu. 7.

8.

9. 10.

11. 12.

13. 14. 15. 16. 17. 18. 19.

[Ans. 8.36 MeV, 13.05 MeV]

Calculate the end point energy of b -particle can have in the following decay 19 Æ 19 0 8O 9F + –1e + n 19 Given that: m(8O ) = 19.003576 amu, m(9F19) = 18.998403 amu, m(–1e0) = 0.000549 amu [Ans. 4.3 MeV] The sun is believed to be getting its energy from nuclear fusion of 4 protons (1H1) to form a helium nucleus (2He4) and a pair of positrons (e+). Calculate the energy released per fusion in MeV given mp = 1.007825 amu, me = 0.000549 [Ans. 25.7 MeV] amu and mHe = 4.002603 amu. Find the nuclear density of 92U235 if R0 = 1.1 ¥ 10–15 m. [Ans. 2.98 ¥ 1017 Kgm–3]. Calculate the packing fraction and average binding energy per nucleon for 8O16 of nuclear mass 15.9949 amu, mp = 1.007825 amu and mn = 1.0088665 amu. [Ans. 129.066392] Calculate packing fraction of a-particle from the given data ma = 4.0028 amu, [Ans. 7.58 ¥ 10–3 amu/N] mp = 1.00758 amu and mn = 1.00897 amu. What is the power output of 92U235 reactor if it takes 30 days to use up to 2 kg of fuel and if each fission gives 185 MeV of usable energy? [Ans. 58.3 ¥ 106 Watt] How many a and b particles are emitted when 92U238 decays to 82Pb206. [Ans. 8a and 6b] 1 14 Show that the nuclear density of 1H is about 10 times greater than the atomic density. Assume the atom to have the radius of first Bohr’s orbit. 1 gm of radioactive material takes 50 sec to lose 1 centigram. Find the decay constant and also half-life? [Ans. l = 2.03 ¥ 10–4 s–1, T = 57 min]. 1 gm of Ra is reduced by 2.1 mg in 5 years through a-decay? Calculate halflife? [Ans. 1672 years] Tritium (1H3) has half-life of 12.5 years. What fraction of a pure sample of 1H3 will remain undecayed for 25 years. [Ans. 25% or ¼]. Calculate the decay constant for an element whose half life is 20 years. [Ans. l = 3.466 ¥ 10–2 year–1] Disintegration constant of uranium (U234) is 8.8 ¥ 10–14 sec–1. What fraction of this element will be left after 2 ¥ 105 years. [Ans.

20.

N = 0.5696]. N0

The half-life of Rn is 3.8 days. After how many days will only 1/20th of this sample remain? [Ans. 16.5 days (approx)]

ZĂĚŝŽĂĐƟǀŝƚLJപ139



21. 22. 23. 24.

25. 26. 27. 28. 29.

The activity of 1 gm sample of fermium (100Fm253) is 11.47 ¥ 104 Ci. Calculate the mass which will decay by this time. [Ans. 3.36 ¥ 107 dis/s] 226 Calculate the activity of 10 mg of Ra which has half-life of 1620 years. [Ans: 3.36 ¥ 108 disintegration/s] How long does it take for 60% of a sample to decay? Half-life of the sample is 3.8 days? [Ans. 5 days] Atomic ratio of 92U238 and 92U234 in a material sample is found to be 1.8 ¥ 104. The half-life of U234 is 2.5 ¥ 105 years. Find half life of U238. [Ans. T = 4.5 ¥ 109 years] 209 Calculate the activity of 1 gm of Bi with half-life 2.7 ¥ 107 years. [Ans. R = 6.36 ¥ 10–5 Ci] 226 1 gm of Ra has an activity of one Curie. Calculate the mean life and halflife? [Ans. Tav = 2298 years, T = 1585 years] 238 The half-life of 92U against the a-decay is 4.5 ¥ 109 years. What is the activity [Ans. 1.235 ¥ 103 Bq] of 1 g of 92U238? Half-life of a radioactive sample is 40 days. Find the time taken for the activity to decay 30% of its initial value? [Ans. 69.49 days] The nuclei P and Q have equal number of atoms at t = 0. Their half-lives are Ê dN ˆ 3 hr and 9 hr respectively. Compare their rate of disintegration Á after Ë dt ˜¯ 18 hours?

Summary 1. The rate of disintegration of radioactive nuclei is not at all affected by the imposed conditions of temperature, pressure, chemical combinations, etc. 2. Radioactivity is a phenomenon of disintegration of nuclei of certain radioactive elements and not orbiting electrons. 3. The radiations emitted by the radioactive nuclei are a-particles, b-particles and g -rays. 4. a-particles, b-particles possessing charge are deflected by electric field and magnetic field but g -rays are the em radiation and are not deflected by E and B. 5. The radioactive process cannot be checked through physical or chemical means. It is a statistical phenomenon. 6. The discovery of artificial (induced) radio activity by Joliot and Irene Curie widened the scope of researching in this field. 7. A variety of radioactive isotopes are found very useful in agriculture, medicine, industry, and food preservation, etc. India earns substantial foreign exchange through the export of radioisotopes. 8. Rutherford and Soddy formally developed the famous law of radioactivity N = N0e–lt.

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9. The decay constant l for a given radioactive species is the reciprocal of the time during which the number of atoms decreases to 1/e times the original value. 10. l is higher for fast decaying specimens and vice versa. Also l = 0 for a stable specimen. 11. The half life (T) and disintegration constant (l) are related by the equation 0.693 . T= l 0.693 1 , Tav = . 12. The average or mean life of a sample related to T and l as T = l l So T = 0.693 Tav. 13. The intensity of radioactivity or simply the activity (R) is measured in terms of curie and rutherford. While 1 Ci = 3.7 ¥ 1010 disintegration/s 1 rd = 106 disintegration/s 14. Radioactive series represent a succession of radioactive stages, forming a chain disintegration. 15. There are three series of naturally radioactive nuclides (Z > 82) namely, the uranium series, the actinium series and Thorium series. In all the three series, the end product (stable) is some isotope of Lead (Pb). 16. Nuclides with Z < 82 can also be radioactive. Typical examples are C14, Na24, P32, S35, Co60, Nd144, Gd152, Hf174, Pt190, etc. 17. Radioactive dating and carbon dating are well established techniques that are used to determine the age of a rock or dead old trees, human or animals (extinct). 18. Geologists use radioactivity for rocks, etc., while anthropologists and archaeologists use carbon dating. 19. Radiation hazards can be due to both natural as well as manmade causes. 20. Radioisotopes, cosmic rays, etc., are the natural causes of harmful radiations. 21. Electronic devices like TV, computers, X-rays and Nuclear reactors, etc., are the manmade sources of harmful radiations. 22. Safety measures against radiations are essential. 23. Maximum permitted dose for human being safety is 5r (roentgen) per year. The unit of radiation dose is roentgen which is the quantity of a radiations that can produce 1 esu of charge in 1 cc of dry air.

Short Answer Questions 1.

Why

92U

238

is not suitable for chain reaction?

Ans. The natural uranium consists of three isotopes namely U233, U235 and U238. The relative abundances of three isotopes are 0.006%, 0.714% and 99.28% respectively. The slow moving neutrons having energy of the order of 0.03 eV

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can cause the fission of U235 nuclei, but the fission of U238 can be carried out by fast neutron having energy more than 1 MeV. Natural uranium contains practically U238 nuclei, therefore when neutrons of energy greater than 1 MeV are used to carry out the fission of U238 nuclei, the secondary neutrons so produced get slowed down on colliding against U238 nuclei and they are no longer capable of causing further fission of U238 nuclei. 2.

What is critical size and critical mass as regards to nuclear fission?

Ans. During the nuclear fission of U235, three neutrons are produced in every fission reaction. Three neutrons produce further three fission in the Uranium block. If the size of the uranium block is small, a few neutrons may be lost from the block without causing fission. In case, the number of neutrons lost per second is just equal to the number of neutrons produced per second then the size of uranium block is called the critical size and corresponding mass of the block is called the critical mass. 3.

What are thermal neutrons?

Ans. The neutrons which have been slowed down as a result of collision the hydrogen nuclei of the moderator are called thermal neutrons. Such neutrons possess energy of about 0.025 eV, which corresponds to the energy of the neutrons at 3 room temperature kT . Such neutron can cause fission of U235 nuclei. 2 4. What is nuclear fusion reaction? Why is nuclear fusion difficult to carry out? Ans. The process in which light nuclei combine together to form a heavy nucleus with the release of large amount of energy is defined as nuclear fusion. In the case of nuclear fusion, nuclei of two atoms have to be brought within the attractive range; hence it requires high temperature ( @ 107 K) conditions. This high temperature cannot be produced by any physical means. Because of this reason it is difficult to carry out. 5.

You are given two nuclei 3X7 and 3Y4. Explain giving reason as to which one of the two nuclei is likely to be more stable?

Ans. In case of 3X7; ratio of neutron to proton =

N 4 = = 1.33 P 3

N 1 = = 0.33 P 3 For stable nuclei, the ratio of neutrons to protons should be close to one. Hence, 7 4 3X is more stable than the nucleus 3Y . In case of 3Y4; ratio of neutron to proton =

6.

Binding energy of 8O16, 17Cl35 are 127.35 MeV and 289.3 MeV respectively. Which of the two nuclei is more stable?

Ans. It is not the total binding energy of the nucleus which tell about its stability but it is the binding energy per nucleon which gives the stability of the nucleus. More the BE/A more be the stability of the nucleus

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For 8O16; BE per nucleon = For

7.

17Cl

35

127.35 = 7.96 MeV/N 16

; BE per nucleon =

289.3 = 8.27 MeV/N 35

As stability of the nucleus a BE/A Therefore, 17Cl35 is more stable than 8O16. Which property of nuclear force is responsible for the constancy of BE/A?

Ans. Nuclear forces are saturated in nature and hence B.E./A is independent of mass number. This property makes BE/A constant for most of the nuclei. 8.

What is the effect on n/p ratio in nucleus when (1) an electron (2) a positron is emitted?

Ans. When a nucleus emits an electron, a neutron is converted in a proton. Therefore, number of neutrons decreases and the number of protons increases. The neutron-proton ratio (n/p) decreases. In the emission of positron (e+) proton is converted into neutron and hence ratio n/p increases. 9.

A chain reaction dies out sometime, why?

Ans. A chain reaction may die sometimes due to any of the following reasons: 1. Size of fissionable material may be less than the critical size. 2. Mass of the fissionable material may be less than the critical mass. 3. Neutron absorbing material might absorb neutrons at a faster rate than the rate at which they are produced. 10.

What is the essence of the theory of a-decay?

Ans. a -particles (2He4 nuclei) exists inside the nucleus. Their energy is much less than the height of nuclear potential barrier. Classically, we cannot explain their escape from a radioactive nucleus. Quantum mechanically, it is possible for a -particle to leak through such a high walled potential barrier. This is defined as quantum tunneling effect. 11.

What is neutrino hypothesis?

Ans. Because of the problem faced in understanding of b-decay process Pauli proposed that the b -particle emission, radio nuclei is also accompanied by a massless, chargeless particle having half integral spin particle known as neutrino (n). 12.

Is a free neutron a stable particle?

Ans. No, a free neutron is not a stable particle. It decays spontaneously into proton, electron and antineutrino, i.e., n Æ p + e– + n Its half-life is 11 minutes.

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13.

Explain one similarity and one dissimilarity between nuclear fission and nuclear fusion?

Ans. Nuclear fission involves splitting of a heavy nucleus into two or more light nuclei. Whereas nuclear fusion in the process in which two or more light nuclei combine together to form a heavy nucleus. Similarity between the two is that both involve mass defect (Dm) and hence, nuclear energy is released in both the processes as per the relation E = Dmc2. 14,

What is natural radioactivity? Can it be controlled?

Ans. It is the process of spontaneous disintegration of an unstable nucleus (Z > 82) with the emission of highly penetrating radiations such as a-particles, b-particles and g -rays. The natural radioactivity cannot be controlled by any means like change of temperature, pressure, chemical combination, etc. 15.

What is the difference between X-rays and g -rays?

Ans. Though both are electromagnetic radiations having wavelength l @ 1 Å and l @ 10–3 Å, respectively, their origins are different. X-rays are high energy photons that are released when an atomic electron undergoes a transition from an outer energy orbit to some inner orbit. A g -ray photon is released due to transition of nucleons (p and n) from higher nuclear process, X-ray emission in extra nuclear. 16.

Differentiate between artificial and natural radioactivities?

Ans. Just like natural radioactivity we can also have an artificial one. We convert any stable nucleus into a radioactive one by bombarding it with a suitable high energy particles (probes like a, b, n, d, etc.) having energy of the order of a few MeV. The radioactive isotopes also disintegrate like natural radioactive element. 17.

What is a radioactive series? Name the various radioactive series?

Ans. Whenever a radioactive nuclide decays through the emission (absorption) of certain subatomic particles one by one, it forms a sort of chain of intermediate elements that are also radioactive. This family or chain of radioactive elements is called radioactive series. Radio nuclides are generally divided into three distinct groups or series: 1. Uranium series (or 4n + 2 series), with 92U238 at its head. 2. Actinium series (or 4n + 3 series), with 89Ac227 at its head. 3. Thorium series (or 4n series), with 90Th232 at its head. 18.

How do you define half-life and average life of a radioactive material?

Ans. Half-life is the duration during which half of the radioactive sample disintegrates or the sample is reduced to exactly half (50%) of its original value.

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0.693 , where l is the disintegration constant. Mean life is the l measure of the time span for which one radioactive atom of a given sample survived. Half-life T =

1 l = 1.44 T

Tav = Average (Mean) life = Therefore, 19.

T = 0.693 Tav or Tav

Find the half-life of a radioactive material if its activity drop to 1 th of its 64 initial value in 30 years.

N Ê 1ˆ Ans. Since =Á ˜ N0 Ë 2 ¯

t/T

N 1 Ê 1ˆ = =Á ˜ . Hence, N0 64 Ë 2 ¯ 6

Ê 1ˆ Ê 1ˆ ÁË ˜¯ = ÁË ˜¯ 2 2

6

t/T

t t = 6 or T = T 6

fi

T = 30/6 = 5 years. 20.

What fraction of atoms is left after ten half-lives of a radioactive sample?

N Ê 1ˆ =Á ˜ Ans. Since N0 Ë 2 ¯

t/T

, t = 10 T or

t = 10 T 10

N Ê 1ˆ 1 =Á ˜ = . N0 Ë 2 ¯ 1024 21.

Define two main units of intensity of radioactivity (activity)?

Ans. The two standard units of activity are: 1) curie 2) rutherford 1. 1 Ci is the activity of a radioactive sample which undergoes 3.7 ¥ 1010 disintegration/s. 2. 1 rutherford is the activity of a radioactive substance which undergoes 10 6 disintegrations per second. 22.

Neptunium series generally is not considered. Comment.

Ans. In Neptunium series, 93Np237 (or (4n + 1) series) at its head and 83Bi209 (stable) at the tail. This series also runs through 11 steps involving a, b decay. As a matter of fact, this series has actually been induced in laboratory. 23.

What is meant by electron capture?

Ans. Because of stronger Coulomb barrier due to protons, the positron emission is not possible in a radioactive nucleus, as it resorts to the electron capture.

