1,869 85 948KB
English Pages 168 Year 2016
PROBLEM SOLUTION MANUAL FOR
Fundamentals of Nuclear Science and Engineering Third Edition
by
J. Kenneth Shultis and
Richard E. Faw Dept. of Mechanical and Nuclear Engineering Kansas State University Manhattan, KS 66506
email: [email protected]
Revised June 2016 (c) Copyright 20072016 All Rights Reserved This typescript is the property of the authors. It may not be copied in part or in total without permission of the authors.
Notice This collection contains solutions to most of the problems in our book Fundamentals of Nuclear Science and Engineering, 3/e (Taylor & Francis, Boca Raton, Florida, 2007. We do not warrant that all the solutions are correct or that other approaches could give equally valid results. This collection is provided to you solely as an aid in your teaching, and we ask that you do not copy this set for others without our permission. If, in your teaching, you develop better solutions than are presented here or find corrections are needed, we would appreciate receiving copies so that, over time, this collection will be improved. A sporadically updated errata for the book can be found on the world wide web at http://www.mne.ksu.edu/~jks/books.htm
Chapter 1
Fundamental Concepts
PROBLEMS 1. Both the hertz and the curie have dimensions of s−1 . Explain the difference between these two units. Solution: The hertz is used for periodic phenomena and equals the number of “cycles per second.” The curie is used for the random or stochastic rate at which a radioactive source decays, specifically, 1 Ci = 3.7 × 1010 decays/second. 2. Advantages of SI units are apparent when one is presented with units of barrels, ounces, tons, and many others. (a) Compare the British and U.S. units for the gallon and barrel (liquid and dry measure) in SI units of liters (L). (b) Compare the long ton, short ton, and metric ton in SI units of kg. Solution: Unit conversions are taken from the handbook Conversion Factors and Tables, 3d ed., by O.T. Zimmerman and I. Lavine, published by Industrial Research Service, Inc., 1961. (a) In both British and U.S. units, the gallon is equivalent to 4 quarts, eight pints, etc. However, the quart and pint units differ in the two systems. The U.S. gallon measures 3.7853 L, while the British measures 4.546 L. Note that the gallon is sometimes used for dry measure, 4.405 L U.S. measure. The barrel in British units is the same for liquid and dry measure, namely, 163.65 L. The U.S. barrel (dry) is exactly 7056 in3 , 115.62 L. The U.S. barrel (liq) is 42 gallons (158.98 L) for petroleum measure, but otherwise (usually) is 31.5 gallons (119.24 L). (b) The common U.S. unit is the short ton of 2000 lb, 907.185 kg, 20 short hundredweight (cwt). The metric ton is exactly 1000 kg, and the long ton is 20 long cwt, 22.4 short cwt, 2240 lb, or 1016 kg. 11
12
Fundamental Concepts
Chap. 1
3. Compare the U.S. and British units of ounce (fluid), (apoth), (troy), and (avdp). Solution: The U.S. and British fluid ounces are, respectively, 1/32 U.S. quarts (0.02957 L) and 1/40 British quarts (0.02841 L). The oz (avdp.) is exactly 1/16 lb (avdp), i.e., 0.02834 kg. Avdp., abbreviation for avoirdupois refers to a system of weights with 16 oz to the pound. The apoth. apothecary or troy ounce is exactly 480 grains, 0.03110 kg.
4. Explain the SI errors (if any) in and give the correct equivalent units for the following units: (a) mgrams/kiloL, (b) megaohms/nm, (c) N·m/s/s, (d) gram cm/(s−1 /mL), and (e) Bq/milliCurie. Solution: (a) Don’t mix unit abbreviations and names; SI prefixes only in numerator: correct form is µg/L. (b) Don’t mix names and abbreviations and don’t use SI prefixes in denominator: correct form nohm/m. (c) Don’t use hyphen and don’t use multiple solidi: correct form N m s−2 . (d) Don’t mix names and abbreviations, don’t use multiple solidi, and don’t use parentheses: correct form g cm s mL or better 10 µg m s L. (e) Don’t mix names with abbreviations, and SI prefix should be in numerator: correct form kBq/Ci.
5. Consider H2 , D2 , and H2 O, treated as ideal gases at pressures of 1 atm and temperatures of 293.2◦K . What are the molecular and mass densities of each. Solution: According to the ideal gas law, molar densities are identical for ideal gases under the same conditions, i.e., ρm = p/RT . From Table 1.5, R = 8.314472 Pa m3 /K. For p = 0.101325 MPa= 1 atm., and T = 293.2◦K , ρm = 41.56 mol/m3. Multiplication by molecular weights yield, respectively, 83.78 , 167.4, and 749.0 g/m3 for the three gases.
6. In vacuum, how far does light move in 1 ns? Solution: ∆x = c∆t = (3 × 108 m/s) × (10−9 s) = 3 × 10−4 m = 30 cm.
Fundamental Concepts
13
Chap. 1
7. In a medical test for a certain molecule, the concentration in the blood is reported as 57 mcg/dL. What is the concentration in proper SI notation? Solution: 123 mcg/dL = 10−3 10−2 g/10−1 L = 1.23 × 10−4 g/L = 57 µg/L. 8. How many neutrons and protons are there in each of the following nuclides: (a) 11 B, (b) 24 Na, (c) 60 Co, (d) 207 Pb, and (e) 238 U? Solution: Nuclide 11
B Na 60 Co 207 Pb 238 U 24
neutrons
protons
6 13 33 125 146
5 11 27 82 92
9. Consider the nuclide 71 Ge. Use the Chart of the Nuclides to find a nuclide (a) that is in the same isobar, (b) that is in the same isotone, and (c) that is an isomer. Solution: (a)
71
As, (b)
59
Ga, and (c)
71m
Ge
10. Examine the Chart of the Nuclides to find any elements, with Z less that that of lead (Z = 82), that have no stable nuclides. Such an element can have no standard relative atomic mass. Solution: Promethium (Z = 61) and Technetium (Z = 43)
11. What are the molecular weights of (a) H2 gas, (b) H2 O, and (c) HDO? Solution: From Table A.3, A(O) = 15.9994 g/mol; from Table B.1 A(H) = 1.007825 g/mol and A(D) = 2.014102 g/mol. (a) A(H2 ) = 2 A(H