ZĂĚŝŽĂĐƟǀŝƚLJപ145



The capture of an orbital electron by the nucleus transforms it into a stable one. Because 1p1 + –1e0 Æ 0n1 + n , effective value of Z reduces by one unit but neutron number increased by 1 unit. 24.

What is the usefulness of radioactivity? Or why geologists generally use uranium dating method?

Ans. It is a technique used by geologists (uranium dating) for estimating the age of a rock (and hence age of the earth) and archaeologists (carbon dating) for estimating the age of trees and animal long dead. 25. What is MPD? Ans. Maximum permitted dose (MPD) is the safe radiation level to which a worker can be exposed without any harmful effect. It is 5 roentgen per year. A radiation that produces 1 esu of charge in 1 cc of dry air = 1r. 26. What is the principle of radioactivity? Ans. Radioactivity/carbon dating are well established techniques that are used to determine the age of a rock/tree or a dead animal. It involves that measurement of the rate of parent and daughter atoms (say U and Pb or C14 and C12). The age is found using t = 27.

1 ÊN ˆ log e Á 0 ˜ . Ë N¯ l

Explain radioactivity is a random phenomenon?

Ans. It has been proved that radioactive decay follows the exponential law. This fact implies that the phenomenon is statistical in nature. Every nucleus in a sample of radioactive substance has a certain probability of decay but there is no way of knowing in advance which nuclei will actually decay in a particular span of time. In other words it mean that radioactivity is a random phenomenon. 28.

The isotopes of 92U238 decay successively to form 90Th234, 91Po234, 92U234, 88Ra226. Mention the radiation emitted in these steps.

Ans. The successive decay of 92U

238

Æ

90Th

92U 234

238

is given as

+ 2He4 (radiation emit a-particle)

Æ

234 + –1e0 (radiation emit b-particle) 91Po 234 Æ 234 + –1e0 (radiation emit b-particle) 91Po 92U 234 Æ 226 + 4 2He4 (radiation emit 2a-particle) 92U 88Ra 90Th

234

Long Answer Questions 1. Discuss the theory of a-decay and obtain the expression for the probability of emission of a a-particle per second. 2. What is b-decay? Develop its theory and explain what role does a neutrino play?

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3. Discuss b-decay and why were neutrinos predicted? How has this concept been successful? 4. Discuss the arguments which led Pauli to postulate the existence of neutrino. Mention the properties of neutrinos. 5. Describe neutrino hypothesis of b -decay. What is the evidence for the existence of the neutrinos? 6. What is the concept of nuclear potential barrier? Explain tunnelling of a-particles subjected to such a potential well. 7. Do all b-particles emitted by radioactive nucleus have same energy? Explain. 8. Give theoretical basis of a-emission and discuss Gamow’s theory of emission of a-particle from a radioactive substance. 9. What is b-decay? Discuss the energy spectrum curve from b-decay of a radioactive nuclide. What is end point energy? Is b-spectrum discrete or continuous? 10. Define chain reaction? Outline the conditions necessary for nuclear chain reaction. 11. Describe the phenomenon of nuclear fission. Explain nuclear fission on the basis of liquid drop model? 12. Explain the phenomenon of nuclear fusion. Give some examples? 13. Distinguish between nuclear fission and nuclear fusion. 14. Explain the source of energy in the sun or stars? 15. What is natural radioactivity? Explain the term half-life and mean life? Derive a relation for them? 16. Explain the laws of radioactivity? Show that radioactive disintegration obey exponential law? 17. Describe the law of successive disintegration? What are secular equilibrium and transient equilibrium? 18. Define disintegration constant and activity. Give the units of activity of a radioactive element? 19. Explain the terms: 1) Half-life 2) Mean life for a radioactive material. How are they related? 20. Obtain the equation representing growth and decay of a radioactive elements? What is (1) Secular equilibrium (2) Transient equilibrium. 21. Deduce statistical law of radioactive decay. What do you understood by activity and disintegration constant of a radioactive material? 22. Establish with arguments that radioactivity is a pure statistical phenomenon? 23. What do you mean by radioactive series? Discuss the changes in the values of Z, A of a radio nuclide for a, b, g emission? 24. What is an artificial or induced radioactivity? What is the need for the same? 25. What are radioisotopes? How are they produced? Where are they used? 26. Discuss radioactive dating?

ZĂĚŝŽĂĐƟǀŝƚLJപ147



27. What is the principle of radio carbon dating? 28. How is the age of the earth estimated from radioactive decay considerations? Derive the formula used? 29. In relation to radioactive growth and decay, what do you mean by the law of successive disintegration? Obtain the conditions for (1). Secular or permanent equilibrium (2). Transient equilibrium? 30. How is the age of rock estimated from the radioactive decay consideration? Derive the relation used. 31. Write a detailed note on radioactive dating. 32. What is radio tracing? Illustrate (with example) the technique employed. 33. What is radioactive series? Mention the important natural radioactive series. 34. Establish the relation lT = 0.693?

Multiple-Choice Questions 1. Radioactivity is confined almost entirely to the elements __ to __ in the Periodic Table (a) 60, 92 (b) 83, 106 (c) 92, 118 (d) None of the above 2. Which of the following rays are emitted during radioactivity? (a) Alpha-rays (b) Beta-rays (c) Gamma-rays (d) All of the above 235 3. When the nuclei of U is split into approximately two equal nuclei, the amount of energy released per nucleon is (a) 0.45 MeV (b) 0.9 MeV (c) 1.35 MeV (d) 1.7 MeV 4. As per radioactive decay law, the small amount of disintegration of the isotope in a small period is equal to (b) lN (a) –lN (d) 2lN (c) –2lN where l = radioactive decay constant, N = number of radioactive nuclei present at any time t 5. The international system of units (SI) of radioactivity activity is (a) Becquerel (b) Curie (c) Fermi (d) Moles 6. The half-life of radioactive nuclei is (b) 0.793/l (a) 0.693/l (c) 0.693l (d) 0.793l where l = radioactive decay constant

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7. The average (mean) life for particle decay is (a) 1.145 times greater than half-life (b) 1.245 times greater than half-life (c) 1.345 times greater than half-life (d) 1.445 times greater than half-life 8. When U238 is bombarded with slow neutrons, it produces (b) 92U234 (a) 92U232 (c) 92U235 (d) 92U239 9. During fission of U235, the number of neutrons released per fission is (a) 1.5 (b) 2 (c) 2.5 (d) 3 235 10. During fission of U , the average kinetic energy per neutron is (a) 1 MeV (b) 2 MeV (c) 3 MeV (d) 4 MeV 11. What force is responsible for the radioactive decay of the nucleus? (a) Gravitational force (b) Weak nuclear force (c) Strong nuclear force (d)Electromagnetic force 12. Which of the following is an alpha particle? (b) –1e0 (a) +1e0 (d) 2He4 (c) 0n1 13. Which of the following is the ߚ+particle? (b) –1e0 (a) +1e0 (d) 1H1 (c) 0n1 14. Which of the following about the gamma ray is true? (a) It carries a positive charge. (b) It carries a negative charge. (c) It can be deflected by a magnetic field. (d) It has zero rest mass and a neutral charge. 15. Which type of radiation is stopped by a sheet of paper? (a) Alpha particle (b) Beta particle (c) Gamma ray (d) X-ray 16. Which of the following is used to detect fission reaction? (a) Mass spectrograph (b) Microscope (c) Through penetration (d) Thermometer 17. Elements which produce natural radioactivity are known as (a) Radio elements (b) Active elements (c) Nuclear elements (d) Radioactive elements



ZĂĚŝŽĂĐƟǀŝƚLJപ149

18. Elements whose atomic number is greater than 82 are called (a) Positive nuclei (b) Stable nuclei (c) Unstable nuclei (d) Negative nuclei 19. Alpha (a) particles are helium (He) nucleus with a charge of (a) e (b) 2e (c) 3e (d) 4e 20. During natural radioactivity unstable nucleus disintegrates to become more (a) Stable (b) Unstable (c) Excited (d) Unexcited 21. During branching of radioactive series (a) Daughter products are the same (b) Some time same some time different (c) Granddaughter product are the same (d) Granddaughter product are missing 22. Radioactive phenomenon associated with (a) Electron emission from atom (b) Transformation of nuclei (c) Fission of nuclei (d) None of the above 23. Various radiations come out of a radioactive material (a) When it is heated (b) When it is in atomic reactor (c) Under high pressure (d) Spontaneously 24. Radioactive isotopes are (a) Always unstable (b) Some time stable and sometime unstable (c) Always found in heavy elements (d) Finally not understood 25. For secular equilibrium between two radioactive nuclides we must have (b) T1 >>>> T2 (a) T1 > T2 (c) T1 < T2 (d) T1 >> T2 (a) T1 > T2 (c) T1 < T2 (d) T1 ϳ͘ϭപůĂƐƐŝĮĐĂƟŽŶŽĨůĞŵĞŶƚĂƌLJWĂƌƟĐůĞƐ Category

Boson (massless)

Lepton (Fermions)

Mesons (Bosons)

Particle Name

Photon

Rest Mass MeV/c2

Mean Life (s)

Spin

g

g

0

Stable

1

Electron Neutrino

ne

ne

0

Stable

1/2

Muon Neutrino

nm

nm

0

Stable

1/2

Tau Neutrino

nt

nt

0

Stable

1/2

–

+

0.51

Stable

1/2

106

2.2 ¥ 10–6

1/2

1784

5 ¥ 10

1/2

140

2.6 ¥ 10

135

8.3 ¥ 10

140

e

Electron

e

Muon

m–

m+

Tau

t

t

Pion

p

Kaon

– +

+

p

–

p

0

p

0

p

–

p

+

K

+

K

0

K

–

0

K

–13 –8

0

–17

0

2.6 ¥ 10

–8

0

494

1.3 ¥ 10

–8

0

498

9 ¥ 10

549

7 ¥ 10

–11

0

–19

0

Eta Meson

h

Proton

p

p

938.3

Stable

½

Neutron

n

n

939.6

Stable

½

2.5 ¥ 10

½

0

h

0

Lambda

L

0

L0

1116

Sigma Hyperon

S+

S–

1189

Baryons (Fermions) Xi Hyperon

Omega Hyperon

7.3

Symbol Antiparticle

–14

10

1197

1.5 ¥ 10

1315

3 ¥ 10

X0

W–

S0

S

–

S

X

–

X

X

0

W–

8 ¥ 10–11

1192

S

0

+ +

–10

½ ½

–10

–10

½ ½

1321

1.7 ¥ 10

–10

½

1672

1.3 ¥ 10–10

½

INTERACTION BETWEEN ELEMENTARY PARTICLES

Four kinds of interaction between elementary particles can account for all known processes in the physical universe on all scales and size. These interactions are

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classified into four groups in accordance with their characteristics and relative strength. These are: (i) (ii) (iii) (iv)

Gravitational interaction Electromagnetic interaction Strong interaction Weak interaction

7.3.1

Gravitational Interaction

It is the weakest of the four types of interaction and effective up to a large distance. Hence, it is a long-range interaction. The gravitational interaction has a measurable influence on objects of macroscopic size, but for sub-nuclear particles it is very small. The strength of inverse square gravitational force is expressed in terms of a constant known as universal constant (G ª 6.6 ¥ 10–11 Nm2 kg–2). As the value of G is very small. Hence, the interaction between the particles in nuclear physics can be neglected. The gravitational force is supposed to be operating through the exchange of yet undetected graviton. A graviton can be called the quantum of gravitational field and should be boson. But it has yet not been experimentally observed as a unique particle. The gravitons are held responsible for assembling (build-up factor) matter into planets, stars and galaxies and all those that goes into the formation of universe. The graviton should be massless, chargeless and stable, and travel with the speed of light. It should have spin 2 and hence a boson. The mutual force of attraction between two bodies is due to the exchange of gravitons. If energy is to be conserved, the uncertainty principle demands that range of the force a 1/mass of graviton (particle exchanged). Hence, range of gravitational force is infinite provided m = 0.

7.3.2

Electromagnetic Interaction

All charged particles are acted upon by electromagnetic interaction. Thus, electromagnetic interaction is charge dependent. This interaction is attractive as well as repulsive. The quantum of electromagnetic interaction is a photon. The inverse square law of Coulomb’s force may be conceived of as due to exchange of photon. The formation of electron-positron pair from a g -ray photon (g Æ e– + e+) is an example of electromagnetic interaction. The binding force between orbital electron and the positive nucleus is due to electromagnetic interaction. The electromagnetic force is characterized by a dimensionless coupling constant: a=

e2 1 = 4pe 0 c 137

which has been known as fine structure constant. Its magnitude is much greater than gravitational constant. As the rest mass of photon is zero. Hence, range of the force

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is infinite. The electromagnetic interaction also affects uncharged particles that have electric or magnetic moment.

7.3.3

Strong Interaction

The concept of strong interaction was introduced in order to explain the existence of stable nuclei. This force is nuclear force having magnitude 100 times greater than electrostatic repulsive force between the densely packed protons in the nucleus. The strong interaction is independent of charge but depends upon spin orientation of nucleons. According to Yukawa theory the strong interaction is due to the fast exchange of p mesons. The strong interaction is very small range, i.e., it is effective only up to the distance of 10–15 m and does not obey the inverse square law. The strong interaction between hadrons should be traceable to an interaction between quarks. The particle that quarks exchange to produce interaction are called gluons.

7.3.4

Weak Interaction

The range of force of weak interaction is less than 10–17 m. The field quantum of weak interaction corresponding to the photon and the pion is called “intermediate vector boson” of which seem to be of two kinds. As the range of weak interaction is very small hence the rest mass of such particles should be more than 30 times the mass of proton. The so called intermediate vector boson has not been experimentally detected as free particle. The comparative chart of four fundamental interactions is given in Table 7.2

d>ϳ͘ϮപŽŵƉĂƌĂƟǀĞŚĂƌƚŽĨϰ&ƵŶĚĂŵĞŶƚĂů/ŶƚĞƌĂĐƟŽŶƐ Interaction Strong

Particles affected Hadrons

Range

Relative strength

10–15 m

1

mesons

d

~ 10–2

photons

Electromagnetic Charged particle Weak

Hadrons and leptons

Gravitation

All

10

–17

d

m

10

–13

10–40

Particle Exchanged

intermediate vector bosons gravitons

Different scientists have tried at different times to establish some sort of similarity (under certain conditions) between various forces. For instance, Einstein (1920) tried to show if gravitational and electromagnetic interactions are same. He did not succeed. Weinberg, Glasgow and Abdul Salam have shown that two are the fundamentals—electromagnetic and weak are the same. Their work has rekindled the hope of unification of these forces of nature and defined as grand unified theory.

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7.4

QUANTUM NUMBERS ASSOCIATED WITH ELEMENTARY PARTICLES

The elementary particles decay and reactions not only satisfy the conservation laws of charge, mass-energy, momentum and spin but also some other conservation laws. These laws pertain to some intrinsic quantum numbers which are assigned to particles (and its antiparticles). The quantum numbers assigned to each entity are determined experimentally. They are as follows.

7.4.1

Charge Quantum Number (q)

In elementary particle decay, charge is not only conserved but is also quantized in units of electronic charge (e = 1.6 ¥ 10–19 C). Conservation of quantized charge can be expressed by assigning a charge quantum number q = charge/e to every particle, for example, the charge quantum number of electron is (– 1), proton (+ 1), positron (+ 1), antiproton (– 1), neutron (0), photon (0) and so on.

7.4.2

Lepton Number

Reference to the classification of elementary particles we note that electron, muon, tau muon and their antiparticles are leptons. Thus, are e–, m–, t , ne, nm, nt and e+,

+ m+, t , n e , n m , nt are leptons. The group of particles e–, ne and their antiparticles + ,n e e , are called e-lepton. For e–, ne the lepton number Le = ± 1 and for e+ ,n e , Le = –1. Similarly, the group of particles m–, nm and their antiparticles μ+, n m are called μleptons. We assign lepton number Lm = + 1 for m– , nm and Lm = – 1 for m+, n m . On the + same line Lt = + 1 for t , nt and Lt = – 1 for t , n e . The lepton number of all other particles like p, n, l , etc., is zero.

7.4.3

Baryon Number (B)

Baryon group includes the nucleons (p, n) hyperons (L0, S+, S–, S0, X–, X0, W–) and their antiparticles ( p , n , p 0 , S–, S+, S 0 , X+, X0 ). The baryon number B = 1 for baryons and B = – 1 for anti-baryons. For all other particles (lepton, meson, photon) B = 0.

7.4.4

Multiplet Number (M)

Some of the particles have nearly same mass but differ only in their charge. A multiplet number M is therefore assigned to particles to indicate the number of their different charge states. Example. Proton and neutrons having nearly the same mass but different charge, are regarded as different charge states of the same particle—the nucleon of mass 939 MeV. Thus, multiplet (M) both for proton and neutron is 2. Similarly, multiple number of p-mesons (p+, p–, p0) M = 3 and so on.

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7.4.5

Isospin (I )

In the multiplicity of atomic energy states due to spin the total number of states is given by (2s + 1). Where s is the spin quantum number for the electron. Similarly, we define iso spin quantum number (or isotropic spin quantum number) I from Ê M - 1ˆ the relation M = 2I + 1 or I = ÁË ˜ . Isotropic spin is considered a vector quantity 2 ¯ having magnitude I (I + 1) like angular momentum however, isotropic spin is a dimensionless quantity.  Further taking I as isotropic spin vector I in isotropic spin space its components along z-axis is given as Iz. Allowed values of Iz for a given value of I are I, (I – 1), (I – 2), …, (– I). For example, for nucleons, the multiplet number, M = 2 2-1 1 Ê M - 1ˆ Therefore, I = ÁË = ˜¯ = 2 2 2 1 1 1 1 Hence, the values of Iz are – , . Hence, Iz = for protons and Iz = – for 2 2 2 2 neutrons. For p–mesons, M = 3, I =

3-1 =1 2

And the value of Iz are + 1, 0, – 1 And for p+, Iz = + 1, for p0, Iz = 0 and for p –, Iz = – 1

7.4.6

Hypercharge (Y)

Another quantum number associated with particles (antiparticles) has been found quite useful in identifying particle families in hypercharge (Y). The average charge q of a particle in a multiplet group having same M is obtained as q=

q1 + q2 + º + qM M

The average charge ( q ) of nucleon =

0+1 1 = 2 2

The charge quantum number, q = Iz + q So charge quantum number of proton = And charge on neutron, q = –

1 1 + =1 2 2

1 1 + =0 2 2

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The quantum number known as hypercharge (Y) is defined as double of average charge ( q ) of the multiplet. i.e.,

Y= q = q= q=

As Therefore,

2 q Y/2 Iz + q = Iz + Y/2 Iz + Y/2

The value of Y for different particles can be obtained.

7.4.7

Strangeness Quantum Number (S)

The quantum number called strangeness was introduced to explain the strange behaviour of K mesons and hyperons also called strange particles. It is found experimentally that K mesons and hyperons were always produced in pair, a phenomenon called associated production. They are produced by strong interaction in high energy nucleon-nucleon collisions but decay only very reluctantly by means of weak interaction. The strange quantum number (S) is defined as the difference of hyper charge (Y) and baryon number (B). S=Y–BfiY=S+B q = Iz + q = Iz + Y/2 = Iz +

Hence,

(S + B) 2

The strangeness quantum number is zero for the particles which are not strange. For L0 and S0 hyperons: q = 0, Iz = 0, B = 1 Therefore, 0 = 0 +

1 B+S (s + 1) fi s = – 1 (using relation q = Iz + ) 2 2

For S –; q = – 1, Iz = – 1, B = 1 – 1 = – 1 + ½ (s + 1) fi s = – 1 For X , q = – 1, Iz = – ½, B = 1 –

– 1 = – ½ + ½ (s + 1) fi s = – 2 For W , q = – 1, Iz = 0, B = 1 –

– 1 = 0 + ½ (s + 1) fi s = – 3 The multiple number (M), average charge ( q ), hypercharge (Y), strangeness number (S), isospin I, z component (Iz) and total charge (q) for various particles is given in Table 7.3.

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d>ϳ͘ϯപ^ƵŵŵĂƌLJŽĨ;DƵůƟƉůŝĐŝƚLJ͕ǀĞƌĂŐĞŚĂƌŐĞ͕,LJƉĞƌĐŚĂƌŐĞ͕^ƚƌĂŶŐĞŶĞƐƐ͕/ƐŽƐƉŝŶ ĂŶĚdŽƚĂůŚĂƌŐĞŽĨsĂƌŝŽƵƐWĂƌƟĐůĞƐͿ

Photon

Multiplicity Average M charge

Hypercharge Y

Strangeness S

Isospin I

Iz

Total charge

0

0

1 –1

¸1 Ô ˝0 Ô ˛ –1

1/2

¸ 1/2 ˝ ˛–1/2

¸1 ˝ ˛0 –1

(g )

1

0

0

0

0

p ¸ Ô p0˝ –Ô p˛

3

0

0

0

1

+

K+¸ ˝ K0˛ Meson

2

1/2

1

1

K– ¸ ˝ K0 ˛

2

–1/2

–1

–1

–1/2

¸–1/2 ˝ ˛ 1/2

h

1

0

0

0

0

0

0

n

2

1/2

1

0

1/2

¸ 1/2 ˝ ˛–1/2

L0

1

0

0

–1

0

0

0 1

–1

¸1 Ô ˝0 Ô ˛–1

Nucleon P

S¸ +

Ô S˝ Ô S–˛ 0

Hyperon X ¸ ˝ X– ˛

3

0

0

–1

0

W

7.5

0

¸ ˝ ˛

Particle

–

2

–1/2

–1

–2

1/2

1

–1

–2

–3

0

¸ 1/2 ˝ ˛–1/2 0

¸1 ˝ ˛0

0 –1 0 –1 –1

CONSERVATION LAWS OBEYED

The familiar conservation laws for (i) mass-energy, (ii) linear momentum, (iii) angular momentum or spin are found to hold whether the process goes by strong, weak or electromagnetic interaction. In addition, we have the following conservation laws for elementary particles.

7.5.1

Conservation of Charge

In any reaction the total charge quantum number is conserved. For example, antiproton produce p + p Æ p + (p + p ) Total charge number, q = 1 + 1 = 1 + 1 + 1 – 1, i.e., initial and final values of total charge quantum number are equal or total charge quantum number is conserved.

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7.5.2

Conservation of Lepton Number

Lepton number is conserved in all reaction and decay schemes. Le and Lm quantum number will also be conserved separately. (1) Consider the reaction: K0 Æ p + + e– + n e Conservation of lepton number Conservation of electron lepton number

L: 0 + 1 – 1 L e: 0 + 1 – 1

(2) Consider the reaction: m– Æ e – + nm + n e Conservation of lepton number Conservation of electron lepton number Conservation of

L: 1 = 1 + 1 – 1 conserved Le: 0 = 0 + 1 – 1 conserved Lm: 1 = 0 + 1 + 0 conserved

7.5.3

Conservation of Baryon Number

Baryon number (B) is conserved in all the possible decay and reaction For example: n Æ p + e– + n e Baryon number, B: 1 + 0 + 0 conserved K – + p Æ L 0 + p ++ p – Baryon number, B: 0 + 1 = 1 + 0 + 0

7.5.4

conserved

Conservation of Iso-spin Quantum Number

For any strong interaction the total isotropic spin is conserved. Thus, isotropic spin quantum number (I) is conserved in strong but not in weak or electromagnetic interaction. Also z component of Iso spin quantum number (Iz) is conserved in strong and electromagnetic interaction. It should also be noted that Iz is conserved in electromagnetic interaction but I itself is not. For example: For strong interaction p0 + p Æ p – + n

  1 = Isospin I for reactants: 1 + 2   1 I for product: 1 + = 2  1 Iz for reactants: 0 + = 2   1 Iz for product: 1 – = 2

3 1 or 2 2 3 1 or 2 2 1 2 1 2

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Thus, both isotropic spin and its Z-component (I and Iz) are conserved. For reaction: K0 Æ p0 + p0 I:

  1 π 1 + 1 = 2, 1, 0 (not conserved) 2

1 π 0 + 0 (not conserved) 2 As I and Iz are not conserved so must be a weak interaction. Iz : -

7.5.5

Conservation of Hypercharge (Y )

For any strong and electromagnetic interaction, the total hypercharge before and after the reaction remains the same. Example: Consider the reaction by strong interaction p + p Æ L0 + K0 + p + p Hypercharge, Y: 1 + 1 = 0 + 1 + 1 + 0 Hence, hypercharge is conserved. However, hypercharge is not conserved for weak interaction.

7.5.6

Conservation of Strangeness (S)

Total strangeness quantum number is conserved in strong and electromagnetic interactions. But the event involving a weak interaction S can change by not more than DS = ± 1 or 0. For example, for strong interaction p+ + p Æ S+ + K+ Strangeness, S = 0 + 0 = – 1 + 1

(conserved)

For example, K0 Æ p+ + p – Strangeness, S = 1 = 0 + 0

(not conserved)

Hence, the decay is a weak interaction. Note that the isospin quantum number (I) is conserved for strong interaction. Whereas Iz, Y, S are conserved in strong as well as in electromagnetic interactions. In strong decays leptons are not produced and therefore lepton conservation law does not apply.

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7.6

CPT THEOREM

Charge, parity and time reversal symmetry are fundamental symmetries of physical laws under the spontaneous transformation of charge conjugation (C), parity transformation (P) and time reversal (T). CPT is the only combination of C, P, T that is observed to be the exact symmetry of the nature at fundamental level.

7.6.1

Charge Conjugation (C)

If the law of physics is also held for antiparticle as for the corresponding particle, then the law is said to be invariant under charge conjugation. For example, if s is the cross section for the reaction p– + p Æ S– + K+ at a given energy, then the law of invariance of charge conjugation means that corresponding reaction p+ + p Æ S+ + K+ also has the same cross-sectional area. The name charge conjugation arises because a given particle and its antiparticle generally carry opposite electric charges.

7.6.2

Parity Conservation (P )

It states that “If a law of physical phenomena is found to be invariant under space   inversion transformation, i.e., r Æ - r , it obeys the law of conservation of parity. Let us discuss the parity conservation for gravitational interaction. If two bodies   of masses m1 and m2 are at the position vector r1 and r2 (Fig. 7.1), then y

m1  r1

 F12

 F 21

m2 B

 r2

O

x

z

‹‰Ǥ͹Ǥͳ౨Force acting on two particles of mass m1 and m2

 F12 = Force on mass m1 due to m2,

 Gm m   F12 = -  1 23 (r2 - r1 ) r2 - r1

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 F21 = Force on mass m2 due to m1,  Gm m   F21 = –  1 23 (r1 - r2 ) r1 - r2 (Negative sign stands for attractive force) Equation can also be written as  d 2 r1 Gm m   m1 2 = -  1 23 (r2 - r1 ) dt r2 - r1  d2r Gm m   m2 22 = -  1 23 (r1 - r2 ) and dt r1 - r2     Applying the parity transformation, i.e., r1 to -r1 and r2 to - r2 , the above equations becomes    d 2 r1 Gm m - m1 2 = -  1 23 (- r2 + r1 ) dt r2 - r1  d2r Gm m   m1 21 = -  1 23 (r2 - r1 ) or dt r2 - r1 2 d r Gm m   m2 22 = -  1 23 (r1 - r2 ) Also, dt r1 - r2 Hence, equation of motion remains invariant. Thus, we find that gravitational law is invariant under parity. Similarly, it can be proved that parity is conserved for electromagnetic interaction and strong interaction but not in the case of weak interaction.

7.6.3

Time Reversal (T )

Time reversal (T) requires replacing t by – t. A physical law which is invariant under such transformation is said to be invariant under time. For example, a body of mass m is allowed to fall freely under gravitational force (Fig. 7.2) then the equation of motion of the body is =0

a=g

=0

a = –g

V V

‹‰Ǥ͹Ǥʹ౨A body of mass m is allowed to full freely

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m

d2 z = - mg dt 2

d2 z = -g dt 2 When it rises in time reversal, the same equation governs its motion (obtain replacing t by – t). It has been observed that gravitational, electromagnetic interaction and strong interaction are invariant under time reversal but weak interactions are invariant. or

Numerical Problems Problem 1 (a)

In the following decay S0 Æ L0 + g

The half-life of S0 is 10–12 s. Discuss the nature of this decay. (b) Classify the following reactions in terms of interaction and explain: (i) p – + p Æ L0 + K0 (ii) p – + p Æ p0 + n (iii) L0 Æ p + p – Solution: (a) In the decay S0 Æ L0 + g Strangeness, S : – 1 = – 1 + 0 i.e., strangeness quantum number is conserved. As the half-life of S0 is 10–12 s it decays by the faster em process and by slower weak interaction (10–10 s). (b) (i) p – + p Æ L0 + K0 Baryon number, B = 0 + 1 = 1 + 0 (conserved) Charge quantum number, q = – 1 + 1 = 0 + 0 (conserved) Hypercharge, (Y): 0 + 1 = 1 + 0 (conserved) (conserved) Isospin, Iz: – 1 + ½ = 0 – ½ The reaction obeys the selection rules (DB = 0, Dq = 0, DY = 0, DIz = 0). Hence, it is a strong interaction. (ii) p – + p Æ p0 + n Baryon number, B = 0 + 1 = 1 + 0 (conserved) Charge quantum number, q = – 1 + 1 = 0 + 0 (conserved) Hypercharge, (Y): 0 + 1 = 1 + 0 (conserved) (conserved) Isospin Iz: – 1+ ½ = 0 – ½ The reaction obeys the selection rules DB = 0, Dq = 0, DY = 0, DIz = 0

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Hence, it is a strong interaction (iii) L0 Æ p + p – Baryon number, B = 1 = 0 + 1 (conserved) Charge quantum number, q = 0 = – 1 + 1 (conserved) Hypercharge (Y): 0 = 1 + 0 (not conserved) (not conserved) Isospin, Iz: 0 = ½ – 1 As the reactions do not follow the law of conservation of hypercharge and z-component of spin quantum number. So it is a weak interaction. Problem 2 violated?

Which of the following reactions can occur? State the conservation laws

(i) p + p Æ n + p + p+ (ii) e+ + e– Æ m+ + p – (iii) p – + p Æ n + p 0 Solution: (i) p + p Æ n + p + p + Lepton Number, L: 0 + 0 = 0 + 0 + 0 (conserved) Baryon number, B: 1 + 1 = 1 + 1 + 0 (conserved) Charge quantum number, q: 1 + 1 = 0 + 1 + 1 (conserved) Hypercharge, (Y): 1 + 1 = 1 + 1 + 0 (conserved) Strangeness, S: 0 + 0 = 0 + 0 + (conserved) (conserved) Isospin, Iz: ½ + ½ = – ½ + ½ + 1 Since all the quantum numbers are conserved the reaction can occur. (ii) e+ + e– Æ m+ + p – Le : – 1 + 1 = 0 + 0 (conserved) (not conserved) Lm : 0 + 0 = – 1 + 0 B: 0 + 0 = 0 + 0 (conserved) :+1–1=1–1 (conserved) Since in Le and Lm the reaction should independently be conserved. Hence, reaction cannot occur. (iii) p – + p Æ n + p 0 L: 0 + 0 = 0 + 0 B: 0 + 1 = 1 + 0 Y: 0 + 1 = 1 + 0 S: 0 + 0 = 0 + 0 Iz: – 1 + 1/2 = – 1/2 + 0 Q: – 1 + 1 = 0 + 0

(conserved) (conserved) (conserved) (conserved) (conserved) (conserved)

All quantum numbers are conserved; the reaction can occur.

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7.7

QUARK MODEL

We have noticed that hadronic phenomena are very complex. It is believed that this complex behaviour is due to the fact that the hadrons, like atoms or nuclei, have composite structure. They are made of small number of structureless objects called quarks. Quarks model provides not only an elegant and compact way of classifying hadronic states but also explains most of the hadronic physics—spectroscopy, weak and electromagnetic decays, collisions with electrons, neutrinos and hadron production in e + e– annihilation. In the high energy (15 to 200 GeV) scattering of electrons, muons and neutrinos with nucleons, deep inelastic events were selected. In these events a large amount of momentum and energy is transferred to the nucleon. It was observed that the incident particles were scattered through angles much larger than those expected from a continuous distribution of charges within a nucleon. From these observations it was concluded that nucleon is not a structureless particle but it is made of small objects. This conclusion is similar to the one derived from the large-angle scattering of a-particle from atom. Since these small particles exist inside the nucleon, their size must be smaller than the size of these particles as predicted to be 10 –16 cm or 103 times smaller than the dimensions of the nucleons. Experiments also show that particles have spin ½. It is not yet certain that the present form of the quark model represents the ultimate theory of hadronic structure. It is also not yet clear whether quarks exist in isolation or only bound inside a hadron. However, the model is very successful in explaining many processes of hadronic physics.

7.7.1

Quarks

Following are the basic assumptions in connection with quarks and quark model: 1. Quarks must be fermions with spin ½ because only with fermions, baryons and mesons can be constructed. 2. The muon is made of quark and anti-quark. 3. The baryon is made of three quarks. 4. Since hadrons (baryons and mesons) are characterized by additive quantum number; Q (charge), B (baryon number), Iz (z component of iso spin), S (strangeness quantum number), C (charm), B (bottomness quantum number or beauty) and T (topness quantum number). Therefore, at least six quarks are needed to account for the observed hadron. Baryon number assigned to a baryon and a meson are +1 and 0 respectively, so we assign B = 1/3 to all quarks. 5. Two quarks up (u) and down (d) carry only isospin and all other quantum numbers are zero (S = C = B = T = 0). They are also called ordinary quarks and are required to account for the non-strange particle of charge 0 and ± 1. 6. To construct strange hadron (strange, baryon or meson) at least one strange quark is required. Thus, the strange quark (S) which carries strangeness, S = – 1 and C = B = T = 0.

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7. To construct charmed hadrons at least one charm quark C with charm number, C = 1 and S = B = T = 0, is also required. 8. To construct hadrons with beauty or bottomness quantum number B at least one bottom quark (b) with bottomness, B = – 1 and S = C = T = 0, is needed. 9. Sixth quark top (t) which carries truth or topness quantum number T = ± 1 and S = C = B = 0 is required to account for the hadrons having topness quantum number. 2 1 10. The charge quantum number assigned to each quark is qu = + e, qd = – e, qs 3 3 1 2 1 2 =– e, qc = + e, qb = – e and qt = + e. The anti-quarks having charge 3 3 3 3 opposite to that of quarks. The values of various quantum numbers for six different quarks are given in Table 7.4.

d>ϳ͘ϰപ^ƵŵŵĂƌLJŽĨŝīĞƌĞŶƚYƵĂŶƚƵŵEƵŵďĞƌĨŽƌ^ŝdžŝīĞƌĞŶƚYƵĂƌŬƐ Up

Quark Q-charge +

2 e 3

Down

-

1 e 3

Strange (S)

-

1 e 3

Charmed (C) +

2 e 3

Bottom (b)

-

1 e 3

Top (t) +

2 e 3

Iz Isospin

½

–½

0

0

0

0

S-Strangeness

0

0

–1

0

0

0

Charm (C)

0

0

0

+1

0

0

Bottomness(B)

0

0

0

0

1

0

Topness (T)

0

0

0

0

0

1

Since the quantum numbers Iz, S, C, B and T distinguish the various quarks, so it is customary to use the ward flavour for them. The quark have a flavour whereas anti-quark have opposite flavour. Quarks and anti-quarks states like uu , dd , ss , etc., are flovourless. Many unsuccessful attempts have been made to identify free quarks. Nowadays, it is argued that free quarks cannot exist. For this reason the masses of quarks estimated from the analysis of the hadron level scheme. For these studies it is concluded that mt > mb > mc > ms > md > mu >

7.7.2

Composition of Hadrons According to Quark Model

Hadrons include baryons and mesons. Baryons include nucleons and hyperons. According to this model, each hadronic particle is made up of quarks. Quarks are supposed to be the ultimate building blocks of these particles and hence, matter.

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In order to see how different elementary particles are composed of quarks, certain quantum numbers have to be taken note of. Each quark has a baryon number, 1 1 B = - . Hence, a baryon must be made of three quarks so that B = 3 ¥ = 1. Anti3 3 baryon must have B = – 1 made of three anti quarks. A meson is made up of one quark and one anti-quark so that its baryon number, 1 1 B = - = 0 . Quarks are fermions having spin ½ units each. This accounts for the 3 3 half integral spin of baryons and zero spin for mesons (bosons). As per quark model how three quarks (anti-quarks) combine suitably to form some well known hadrons (mesons and baryons). The structure of five hadrons (p+, K+, p, n and W–) with the quantum number is shown in Table 7.5.

d>ϳ͘ϱപYƵĂƌŬŽŵƉŽƐŝƟŽŶŽĨ,ĂĚƌŽŶƐ Hadrons

Quarks

Charge

B(Baryons)

S(Strangeness)

Spin

p

ud

2 1 + =1 3 3

1 1 - =0 3 3

0+0=0

≠Ø = 0

K+

us

2 1 + =1 3 3

1 1 - =0 3 3

0+1=1

≠Ø = 0

p

uud

2 1 1 + - =1 3 3 3

1 1 1 + + =1 3 3 3

0+0+0=0

n

udd

2 1 1 - - =0 3 3 3

1 1 1 + + =1 3 3 3

0+0+0=0

W–

sss

1 1 1 - - - = -1 3 3 3

1 1 1 + + =1 3 3 3

–1–1–1=–3

7.7.3

+

≠≠Ø =

1 2

≠ØØ =

1 2

≠≠≠=

3 2

Coloured Quarks

The quark model has some serious problems. It was the presence of two or three quarks of the same kind in a particular particle. For example, a proton with two u-quarks, a neutron with two de-quarks and W– (omega) has three s-quarks, respectively violating Pauli’s exclusion principle which must be obeyed by quarks as those are fermions with spin ½. To resolve this problem, it was suggested that quarks and anti-quarks have an additional property designated as ‘colour’ which has three possibilities called red, green and blue for quarks and antired, antigreen, antiblue for anti-quark. According to colour hypothesis all the three quarks in a baryon have different colours which satisfies Pauli’s exclusion principle. Since all of them are in different states even if two or three of them are otherwise identical. Such a combination can be considered to be

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white or colourless in the same way in which red, green and blue colours combine to make white light. Again according to colour hypothesis, a meson is supposed to consist of a quark of one colour and an anti-quark of corresponding anti-colour thus cancelling the colour effect as it is supposed that a colour and its anticolour combine to form white. The W– hyperon consists of three quarks SR, SG, SB, i.e., 3s quarks of different colours (red, green and blue). Thus, we find that both baryons and mesons are always colourless. The quark colour is a property which is significant only within the hadrons but is never directly observable.

7.7.4

Gluons and Quantum Chromodynamics

Since we know that hadrons are seen to be composed of quarks. So the strong interaction between hadrons should somehow finally be linked to an interaction between quarks. The strong force between the quarks works on the line of an exchange force. The particles that exchange quark is called gluons. The gluons are massless with spin equal to 1 and travel with speed equal to the speed of light. Each gluon carries colour and an anticolour. The exchange of gluons by a quark change the quarks colour. The gluons must therefore be represented as a combination of a colour and a different anticolour. Gluons are said to be eight in number. Thus, the field that binds the quarks is a colour field. The theory of mechanism of interaction of quarks is known as quantum chromodynamics. It is because this has been modeled on quantum electrodynamics which deals with how charged particles interact with each other. The charge has been substituted by the colour. Quantum chromodynamics is highly complex and obeys quantum principles not classical ones. It explains the behaviour of quarks within hadrons and predicts certain effects.

Cosmic Rays It has been seen that a well insulated gold leaf electroscope slowly loses its charge even when apparently no ionizing agents are present. At the first, it was assumed that the loss of charge of the electroscope was due to ionizing radiations coming from radioactive substances present in the earth. But experiments made by Millikan on the loss of charge of an electroscope in lakes, about few kilometer below the water level, revealed that there was no decrease or increase in the rate of discharge. On the other hand, further experiments indicated that the rate of loss of charge increases as we go higher and higher above the surface of the earth. From these observations, the loss of charge is not due to the ionizing radiation coming from the radioactive.

Numerical Problems Problem 1 Given the charge number, baryon number (B), isospin (I), z component of isospin (Iz), hypercharge (Y) and strangeness number (S) to three basic quarks. Hence, give the quark model of L, S, X and W.

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Solution: Three basic quarks are up (u), down (d) and strange (S). The various quantum numbers assigned to them are given below: Quark

q

B

I

Iz

Y

S

u

2 e 3

1 3

1 2

1 2

1 3

0

d

1 - e 3

1 3

1 2

1 3

0

s

1 - e 3

1 3

0

-

1 2

0

-

–1

2 3

For the corresponding anti-quarks the value of different quantum numbers is same in magnitude but different in signs. The quark model of L, S, X and W the different values of quantum number is given as; Particle

Quark Model

q

B

I

Iz

Y

S

L0

uds

0

1

0

0

0

–1

S

+

uus

1

1

1

1

0

–1

S

0

uds

0

1

1

0

0

–1

S

–

dds

–1

1

–1

1

0

–1

X

0

uss

0

1

½

½

–1

–2

X

–

uss

0

1

½

½

–1

–2

sss

–1

1

0

0

–2

–3

W

–

Problem 2. (a) Which particles correspond to quark composition u s , ddu, sss, ds and uds? (b) Given quark composition of p – and X –. Solution:

2 1 + = 1, spin (≠Ø) = 0 and s = 0 + 1 = 1. Hence, the 3 3 particle is K+.

(a) u s ; q =

1 1 2 + = 0, spin (≠Ø≠) = 1/2 and s = 0 + 0 + 0 = 0. Hence, – 3 3 3 the particle is neutron. ddu; q = -

1 1 1 – = – 1, spin (≠≠≠) = 3/2 and s = – 1 – 1 – 1 = – 3. – 3 3 3 Hence, the particle is W– hadron.

sss; q = -

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1 1 ds ; q = - + = 0, spin (≠Ø) = 0 and s = 0 + 1 = 1. Hence, the particle 3 3 is K0. 2 1 1 uds; q = – – = 0, spin (≠Ø≠) = 1/2 and s = 0 = 0 – 1 = – 1. Hence, 3 3 3 the particle is S0 hadron. 2 1 (b) p – is represented as u d; q = - = – 1, spin (≠Ø) = 0 3 3 1 1 1 X– is represented as dss; q = - - - = – 1, spin = 0 – 1 – 1 = – 2 3 3 3

Summary 1. For a long time electrons, protons, neutrons and photons were regarded as fundamental particles. 2. Subsequently, many more elementary particles and their antiparticles have been discovered using accelerating machines and particle detection devices. 3. According to Dirac, corresponding to every particle, there is an antiparticle in our world. 4. Elementary particles have been classified on the basis of their half-lives as stable and short-lived particles. 5. The stable particles (about 35) have been further classified into four groups (1) photons and gravitons (2) leptons (3) mesons (4) baryons. 6. While photons and mesons are bosons (obeying B-E statistics) whereas lepton and baryons are fermions (obeying F-D statistics). 7. Protons and neutrons are the two baryons which are stable. Rest are unstable and are known as hyperons with mass more than of neuclons. 8. The rest mass of elementary particles is expressed either in MeV/c2 or, in turn, of the electron mass (me). 9. As we proceed from photons, leptons, mesons and baryons the particle’s rest ± mass increases. The only exception is the tau meson (t ) with rest mass 1784 MeV/c2 which may be the largest one. 10. Many of unstable hadrons (mesons and baryons are collectively called hadrons) with lifespan @ 10–23 s are nothing but the resonate states of the more easily observable particles. 11. While stable particles are stable against the decay, the unstable ones undergo decay. The possible modes of decay depends upon the various conditions. 12. The probability for various modes of decay for even the same particle (antiparticle) may be different. A large amount of energy involved in such decay reactions.

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13. Certain quantum numbers have been introduced or defined for simplicity of understanding such large number of elementary particles. They have been determined by experiments. 14. All the quantum numbers such as B, L, S, Y, I (IZ) may not necessarily represent precisely measurable physical notions. 15. Just like chemical reactions, nuclear reactions obey certain conservation laws. These are related to the conservation of (i) mass and energy, (ii) linear momentum, (iii) angular momentum, and (iv) charge. 16. Nuclear reactions obey certain additional laws. These are related to the conservation of (i) baryon number (B) (ii) lepton number (L) (iii) strangeness number (S) and (iv) CPT; CPT stands for the charge conjugation, parity and time reversal. 17. If a law governing a physical phenomenon is invariant under space inversion   (r Æ - r ) , it surely obeys the law of conservation of parity. 18. While gravitational and electromagnetic forces are invariant under parity, a weak interaction is not so. 19. Conservation laws have some kind of symmetry involved. 20. Parity of a system of particles may be odd or even. Parity is a scalar quantity. 21. The four fundamental interactions are well known. Since their range is markedly different, a comparison of their strength is not of physical importance. 22. The strong interaction among the nucleons is due to fast exchange of pions. 23. The electromagnetic interaction is due to exchange of photons. 24. Weak interaction operates through the exchange of photons. 25. The gravitational interaction is supposed to involve exchange of massless, and chargeless particles (known as gravitons). 26. To simplify the understanding and grouping of about 200 particles and antiparticles known so far, Murray Gell-Mann proposed the quark model. 27. As many as six quarks have been proposed. These quarks are: up, down, strange, charm, beauty (or bottom) and truth (or top) quarks. 2 28. The quarks u, c, t carry a positive charge of e each, the quarks d, s, b carry 3 1 negative charge of e each. 3 29. The antiquarks of these quarks are u*, d*, s*, c*, b*, t*. 30. Baryons are made of three quarks whereas mesons are made of a quark and anti-quark. 31. The six quarks are considered to be the fundamental or elementary particles. The quarks have no internal structure. 32. Each quark has a baryon number (B) equal to 1/3 and anti-quark has baryon number – 1/3. 1 33. Quarks are fermions having spin units each. 2

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34. Two or more quarks (anti-quarks) of the same kind (fermions) could violate Pauli’s exclusion principle. When present together in a particle (e.g., p = uud and n = udd). But it has been avoided through another property given to the quarks and anti-quarks. This property is the colour (blue, green, red). This colour is different from physical colours but it is the quantum number associated with each quark or anti-quark. 35. The concept of quark colour not only provides a way out as regards the likely violation of Pauli’s principle but it provides a possible link with the strong interaction. 36. The theory of mechanism of interaction of quarks is known as quantum chromodynamics. While the quantum electrodynamics deals with how charged particles interact with each other, quantum chromodynamics deals with colour of the interacting particles.

Short Answer Questions 1. What are elementary particles? 2. What are antiparticles? Do they exist? Or most elementary particles have antiparticles. Comment on the statement. 3. Name the four main groups into which elementary particles have been divided? 4. What are fermions? Name their subgroups? 5. Give a comparison between leptons and baryons? 6. Name the various conservation laws obeyed by the elementary particles/decay modes. 7. What are quarks? 8. Explain charge conjugation? 9. Which quarks make a proton and a neutron. 10. What do you mean by isospin?

Long Answer Questions 1. Give a classification of elementary particles? Give their properties in brief? 2. What are leptons? Discuss their properties? 3. What are the various quantum numbers associated with the elementary particles. Discuss their corresponding conservation laws? 4. Write notes on strangeness and baryon number? 5. Describe various conservation laws for elementary particles. 6. What is invariance under CPT? Explain. 7. Write notes on: Parity, baryon number, lepton number? 8. What is quark model? Explain it.

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9. Explain the four fundamental interactions among the elementary particles? 10. Explain strong and weak interactions as applied to elementary particles? 11. Explain why relations are forbidden (2) m+ Æ e+ + e+ + e– (1) n + n Æ L + L (3) S+ Æ p + g (4) X– Æ h0 + p – (5) p Æ e+ + g (6) p++ n Æ p – + p 0 –Æ n +p (8) p0 Æ 3g (7) p + p 12. Explain why the following relations are allowed: (2) X– Æ L0 + p – (1) S+ Æ L0 + g – 0 0 (3) p + p Æ S + K (4) p + p Æ S+ + K0 + p+ + n (6) p0 + n Æ p – + p (5) n + n Æ p0 + p+ + p – (8) p + p Æ p + n + p+ (7) p+ Æ p0 + e+ + ne 0 (9) S – Æ L + e – + n e (10) K– + p Æ X– + K0 13. Show that decay of K + Æ p0 + e+ + ne is a weak decay? 14. Show that the decay L0 Æ p + p – is a weak decay? 15. Find the energy produced in the anhilation of proton-antiproton pair (p – p )? [Ans. 1876 MeV]

Multiple-Choice Questions 1. What is the atomic number of transuranic elements (a) greater than 92 (b) greater than 89 (c) less than 92 (d) greater than 112 2. Matter is built of small particles called (a) radiants (b) atoms (c) isotopes (d) ions 3. Temperature at centre of the sun is nearly (a) 20 million kelvin (b) 10 million kelvin (c) 30 million kelvin (d) 25 million kelvin 4. Ashes from a fire in forest show carbon-14 activity of only 1/8 activity of fresh wood. How long ago forest had fire? (a) 18590 years (b) 17190 years (c) 16580 years (d) 14820 years 5. Release of energy from the sun is due to (a) Nuclear fission (b) Nuclear fusion (c) Burning of gases (d) Chemical reaction 6. When alpha and beta particles slow down by collisions, they become (a) Harmful (b) Harmless (c) Useful (d) Expensive



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7. Cosmic radiations interact with atom’s atmosphere to create (a) Primary radiation (b) Secondary radiation (c) Tertiary radiation (d) No radiation at all 8. Atoms of an element which have same number of protons but different number of neutrons in their nuclei are known as (a) Ions (b) Isotopes (c) Radiants (d) Electrons 9. To create a shower of secondary radiation, cosmic radiation interacts with (a) Atoms (b) Electrons (c) Protons (d) Neutrons 10. Cosmic radiation consists of (a) Protons and electrons (b) Alpha particles (c) Larger nuclei (d) All of the above 11. Half-life of 7N15 N is 6.5 s. A sample of this nuclide of hydrogen is observed for 32.5 s. fraction of original radioactive isotope remaining after this time is (a) 1⁄32 (b) 1⁄16 (c) 1⁄8 (d) 1⁄4 12. Radiation which does not change its direction is (a) b radiation (b) a radiation (c) g radiation (d) No radiation 13. For curing cancerous tumors and cells, doctors use (a) Radioactive cobalt-60 (b) Radioactive cobalt-80 (c) Radioactive cobalt-90 (d) Radioactive cobalt-50 14. Time during which half of the unstable radioactive nuclei disintegrate of sample of radioactive element is called the (a) Full life (b) Half-life (c) Double life (d) Quarter life 15. C-14 : C-12 ratio in a fossil bone is found to be 1/4th that of ratio in bone of an animal. half-life of C-14 is 5730 years, approximately age of fossil is (a) 12460 years (b) 11460 years (c) 13590 years (d) 14580 years 16. Radiation emitted from uranium salt has ability to (a) Destabilize the electrons (b) Break the protons (c) Make the protons (d) Ionize the gas 17. Temperature at the centre of the sun is (a) 50 million kelvins (b) 20 million kelvins (c) 30 million kelvins (d) 40 million kelvins

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18. Word atom is derived from Greek word (a) Antimum (b) Atomis (c) Otomos (d) Optium 19. A conservation law that is not universal but applies only to certain kinds of interactions is conservation of (a) Lepton number (b) Baryon number (c) Spin (d) Strangeness 20. Conservation laws that describe events involving the elementary particles include the conservation of (a) Energy (b) Linear and angular momentum (c) Electric charge (d) All of these are correct 21. A particle with no charge and no rest mass is (a) Proton (b) Neutrino (c) Photon (d) Positron 22. The decay modes are applicable only to a (a) Stable elementary particle (b) Unstable elementary particle (c) Electromagnetic radiation (d) All types of particles 23. The antiparticle and the corresponding particle have between them (a) All characters common (b) Not all but one character common (c) Nothing at all common 24. The half-life of stable elementary particle is (b) ≥ 10–16 s (a) @ 10–10 s (d) @ 10–12 s (c) @ 10–23 s 25. The elementary particles with masses greater than the nucleonic masses are known as (a) Mesons (b) b-particles (c) Baryons (d) Leptons 26. The earliest known lepton is (a) Proton (b) Neutron (c) p-meson (d) Electron 27. All mesons are (a) Fermions (b) Bosons (c) Gravitons (d) Gluons



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28. Elementary particles with heaviest mass are termed (a) Baryon (b) Lepton (c) Mesons (d) Hyperons 29. Parity is not conserved in (a) Gravitational interaction (b) Weak interaction (c) Electromagnetic interaction (d) Strong interaction 30. Hyperons are essentially (a) Photons (b) Leptons (c) Mesons (d) Baryons 31. Hyperons are particles which are (a) Heavier than a nucleon (b) Lighter than a nucleon (c) Heavier than electron but lighter than proton (d) Electromagnetic radiation 32. Quarks are (a) Bosons (b) Fermions (c) Zero mass objects (d) Still not known 33. A baryon is a bound state of (a) 2 quarks (b) 3 quarks (c) 4 quarks (d) None of the above 34. A meson is a bound state of (a) 2 quarks (b) 3 quarks (c) 4 quarks (d) None of the above 35. The process p0 Æ g + g is an example of (a) Weak interaction (b) Electromagnetic interaction (c) Gravitational interaction (d) Strong interaction 36. The CPT conservation is (a) Not essential for elementary particle decay (b) Irrelevant in the study of elementary particles (c) Gravitational interaction (d) Strong interaction 37. A quark model postulates as many as (a) Three quarks (b) Four quarks (c) Five quarks (d) Six quarks 38. In the light of the quarks model our notion of charge quantization will (a) Remain unchanged (b) Be slightly modified (c) Totally collapse (d) All of the above

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39. The grand unification theory has been triggered by the idea of the fusing together (a) Weak and electromagnetic interaction (b) Electromagnetic and strong interaction (c) Gravitational and electromagnetic interaction (d) All the four fundamental interactions 40. The decay L0 Æ n + p0 is an example of (a) Electromagnetic interaction (b) Weak interaction (c) Strong interaction (d) Gravitational interaction 41. p Æ e– + p+ + p+, the decay is (a) Allowed (b) Forbidden (c) Charge non conserving process (d) None of them 42. What particle is made up of two down (d) quarks and one up quark (u) (a) Atom (b) Electron (c) Neutron (d) Proton 43. What theoretical particle travels faster than the speed of light (a) Quark (b) Boson (c) Fermions (d) Tachyon 44. Light is made of what type of particle (a) Quarks (b) Leptons (c) Photons (d) Gluons 45. What type of elementary particles are electrons? (a) Leptons (b) Quarks (c) Photons (d) Mesons (e) Baryons 46. Which particle is responsible for strong interaction between quarks? (a) Leptons (b) Photons (c) Gravitons (d) Gluons 47. The charge on the particle dds is (a) e

(b)

1 e 3

2 (c) - e (d) – e 3 48. The conservation law violated by the reaction p Æ e+ + p0 is the conservation of (a) Charge (b) Energy (c) Angular momentum (d) Lepton and baryon numbers

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49. Which of the following choices lists the four forces in nature in order of decreasing strength (a) Strong nuclear, electromagnetic, weak interaction, gravitational (b) Electromagnetic, strong nuclear, weak interaction, gravitational (c) Strong nuclear, electromagnetic, gravitational, weak interaction (d) Strong nuclear, weak interaction, electromagnetic, gravitational 50. Particles which participate in the strong interaction are called (a) Neutrinos (b) Hadrons (c) Leptons (d) Protons

Answers to Multiple-Choice Questions 1. 8. 15. 22. 29. 36. 43. 50.

(a) (b) (b) (b) (b) (c) (d) (b)

2. 9. 16. 23. 30. 37. 44.

(b) (a) (d) (b) (d) (d) (c)

3. 10. 17. 24. 31. 38. 45.

(a) (d) (b) (c) (a) (b) (a)

4. 11. 18. 25. 32. 39. 46.

(b) (a) (c) (c) (b) (d) (d)

5. 12. 19. 26. 33. 40. 47.

(b) (c) (d) (d) (b) (b) (d)

6. 13. 20. 27. 34. 41. 48.

(b) (a) (d) (b) (a) (b) (d)

7. 14. 21. 28. 35. 42. 49.

(b) (b) (c) (a) (b) (c) (a)

8 Cosmic Rays

8.1

INTRODUCTION

It is seen that a well insulated gold leaf electroscope slowly loses its charge even when apparently no ionizing agents are present. At the first, it was assumed that the loss of charge of the electroscope was due to ionizing radiations coming from radioactive substances present in the earth. But experiments made by Millikan on the loss of charge of an electroscope in lakes, about a few kilometres below the water level, revealed that there was no decrease or increase in the rate of discharge. On the other hand, further experiments indicated that the rate of loss of charge increases as we go higher and higher above the surface of the earth. From these observations, the loss of charge is not due to the ionizing radiation coming from the radioactive substances. Finally, it was concluded that a penetrating radiation was incidenting on the earth from all directions and are coming from the interstellar space.

8.2

COSMIC RAYS

Cosmic rays are extremely penetrating radiations coming from very far away from the earth and continuously bombarding the earth on all sides with more or less uniform intensity. Although the investigations have been directed towards understanding the nature and origin of these rays, yet it is clear that their studies can add a considerable information to the understanding of the nuclear structure. Two new particles, the positron and the negatron, which are as fundamental to the nucleus as electrons, protons and neutrons, have been discovered. Also the cosmic rays, particles have been found with large amount of energy of the order of 1012 eV and more which are capable of exploring the interior of nuclei with greater effectiveness.

8.3

TYPES OF COSMIC RAYS

Cosmic rays consist of high energy atomic nuclei, mainly protons reaching our earth as well as secondary radiations produced by these nuclei in the earth’s atmosphere. Most of these particles have energy of the order of 15 GeV but occasionally some particles are found to posses energies as high as 1011 GeV. A distinction between

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the cosmic radiations coming from the outer space and entering the top of earth’s atmosphere (primary) and the radiations found in the space between the top of the atmosphere of the earth and its surface or below (secondary) should be made.

8.3.1

Primary Cosmic Rays

The primary cosmic rays are those which initially incident upon the outer boundaries of the earth’s atmosphere. The primary cosmic rays approaching the earth’s atmosphere contains 92% proton, 7% a -particles and 1% heavier nuclei like carbon, nitrogen, oxygen, iron, etc. The relative abundance of the various nuclides present in the primary cosmic rays is almost same as the relative abundance of elements in the earth’s crust as well as in the stars. From the experiments it has been found that particles moving with relativistic velocities have their ionization proportional to Z 2.

8.3.2

Secondary Cosmic Rays

Upon entering the atmosphere, high energy particles soon collide with another nucleus, splitting one or both particles into a number of smaller nuclear fragments, each one of which carries away some of the primary energy. These high speed particles, in turn, collide with other nuclei further dividing their energy to produce other high energy particles. All these particles, with the exception of the particles of primary rays are called secondary cosmic rays. The process repeats with the result that energy content of the secondary cosmic rays gets greatly reduced. It has now been established that at sea level, these rays consist of 70% mesons, 29% b-particles, about 1% heavy nuclear fragments including a -particles, protons, neutrons, etc., and even photons. It must be stated that the composition of secondary cosmic rays is indeed very complex. The relative percentage of the various particles, as listed above does vary from place to place, due to a variety of complicated processes of which they are produced. The secondary cosmic rays divide themselves into two components. The ‘hard component’ of secondary cosmic rays consists of mesons and very small number of protons and may be very small number of highly energetic electrons or photons. The ‘soft component’ of secondary cosmic rays consists of electrons and photons and possibly some very slow moving mesons and/or protons. The hard component is highly penetrating (more energetic) while the soft component is easily absorbed in the material like lead (Pb).

8.4

NATURE OF COSMIC RAYS

Investigations of the nature of cosmic rays were initiated in 1927 with the observation of the latitude effects. Measuring the intensity of cosmic rays at different latitudes, a definite variation in the intensity, being distinctly less in the equatorial regions than in higher latitudes. This phenomenon was explained by assuming the presence of charged particles in the cosmic rays.

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If these rays consist of uncharged particles only like photons, they would not be affected by the earth’s magnetic field and should therefore come freely to the equator as to polar regions of the earth. If the cosmic rays were made up of charged particles, they would be influenced by the earth’s magnetic field. The particles which approach the earth at the equator, with the direction of motion right angled to the direction of earth’s magnetic field (BH which is along south to north) experience the maximum magnetic force (Fm = qvB sin 90° = qvB = max) and are deflected from the earth (Fig. 8.1). But the particles approaching near the poles and moving along the earth’s magnetic lines of force would have no difficulty in reaching the surface of the earth (Fm = qvB sin 0° = 0). Hence, a decrease in the intensity from pole to equator would be expected, if the cosmic rays consist of high energy charged particles.

N W

Cosmic ray E

Cosmic ray

S

‹‰ǤͺǤͳ౨‡ϐŽ‡…–‹‘‘ˆ…‘•‹…”ƒ›„›ƒ‰‡–‹…ϐ‹‡Ž†‘ˆ–Š‡‡ƒ”–Š

Clay’s observation received further confirmation by the experiments of Bothe and Kolhorster with coincidence counters and an interposed absorber which gave a strong indication of the presence of charged particles among cosmic rays observed within the earth’s atmosphere. These evidences for a charged particle composition of cosmic rays started a series of researches on geographical distribution of cosmic rays such as latitude, longitude azimuth and north-west effect are discussed below.

8.4.1

Latitude Effect

The fact that primary cosmic rays consist of charged particles and not photons is established by the latitude effect. This effect was discovered by Compton in 1933 and it shows that as we move from the magnetic north to the magnetic south, the intensity of cosmic rays remains steady until a magnetic latitude of 42° is reached. The intensity then begins to drop, reaches a minimum value about 11% less than the value beyond 42° latitude, at the magnetic equator and then rises to symmetrical values in the southern hemisphere as shown in Fig. 8.2.

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100 98 Relate intensity

96 94 92 90 50 40 30 20 10 0 10 20 30 40 50 Geometrical latitude

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8.4.1.1

Cosmic Intensity at the Equator

The decrease in the intensity of cosmic rays at the earth’s magnetic equator is now explained as being due to earth’s magnetic field. The earth’s magnetic field extends far out of earth’s atmosphere and can effect the path of incoming charged particles even at large distances from the earth. The paths of all charged particles crossing the earth’smagnetic field are bent by a force that is perpendicular to the direction of   the field ( Fm = q v ¥ B ). The amount of deflection depends upon the momentum and charge on the particles. All the particles with energy less than 10 4 MeV are so strongly deflected by the earth’s field at the equator (q = 90°) that they are unable to enter the earth’s atmosphere. In fact, at low energies, the particles which do not travel parallel to the magnetic lines of force, spiral around these lines. As the particles approach the earth, the field strength increases, the radii of their spiral path become smaller and mv ˆ Ê till ultimately they are so sharply bent that they get reflected and smaller Á r = qB ˜¯ Ë spiral around the lines of force again approaching the earth on the other side. Such charge particles travelling to and fro constitute the van Allen belts, which surround the earth except the magnetic poles. The outer belt is constituted by the low energy particles like protons and electrons from the sun and the inner belt is due to more energetic one from outer space. The particles approaching the earth in the direction of its magnetic axis are not deflected by the magnetic field (Fm = 0 as q = 0). Hence, primary cosmic particles of any energy can enter the earth’s atmosphere near the magnetic poles. But particles with only the highest of energies can get down to the earth’s atmosphere near the magnetic equator. This explains the latitude effect.

8.4.2

Altitude Effect

The study of the variation of cosmic ray intensity with altitude and depth was undertaken to know more of the nature and origin of the rays.

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Ionization produced by cosmic rays at higher and higher altitudes reaching practically up to the top of the earth’s atmosphere has been studied. It is seen that the intensity increases slowly up to the height of about 8 km and thereafter rapidly up to 20 km. At a height beyond 24 km, the intensity starts decreasing slightly. All the four experimental curves corresponding to magnetic latitude 30° N, 38° N, 51° N and 60° N, however, show that nearing the magnetic equator the cosmic ray intensity decreases at high altitudes as well as at sea level. The observed variation of cosmic ray intensity with altitude is explained as under. At an altitude of about 25 km, the secondary cosmic ray particles outnumber the primary particles by almost fifteen times. But as these particles travel downwards, they progressively lose energy by breaking the nuclei and ionizing the air until they are in thermal equilibrium with the air molecules and ceases to exist as charged particles. However, high speed secondary and a few primaries do reach the sea level and some of them still have sufficient energy to penetrate several thousand metres of the earth and water.

8.4.3

Longitude Effect

The variation in the intensity of the cosmic rays with longitude (d) a fixed geometrical latitude and altitude was discovered by Clay and Millikan. While crossing the equator at different longitudes they found that the apparent equatorial intensities vary appreciably indicating that at the equatorial belt, the drop in intensity is different at different longitudes, being greater in the eastern hemisphere than in the western. The effective amount roughly is 5% at the equator and diminishes to zero on higher latitudes. The variation is due to the fact that earth’s magnetic field is not symmetrical with respect to the axis passing through the centre of the earth.

8.4.4

East-West Effect

It has been observed by Rossi and Johnson that the number of particles arriving per second from west is greater than from the east. This effect is most pronounced at the equator where the excess is about 14.2%. There is an asymmetry involved and the effect is rightly named as east-west asymmetry or azimuth effect. It has direct bearing on the nature of charge on the cosmic ray particles. If the particles of cosmic rays are positively charged and coming down vertically towards the magnetic equator, these are deflected by the earth’s magnetic field. The magnetic lines of forces of the earth’s magnetic field have the direction from geographical south to north outside the earth. According to Flemming’s left hand rule it is seen that the particles are deflected towards east and thus appear to come from the west of the vertical as shown in Fig. 8.3.

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 +

+

+

N W

E S

‹‰ǤͺǤ͵౨ƒ•–Ǧ™‡•–‡ˆˆ‡…–

Similarly, if the particles are negatively charged they will be deflected towards the west and will appear coming from the east. Thus, if the particles are positively charged more will arrive from the east than from the west. Actual measurements show that at the equator the intensity of cosmic rays coming from west is about 14.2% more than that from the east direction. Hence, the primary cosmic rays are composed predominantly of positively charged particles and their deflection by the earth’s magnetic field which accounts for the east-west asymmetry in cosmic ray intensity.

8.5

COSMIC RAY SHOWER

The primary cosmic ray particles mostly protons, some a-particles and a few heavier nuclei interact with the nuclei of the earth’s atmospheric gases giving rise to secondary cosmic rays. For example, when cosmic ray proton strikes against the nuclei of nitrogen or oxygen, then it will disintegrate into a fast moving proton, neutron and a charged as well as neutral p-meson or pions as shown in Fig 8.4. Other particles such as antiprotons, antineutrons and heavy particles are known as hyperons are also produced. p

Nuclear p+ p

n

ve

p–

‹‰ǤͺǤͶ౨‘•‹…”ƒ›•Š‘™‡”

All these particles leave the site of collision with higher energies. If these particles are observed by their tracks in photographic emulsion, they present the appearance

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of a star. The fast protons and neutrons interact with atmospheric nuclei and produce a nucleonic shower of low energy neutrons and protons that reach the earth’s surface or sea level. The charge p+ and p – mesons being short-lived (half-life 2 ¥ 10–8 s) decay to give high energy μ-mesons or muons of the same sign and a massless, chargeless particle called neutrino. The μ-meson interact weakly with the nuclei. Because of this reason they have large penetration power in the matter and can reach the sea level after travelling a large distance. Thus, at the sea level, the hard component secondary cosmic rays consist mainly of strongly penetrating muons and some low energy nucleons (p and n). The charged muons μ+ and μ–, as such exhibit a small tendency to interact with nuclei and most of them decay as under the following scheme. μ+ Æ e+ + ne + n m μ– Æ e– + n e + nμ When ne and n e are electron neutrino and anti-neutrino and nμ and n m are muon neutrino and anti-neutrino, respectively. Thus, the decay of p+ and p – into the corresponding muon and neutrino and further decay of μ+ and μ– into the electron and positron and a neutrino and anti-neutrino is shown in Fig. 8.5. p0

p+ m+

g

e+ ue

p-

g

g

u

m– e+

e–

nm

e–

e+

(a)

e–

(b)

n e um

(c)

‹‰ǤͺǤͷ౨‘•‹…”ƒ›•Š‘™‡”•

The neutral p0 decay to two photons. If the energy of the photons is more than 100 MeV, then high energy g -rays photon interact with matter and give rise to an electron (e–) and positron (e+) pair. Thus, for a single proton we get a whole series of small particles. This is defined as cosmic showers. Thus, at low altitudes the soft component of secondary cosmic rays mostly consist of e–, e+ and g -rays. The hard component accounts for nearly 75% and soft component about 25% of the cosmic rays at the sea level.

8.6

ORIGIN OF COSMIC RAYS

As we know, our earth is continuously bombarded from all the sides with almost uniform intensity by the cosmic rays. It is now well established that cosmic rays originate outside the earth’s atmosphere. Initially, it was supposed that source of

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cosmic rays is the sun, but intensity of cosmic rays remains almost constant at all hours of day or night. If, therefore, the sun would have been the origin of cosmic rays, then the intensity must vary with the time of the day and also must have a higher value in the direction of the sun. This is, however not observed and hence we assume that if at all only a small fraction of cosmic rays is coming from the sun. The majority of primary cosmic rays consists of high energy protons. The origin of these rays might be: 1. Other gases 2. Our own galaxy. The presence of high energy protons in the cosmic rays leads Fermi to suggest that cosmic rays originated in our own galaxy. Some stars like Super Novas undergo explosions once in 250–300 years and throw out stellar matter. According to Fermi this diffuse material is our galaxy—the Milky wayphoto ionized by the stars nearby is a source of cosmic rays. 3. Interstellar Space. Observations on the polarization of light from the stars indicate that the interstellar space is not a perfect vacuum but contains on an average one proton per centimetre cube (cc). The interstellar space is not evenly populated and at some places clouds of protons containing about 100 protons per cc are present. These clouds move through the space with average velocity 30 km s–1. Thus, moving clouds give rise to magnetic field in the interstellar space. The protons of the primary cosmic rays suffer repeated acceleration in the vary magnetic field and hence gain large energies. 4. Big-Bang Explosion Theory. Initially, the whole mass of the universe was supposed to be concentrated in a single nucleus, which exploded forming galaxies. During this explosion, a huge amount of radiations created protons and other nuclei with a large amount of energy in all directions. The cosmic rays are the residual particles and radiations left out after the big-bang explosion. Thus, we concluded that cosmic rays owe their origin to a little extent to the sun and to a large extent to galaxies, stars of our own galaxy, interstellar space and big bang explosion.

Numerical Problems Problem 1 Find the energy of two photons that are produced when annihilation occurs between a electron and a positron that are initially at rest? Solution: The annihilation process of electron and positron is given as e– + e+ Æ 2g As electron and positron are initially at rest, their KE is equal to zero. The whole rest mass energy of electron and positron is carried by the photons.

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Rest mass energy of electron and positron, E = 2m0c2, where

m0 = 9.1 ¥ 10–31 kg E = 2 ¥ 9.1 ¥ 10–31 ¥ (3 ¥ 108)2 = 1.638 ¥ 10–13 J =

1.638 ¥ 10 -13 = 1.02 MeV 1.6 ¥ 10 -13

Hence, the energy of each photon = 1.02/2 = 0.51 MeV. Problem 2 An electron-positron pair is produced when g -ray photon of energy 2.36 MeV passes close to a heavy nucleus. Find the KE of each particle produced as well as the total energy with each. Solution: Reaction is given as g Æ e– + e+ E = m0c2 + K.E. (e–) + m0c2 + K.E. (e+)

and

But 2 m0c2 = Rest mass energy of e– and e+ = 1.02 MeV Therefore, 2.36 = 1.02 + K.E. (e–) + K.E. (e+) = 1/2(2.36 – 1.02) = 0.67 MeV Total energy carried by each particle (e– or e+) = m0c2 + K.E. = 0.51 MeV + 0.67 = 1.18 MeV Problem 3 A gamma ray photon of energy 1896 MeV annihilates to produce proton and anti-proton. If the rest mass of each particle is 1.007276 amu. Find the KE carried by these particle. Solution: The reaction is given as g Æ p+ p and E = m0c2 (rest mass energy of p) + KE (proton) + m0c2 (rest mass energy of antiproton) + KE (p–) = 2 m0c2 + KE (p) + K.E. ( p ) But

2 m0c2 = 2 ¥ energy equivalent to 1.007276 amu = 2 ¥ 1.007276 ¥ 931 = 1876 MeV (1 amu = 931 MeV)

Therefore, KE of each particle = ½ (1896 – 1876) = 10 MeV.

Summary 1. Cosmic rays are highly penetrating radiations, coming from regions very far away from the earth. 2. Cosmic rays are composed of primary and secondary cosmic rays. 3. Primary cosmic rays consist of 90% protons, 9% a-particles and remaining 1% heavier elements like carbon, nitrogen, oxygen, etc.



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4. On entering the earth’s atmosphere, primary cosmic ray particles undergo inelastic collision with the atmospheric gases. The secondary cosmic rays so produced consist of mesons, protons, neutrons, etc. 5. Elastic collision of secondary particles with themselves produces a nucleonic cascade or shower. 6. The composition of the secondary rays, as they reach the ground, has been established. They contain 70% μ-mesons and the remaining are protons, neutron, etc. 7. The exact origin of cosmic rays (primary) is not definite and precisely known. 8. Cosmic rays have extremely large energies associated with their particles about 103 MeV to 1014 MeV. 9. Studies of cosmic rays led the discovery of some elementary particles. The positrons were discovered in 1938 by Anderson and Neddermeyer while pions were discovered in 1947 by Powell. 10. According to Dutch physicist Clay, the intensity of cosmic rays is different at different places (latitudes). While the intensity was lower in the equatorial regions (for small value of l), it was higher in the polar regions (l @ 90°). Hence, it drops by about 10% from latitude angle 50° to the equator. This is called “latitude effect”. 11. Variation of cosmic ray intensity with altitude (height) is called altitude effect. Intensity peak occur at about 20 km above the surface of the earth. 12. Altitude effect could be distinguished in two components of cosmic rays 1. Soft component: very rapidly absorbed and most effective in the upper part of the atmosphere 2. Hard component: more penetrating and effective even in the lower part of the atmosphere. 13. Another observation, having a direct bearing on the nature of cosmic rays, is that the number of such particles arriving per second from west is greater than from the east. This shows that majority of cosmic ray particles is positively charged. This is known as “east-west asymmetry”. 14. The primary cosmic rays are believed to be originated from the explosions of supernova the exploding stars which are million times brighter then our sun. The pulsars (neutron star of extremely high density) have also been seen as the origin of primary cosmic rays. 15. Very high levels of cosmic ray intensity can possibly be maintained if either a supernova explodes every 50 years, or pulsars are born once every 50 years. 16. Wilson, Elstar and Geital, Rutherford, Cook and Hess were associated with the studies of cosmic rays. Hess was awarded Nobel Prize in 1935 for his detailed study.

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Short Answer Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

What are cosmic rays? How were cosmic rays discovered? What is the composition of primary cosmic rays? What are soft and hard components of secondary cosmic rays? Discuss latitude effect on the intensity of cosmic rays? What is east-west asymmetry (effect)? What is the cause of east-west asymmetry? What are cosmic ray showers? What is cosmic ray shower and what are its constituents? Give the origin of cosmic rays? What happens when a particle combines with an antiparticle?

Long Answer Questions 1. Describe the discovery of cosmic rays? 2. What are cosmic rays? What is the origin of cosmic rays? 3. Explain the latitude effect on cosmic rays? Is the intensity of cosmic rays same at the equator. 4. What is the effect of altitude on the intensity of cosmic rays? 5. What is east-west asymmetry? 6. What do you understand by cosmic shower? How are they produced? 7. Describe the origin of cosmic rays? 8. What are soft and hard components of cosmic rays? 9. What are cosmic rays? Distinguish between primary and secondary cosmic ray? Discuss their origin?

Multiple-Choice Questions 1. Cosmic shower consists of (a) Baryons and mesons (b) Electrons and protons (c) Electrons, positrons and photons (d) Muons and baryons 2. At a given latitude, the number of particles arriving per second is (a) Move from west then east (b) Move from east then west (c) Just the same from all sides

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3. The mechanism of cosmic ray shower was explained by (a) S.M. Bose (b) H.J. Bhaba (c) H.G. Khurana (d) M.N. Saha 4. The mechanism of a cascade shower involved the phenomenon of (a) Compton effect (b) Pair production (c) Annihilation process (d) Yet unknown process 5. Primary cosmic rays mostly consist of (a) Alpha particles (b) Protons (c) Electrons (d) Neutrons 6. Cosmic rays consist of (a) High penetrating radiations (b) Only charged particles (c) Electromagnetic radiations (d) Less penetrating radiations 7. Cosmic rays were discovered by (a) Thomas and others (b) Hess and others (c) Bohr’s and others (d) Anderson and others 8. Cosmic rays are (a) X-rays originating from the sun (b) g -rays originating from stars (c) Electromagnetic waves coming from cosmos (d) Charged/uncharged high energy particle 9. Cosmic ray penetration power is (a) Low (b) Very low (c) High (d) Very high 10. Hard secondary cosmic rays consist of (a) Lighter particles like electrons, positrons and photons (b) Heavier (charged) fast particles like 1H2, 2He4 (c) Purely high energy electromagnetic radiations (d) Electron proton pairs

Answers to Multiple-Choice Questions 1. (c) 8. (d)

2. (a) 9. (d)

3. (b) 10. (b)

4. (b)

5. (b)

6. (a)

7. (b)

9 Experiments of Nuclear Lab

EXPERIMENT 1 Aim. To determine the optimal voltage and plateau of a G-M counter. Apparatus used. Geiger-Muller counting system with A.C. mains, source holder bench, connecting cables, sources (Co-60, Cs-137, Sr-90). Theory. Geiger and Muller developed the counter which was named after them as G-M counter. It has been one of the most widely used nuclear detectors in early days of nuclear physics. It is a gas filled counter. It is based upon the principle of ionization of a gas whenever a stream of charged particles is passed through it. Actually, at a certain higher potential difference applied between the electrodes, the secondary ionization (or indirect ionization) is caused by the electrons. A G-M counter is a very useful, extremely sensitive and popular detection device for charged particle. It operates at a potential of the order of 300–900 V. Furthermore, while a proportional counter can, a G-M counter cannot distinguish the ion sources (from one another) from the pulse heights produced. The advantage of G-M counter is that it produces pulses of much larger size and the pulse height is independent of the amount of primary ionization (originally produced by the particle to be detected). It consists of a multi-chamber with thin central tungsten wire insulated from the outer chamber. The central wire is at positive potential with respect to the outer chamber and hence the central wire acts as anode while the outer cylinder acts as cathode. If the outer chamber is made out of glass, then its inner surface is coated with some conducting material to serve as cathode. G-M counters are usually filled with noble gases such as argon, neon, etc. G-M counter operates in that potential region in which the charge collected becomes independent of the ionization initiating it. Also the pulse size is completely independent of the initial ionization and all particles produce pulses of same height irrespective of their energy and primary ionization. If every initial electron gives rise to n-secondary electrons by collisions would also give rise to photons, which in turn, could produce photo electrons in the counter.

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Procedure: 1. First the G-M tube is connected to one of the pre-amp inputs. The pre-amp is also taking input from the high voltage supply. 2. Then enter the HV (high voltage) mode and then set the voltage to the recommended value (360 to 400 V approx.). 3. Now place the beta or gamma source close to the G-M counter window in the holder. 4. Switch on the main power button and set the preset time to 30 sec ( it can vary up to 60 sec). 5. After setting the preset time press the start button and you will get the number of counts for the required voltage. 6. Now gradually increase the HV and plot a table as shown in Table 9.1. 7. Also calculate the background counts (without source) for all voltages. 8. Tabulate all the readings and do the desired calculations.

TABLE 9.1 S.No.

Voltage

1

330

2

360(V1)

3

390

4

420

5

450

6

480

7

510

8

540

9

570 (V2)

10

Counts for 30 sec (N)

Background counts for 30 sec (Nb)

600

Calculations: As from Table 9.1 we get Voltage, V1, i.e., starting voltage of plateau @ 360 V Voltage, V2, i.e., upper threshold voltage of plateau @ 570 V Length of the plateau region = V2 – V1 Operating Voltage = (V2 – V1)/2 Slope =

N 2 - N1 100 ¥ ¥ 100 = % N1 V2 - V1

Corrected Counts (N – Nb)

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Counts per 30 sec/60 sec

where N2 and N1 are the numbers of counts of upper and lower regions respectively. V1 = Lower threshold voltage V2 = Upper threshold voltage

V1

V2

Voltage

‹‰ǤͻǤͳ౨Voltage vs counts graph

Viva Questions 1. 2. 3. 4. 5.

Why G-M counter cannot detect energy? What is quenching in G-M counter? What radiation does G-M counter detect? Can G-M counter detect neutrons? What is ionization?

EXPERIMENT 2 Aim. To determine the dead time and recovery time in G-M counter. Apparatus. A G-M counter consists of a cylindrical gas filled tube, a high voltage supply, source, a counter and a timer. Theory. In this large potential difference is applied between the G-M tube which acts as a cathode and a wire down the tube axis which acts as an anode. The sensitivity of the instrument is such that any particle capable of ionizing a single gas molecule in the GM tube (thus, producing an electron-ion pair) will initiate a discharge in the tube. In detector systems, there is a minimum amount of time that separates two events that may be recorded as two separate pulses. In some cases the limiting time may be set by processes in the detector itself, while in other cases the limit may arise due to the delays associated with the electronics. This minimum time separation is usually called dead time. Geiger-Mueller tubes exhibit dead time effects due to the recombination time of the internal gas ions after the occurrence of an ionizing event. The actual dead time depends on several factors including the active volume and shape of the detector and can range from a few microseconds for miniature tubes, to over 1000 microseconds

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for large volume devices. When making absolute measurements it is important to compensate for dead time losses at higher counting rates. Procedure: 1. First the G-M tube is connected into one of the pre-amp inputs. The pre-amp is also taking input from the high voltage supply. 2. Then enter the HV(high voltage) mode and then set the voltage to the recommended value. 3. Now place the radioactive source close to the G-M counter window. 4. Switch on the main power and press the count button to calculate the data. 5. To calculate dead time two gamma sources are needed. 6. In the first case keep the first source say G1 in the holder and take the counts for say 300 seconds. 7. Then place the second source say G2 in the holder and calculate the total number of counts for both G1 and G2. 8. Now remove G1 and take total number of counts using G2. 9. Now remove both G1 and G2 and take the background counts. Calculations. Let n1, n2 and n12 be the number of counts for source G1, G2 and combined G1 + G2. Let nt and nb be the true and measured background counts. Then dead time is given as t=

A 1- B C

A = n1n2 – nb n12 B = n1n2 (n12 + nb) – nb n12 (n1 + n2) C = B (n1 + n2 – n12 – nb )/A2

where

TABLE 9.2 Source

n

nb

t

GI G2 G1 + G2

Result The dead time of the given radio isotope is = _________ seconds Viva Questions 1. What do you understand by dead time?

Error

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2. What is the best operating voltage in G-M counter? 3. How do you test a G-M counter? 4. What is recovery time?

EXPERIMENT 3 Aim. To determine the half-life of a radioisotope using G-M counter. Apparatus. A G-M counter, sources, a counter and a timer. Theory. The radioactive half-life for a given radioisotope is the time for half the radioactive nuclei in any sample reduces half of its original value. N t N0/2 elt T

Therefore,

= = = = =

N0 e–lt T, N = N0/2 N0 e–lt 2 or lt = loge 2 0.6931/l

It is the relation between half life and decay constant.

TABLE 9.3 Time

Counts

Corrected counts

Net counts

30 60 90 110

600

Procedure 1. First the G-M tube is connected into one of the pre-amp inputs. The pre-amp is also taking input from the high voltage supply. 2. Then enter the HV (high voltage) mode and then set the voltage to the recommended value. 3. Now place the radioactive source close to the G-M counter window. 4. Switch on the main power and press the count button to calculate the data. 5. Calculate the number of counts and find the required result.

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Result The half-life of the given radio isotope is = _________ seconds Viva Questions 1. 2. 3. 4.

What is the half life of radioactive sample? Is half-life of radioactive sample changes from time to time? What is average life of atoms? Define activity of radioactive substance?

EXPERIMENT 4 Aim. To prove inverse square law using G-M counter. Apparatus. A G-M counter, sources, connecting cables, a counter and a timer. Theory. A point source of gamma rays emits in all directions. As the intensity of the gamma rays decreases with distance from the source the rays are spread over greater areas as the distance increases. The inverse square law is quality which starts out from a point source and travels in straight lines without getting lost. Light and sound intensity both behave according to an inverse square law when they spread out from a point source. For example, if we move away from a point source of light like bulb, the light intensity becomes smaller as the distance from the bulb becomes larger. This is also applicable for sound signals. But if we move twice as far from either of these sources, the intensity reduces to one fourth as great, not half as great. That’s why the law is defined as inverse square law. Consider a point source of gamma rays situated in a vacuum. The radiation spreads in all directions about the source, and therefore when it is at a distance x from the source it is spread over the surface of a sphere of radius x and area 4px2. If E is the energy radiated per unit time by the source, the intensity (energy per unit time per unit area) is given by I = E/4px2 or simply as I μ 1/x2. Thus, the energy varies as the inverse square from the source. Procedure 1. Make the connections as mentioned in the previous experiments. 2. First take the background counts for preset time of 60 sec. 3. Now place a gamma source in the holder at a distance of about 2 cm from the window of G-M tube. 4. Switch on the main power and press the count button to calculate the number of counts. 5. Now slowly increase the distance by 0.5 cm and record the number of counts for each distance.

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6. Tabulate the number of counts at all distances at a gap of 0.5 cm till 10 cm. 7. Calculate the net count rate by reducing the count rate with the background counts.

TABLE 9.4 S. No.

Distance (d)

1.

2

2.

2.5

3.

3

4.

–

5.

–

6.

–

7.

10

Counts (N)

Background counts (NB)

Net count rate (N0)

N 0d 2

1/d2

Calculations. Plot a graph between net count rate and the inverse of the square distance. If it follows the inverse square law it will be a straight line. Viva Questions 1. 2. 3. 4.

Define inverse square law. Does alpha and beta ray also follow the inverse square law? What is the first measurement to be taken during radiation leakage? Does sound follows inverse square law?

EXPERIMENT 5 Aim. To estimate the efficiency of gamma source. Apparatus. A survey meter, sources, a G-M counter, connecting cables and a timer. Theory. As we know different types of radiations have different properties. Efficiency is one of the most important characters of any detector. G-M counter cannot distinguish different types of radiations. As for each type of radiation it only generates an output pulse which does not depends on the incident energy. The detection efficiency measures the percentage of radiation that a given detector detects from the overall yield emitted from the source. It can vary with the volume and shape of the detector material, absorption cross section in the material, attenuation layers in front of the detector, and distance and position from the source to the detector. Procedure 1. Make the connections as mentioned in the previous experiments.

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2. Now place the gamma ray source at a distance of 10 cm from the source holder facing the counter window. 3. First record the counts for preset time say 100 sec. 4. Now take the background counts for same preset time. 5. Now do the required calculations. Calculations

TABLE 9.5 S. No.

Distance d

Background counts (nb)

Counts (with source) n

Corrected counts n0

Let D be the diameter of the end window then the net count rate is given as n0 = (n – ns)/100 As the gamma radiations are emitted in all directions therefore, fraction received at the end window is f = D2/16 d2 If a is the activity the fraction of gamma ray entering the detector, r = a ¥

D2 16 d 2

Therefore, efficiency, n0/r @ 2.3% Viva 1. 2. 3.

Questions What is the efficiency for beta particles? What are corrected counts? What do you understand by penetrating properties of different types of radiation?

EXPERIMENT 6 Aim. To calibrate radiation survey meter. Apparatus. A survey meter, sources, a counter and a timer. Theory. There are about 2450 known isotopes of 100 odd elements in the Periodic Table. There are few unstable nuclei which are above as well as below the nuclear stability curve. These unstable nuclei in order to achieve more stability break into smaller fragments and this phenomenon is known as nuclear fission and the whole process is called radioactivity. This is the phenomena of spontaneous disintegration of nuclei with the emission of electromagnetic radiations or other particles like a particle, b particle and g rays.

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In this experiment we consider a beta decay source, for example, Cs-138. b-particles are the stream of fast moving electrons having one unit negative charge and mass 1 equal to 9.1 ¥ 10–31 kg. When they come out of the nucleus, they have velocity th 10 the velocity of light. The radioactive nuclei in which ratio of neutrons to protons (n/p) is greater than that of stable nuclei it emits b-particles. Since we know that electrons do not exist inside the nucleus. It was proposed that neutrons inside the nucleus are unstable and decay into proton and electron. An electron so produced by the spontaneous conversion of neutron into proton comes out of the nucleus in the form of b-particle. Thus, in b-decay, the mass number (A) remains the same but atomic number increases by one, i.e., from Z to (Z + 1). The b-decay of a neutron is given as n Æ p + e– and b-decay of radioactive nucleus is given as ZX

A

Æ

Z+1Y

A

+–1e0 + Q

The Cs-137 source is small in size and may be treated as a point source for dose calculations. The exposure rate is decreased by factor of 1/x2, as the distance x increases. The source is always kept inside a heavy lead box. A window, when opened, allows collimated gamma rays to pass through the shielding without attenuation. Small lead attenuators, each of ½” thick, may be added to reduce the gamma exposure rate to desired levels at different distances. In this experiment measurements are made at different distances. Procedure 1. First gather all the information distance vs. dose rate and check the survey meter functioning, i.e., battery, zero error, etc. 2. Now put the radiation survey meter at a suitable distance from the attenuators. 3. Switch on the survey meter on proper range and make sure everyone is behind the shield. 4. Switch on the source and measure the survey meter readings. 5. Again repeat the above steps for different distances. 6. Make the readings in a tabular form for different distances and in the end switch off the source. Calculations. Plot the graph between exposed dose rate vs. the meter response. Radiation survey meter helps in operations related to health physics. Viva Questions 1. 2. 3. 4.

Discuss the uses of radiation survey meter. What precautions should be taken during radiation leak? What do you understand by exposed dose rate? Which radionuclide has the greatest impact on the dose?

University Question Paper - I Nuclear and Particle Physics Time: 3 hrs Max Marks: 70 Note: Attempt all questions as instructed

Section A Attempt any five questions: Q. 1. Discuss the principle and basic components of a nuclear reactor. What is the critical size of a nuclear reactor. Q. 2. Prove that electron does not exist inside the nucleus. Q. 3. Find the binding energy and binding energy per nucleon in case of 15P31 of mass 30.973763 mass units. Given that MH = 1.008665 amu. Q. 4. Discuss compound nucleus reaction and write two examples of compound nucleus reaction. Q. 5. Explain the ground state properties of deutron. Q. 6. One gm of radioactive material having a half life period of 2 years is kept in store for a duration of 4 years. Calculate how much of the material remains unchanged. 5 ¥ 6 = 30

Section B Attempt any two questions: Q. 7. What are different types of alpha-particle spectra observed for radioisotopes? How are these spectra explained from the energy level diagram of nuclei? Q. 8. (i) Discuss briefly nuclear waste disposal and radiation hazards from nuclear explosion. [5] –9 (ii) The radioactive disintegration constant l of radium D is 1.13 ¥ 10 sec–1. How much time given sample would take to reduce 1/10 of its original value? [5] Q. 9. Discuss the application of semi-empirical nuclear mass formula to explain the nuclear stability against beta decay. 2 ¥ 10 = 20

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Section C Attempt all questions: Q. 10. (i) Explain alpha-decay paradox. Discuss the quantum mechanical theory of alpha-decay and obtain a relation for barrier penetrability of alpha-decay. [8] (ii) Explain salient features of a liquid drop model and give various assumptions made for this model. [7] (iii) In beta decay process explain the phenomenon of Pauli’s neutrino hypothesis in detail. [5]

University Question Paper - 2 Nuclear and Particle Physics Time: 3 hrs Max Marks: 70 Note: Attempt all questions as instructed

Section A Attempt any five questions: Q. 1. Find the binding energy and binding energy per nucleon for 26Fe56. Given that the experimental mass of 26Fe56 is 55.934932 mass units, mass of neutron is 1.008665 mass units and mass of 1H1 is 1.007825 mass units. Q. 2. Discuss the problem to develop controlled fusion. Q. 3. What is Q-value of a nuclear reaction? Derive an expression for Q-value of a reaction in terms of kinetic energy of incident and product particle scattered through an angle q. Q. 4. Define binding energy and packing fraction of nuclei. Q. 5. Explain in brief the nuclear waste disposal and radiation hazards from nuclear explosion. Q. 6. Explain the term: Breeder reactor. 5 ¥ 6 = 30

Section B Attempt any two questions: Q. 7. Define binding energy and packing fraction of nuclei. How does the binding energy per nucleon for light, medium and heavy nuclei vary with mass numbers of nuclei. Q. 8. (i) Find the mass of a radioactive source Ra226 of 1.0 Curie strength whose lifetime is 1620 years. [5] (ii) Explain nuclear fission in terms of liquid drop model. [5] Q. 9. What is Pauli’s neutrino hypothesis for explanation of beta decay process? Write the properties of neutrino. 2 ¥ 10 = 20

308പhŶŝǀĞƌƐŝƚLJYƵĞƐƟŽŶWĂƉĞƌ

Section C Attempt all questions: Q. 10. (i) Discuss various types of nuclear reactions and conservation laws obeyed in the nuclear reactions. [7] (ii) What are magic and semi-magic numbers? Give the experimental evidence in support of magic numbers and shell structure of nucleons in nuclei. [8] (iii) State clearly what do you understand by terms: (i) electric quadrupole moment and (ii) mass defect with respect to a nucleus. [5]

University Question Paper - 3 Nuclear and Particle Physics Time: 3 hrs Max Marks: 70 Note: Attempt all questions as instructed

Section A Attempt any five questions: Q. 1. Explain briefly neutrino hypothesis of beta-decay. Q. 2. What are different types of nuclear reaction? Give Bohr’s hypothesis of compound nucleus for nuclear reaction. Q. 3. The radioactive disintegration constant l of radium D is 1.13 ¥ 10–9 sec–1. How much time given sample would take to reduce 1/10 of its original value. Q. 4. State clearly what do you understand by: (i) electric quadrupole moment, and (ii) mass defect with respect to a nucleus. Q. 5. Explain in brief nuclear waste disposal and radiation hazards from nuclear explosion. Q. 6. Explain nuclear fission in terms of liquid drop model. 5 ¥ 6 = 30

Section B Attempt any two questions: Q. 7. What are magic and semi-magic numbers? Give the experimental evidence in support of magic numbers and shell structure of nucleons in nuclei. Q. 8. (i) Find the mass of a radioactive source Ra226 of 1.0 Curie strength whose lifetime is 1620 years. [5] (ii) What are internal conversion and pair conversion phenomena? [5] Q. 9. What is Pauli’s neutrino hypothesis for explanation of beta decay process? Write the properties of neutrino. 2 ¥ 10 = 20

310പhŶŝǀĞƌƐŝƚLJYƵĞƐƟŽŶWĂƉĞƌ

Section C Attempt all questions: Q. 10. (i) Calculate the binding energy per nucleon in case of 6C12 atom having mass equal to 12.00 units. Given the masses of proton, neutron and electron are 1.00728, 1.00867 and 0.00055 units of mass, respectively. [6] (ii) Discuss various types of nuclear reactions and conservation laws obeyed in the nuclear reactions. [8] (iv) Explain with proof why electrons do not exist inside the nucleus. [6]

University Question Paper - 4 Nuclear and Particle Physics Time: 3 hrs Max Marks: 70 Note: Attempt all questions as instructed

Section A Attempt any five questions: Q. 1. Explain liquid drop model to obtain semi-empirical mass formula and discuss the condition on nuclear stability? Q. 2. Explain why photoelectric effect is not possible with free electrons? Q. 3. Discuss: (i) Fermi-Dirac statistics (ii) Packing fraction Q. 4. Explain pairing energy term of semi-empirical mass formula? Q. 5. What is stopping power? Explain range straggling? Q. 6. Discuss Geiger Nuttal law? Also explain the theory of alpha decay? 5 ¥ 6 = 30

Section B Attempt any two questions: Q. 7. Discuss the principle, construction and working of linear accelerator (LINAC)? Q. 8. Write notes on: (i) Scintillation counter (ii) Bubble chamber (iii) Ionization chamber Q. 9. Discuss the laws of disintegration, activity, and its unit, half-life and mean life of radioactivity? 2 ¥ 10 = 20

312പhŶŝǀĞƌƐŝƚLJYƵĞƐƟŽŶWĂƉĞƌ

Section C Attempt all questions: Q. 10. (i) Write short notes on (a) Baryon Number (b) Parity [7] (ii) Discuss four fundamental interactions? Give a brief account of the discovery and properties of elementary particles. [8] (iii) What are the limitations of cyclotron? How have these been overcome in betatron? [5]

Index

D-decay, theory of 123 D-particle properties of 102 scattering of 76 transmutation by 68 E-decay, theory of 128 E-particles, properties of 102 J-radiation 70 J-rays, properties of 103 K-meson 256

A Altitude effect 287 Angle of scattering 77 Annihilation process 254 Anomalous scattering 76 Asymmetry energy 24 Atomic nucleus, properties of 6

B Baryon 256 Baryon number 260 conservation of 264 Betatron 239 Biasing 205 Binding energy 9, 22 Binding energy per nucleon 10 Bose-Einstein statistics 14 Breaking radiation 158 Bremsstrahlung 158 Bubble chamber 203-205

C C.F. Weizsacker formula 22, 25 Carbon dating 117 Carbon-nitrogen cycle 89 Chain reaction 86 Charge conjugation 266 Charge quantum number 260 Charged particle, range of 157 Cherenkov detector 209 Cherenkov radiation 160, 209 Coloured quarks 272 Compound nucleus 60 decay scheme of 61 energy levels of 61 Compound nucleus reaction 50 Compton effect 167 experimental setup of 170 Compton scattering, theory of 168 Conservation laws 51 Conservation of angular momentum 52 Conservation of charge 263 Conservation of linear momentum 52 Conservation of mass-energy 51 Conservation of parity 52 Conservation of spin and statistics 52 Controlled chain reaction 87 Cosmic intensity 287 Cosmic ray shower 289 Cosmic rays 273 nature of 285 origin of 290 types of 284

314 Index Coulomb’s energy 23 Coulomb’s scattering 76 CPT theorem 266 Current ionization chamber 187 Cyclotron 229

Geiger-Nuttal law 113 Gluons 273 Gravitational interaction 258 Graviton 254

H D Deuteron, transmutation by 65 Differential cross section 59 Diffused junction detector 206-207 Diffusion cloud chamber 198 Disintegration 50 Disintegration constant 105 Displacement law 112

Hadrons, composition of 271 Half-life 105 Heavy ion 50 transmutation by 71 High energy quanta 153 High energy reaction 50 High purity germanium detector 209 Hypercharge 261 conservation of 265

E East-west asymmetry 288 East-west effect 288 Electric dipole moment 14 Electromagnetic interaction 258 Electron synchrotrons 235 Electron-proton hypothesis 2 angular momentum of 13 failures of 3 wave mechanical properties of 14 Elementary particles classification of 254 interaction between 257 quantum numbers associated with 260 Energetic charged particles 152 Energy loss 153 Even-odd effect 24 Excitation 152

F Fermi gas model 31 Fermi-Dirac statistics 14 Fission fragments 153

G Gamma decay, theory of 130 Gamma rays, properties of 131 Geiger-Muller counter 189-192, 195

I Impact factor 77 Induced fission 50 Internal conversion 131 Internal conversion coefficient 133 Intrinsic angular momentum 253 Intrinsic magnetic moment 253 Ionization 152 Ionization chamber 187 Iso-spin quantum number 261 conservation of 264

K Kaon 256

L Latitude effect 266 Law of conservation of charge 51 Law of conservation of mass number 51 Law of radioactive disintegration 103 Law of successive disintegration 109 Lepton 255 Lepton number 260 conservation of 264 Level width 62 Linear accelerator 232 Liquid drop model 22 success of 25

Index

Lithium ion drifted junction detector 208 Longitude effect 288

M Magic number 26 Magnetic dipole 14 Mass defect 8 Matter interaction of heavy charged particle with 153 interaction of neutrons with 162 interaction of J-ray with 164 Meson 255 Multiplet number 260

Nuclear radiation nature and properties of 101 types of 100 Nuclear reaction 163 kinematics of 52 types of 48 Nuclear shell model 27 Nuclear size 7 Nuclear stability 20 Nuclear transmutation 63 Nucleus, constituents of 2

O Orbital angular momentum 13

N

P

Neutral or uncharged particles 152 Neutrino hypothesis 130 Neutrino theory 128 Neutron detection of 73 discovery of 71 importance of 76 mass of 73 properties of 75 transmutation by 66 Neutron reproduction factor 88 Nuclear cross section 57 determination of 58 Nuclear density 7 Nuclear emulsion 187 Nuclear energy 13 Nuclear fission 83 Bohr and Wheeler theory of 84 source of energy in 85 Nuclear force properties of 17 Yukawa theory of 19 Nuclear fusion 88 Nuclear mass 6 Nuclear models 21 Nuclear particle reaction 48 Nuclear potential barrier 121

Packing fraction 9 Pair production 171 Pairing energy 24 Parity 15 conservation of 266 Permanent equilibrium 111 Photo disintegration 49 Photoelectric effect 165 experimental verification of 166 Photon 254 transmutation by 70 Pion 255 Positron annihilation 173 Potential barrier, height of 121 Potential well, depth of 121 Primary cosmic rays 285 Product nucleus, energy levels of 61 Proportional counter 192-195 Proton synchrotrons 236 Proton-neutron hypothesis 4 Proton-proton cycle 89 Protons, transmutation by 64 Pulse ionization chamber 187

Q Quantum chromodynamics 273 Quark model 270, 271

315

316 Index Quarks 270 Quenching 191 Q-value of a reaction 52 physical significance of 54

R Radiation hazards 119 genetic damage 120 somatic (general body) damage 120 Radiation length 162 Radiation loss 160 Radioactive capture 49 Radioactive dating 116 Radioactive decay, modes of 100 Radioactive growth and decay 109 Radioactive sample activity of 107 average life of 107 mean life of 107 Radioactive series 112 Radioactive tracer 114 Radioactivity 99 uses of 114 Radioisotopes, handling of 119 Range straggling 159 Reaction of transmutator 48 Rutherford scattering 76 Rutherford scattering formula 81

Spin angular momentum 13 Spontaneous decay 50 Strangeness quantum number 262 Strangeness, conservation of 265 Strong interaction 259 Surface barrier detector 207 Surface energy 23 Synchrotrons 235

T Thermonuclear reaction 88 and the stellar energy 88 Threshold energy 212 Threshold velocity 211 Time reversal 267 Transient equilibrium 112 Transmutation by deuteron 65 heavy ion 71 neutron 66 photon 70 protons 64 tritium 71 Tritium, transmutation by 71

U Uncontrolled chain reaction 86 Uranium dating 116

S

V

Scattering 48, 76 Scintillation counter 199 Scintillation detectors 187 Secondary cosmic rays 285 Secular equilibrium 111 Semiconductor detector 205 Semi-empirical mass formula 22

VAN de Graaff electrostatic generator 227 Volume energy 23

W Weak interaction 259 Wilson’s cloud chamber 196-198

This book includes to-the-point description of titles, related theories, labelled diagrams, tables, examples, etc. This book is beneficial for B.Tech. Physics, B.Sc. Physics and for M.Sc. Applied Physics students. It helps students to prepare for JAM, NET SET exams. This book has been divided into nine chapters. Topics like nuclear structure, nuclear reactions, radioactivity, elementary particles, accelerators, radiation detection, interaction of radiation, cosmic rays are covered in detail. The discussion on all the topics is simple and mathematical derivations are less complex. Data, graphs and figures are mentioned wherever required. It also includes the basic lab experiments required in the nuclear lab for both B.Sc. as well as M.Sc. students. Salient Features ? Discussion on all the topics is simple and mathematical derivations are less complex. ? Data, graph and figures are mentioned wherever required. ? Short and long questions from various examinations of different universities. ? All concepts of nuclear physics are supported with relevant diagrams. Shefali Kanwar is Assistant Professor, Department of Applied Physics, Amity University, Noida, U.P. She obtained her Ph.D. in Nuclear Physics from Thapar University, Patiala. She is a lifetime member of Indian Association for Engineers (IAENG). She has more than 12 years of teaching experience.

Shivani is Assistant Professor, Department of Applied Physics, Amity Institute of Applied Sciences, Amity University, Noida. She has more than 12 years of teaching experience. She is a Gold Medallist in M.Sc. from the University of Allahabad in 2002. She obtained her Ph.D. in 2009 from the University of Allahabad. She has published more than 15 research papers in national and international journals.

978-93-91029-13-5

Nuclear and Particle Physics Shefali Kanwar Pramila Shukla Shivani

• Shefali Kanwar • Pramila Shukla • Shivani

Pramila Shukla is Assistant Professor, Department of Applied Physics, Amity University, Noida, U.P. She obtained her Ph.D. in Physics from the University of Allahabad. She is editor of International Journal of Scientific Research Letters and member of editorial board of International Journal of Engineering Science & Technology Letters. She is a lifetime member of Indian Association for Physics Teachers (IAPT) and has been a member of Optical Society of America. She has more than 14 years of teaching experience.

Nuclear and Particle Physics

Nuclear and Particle Physics

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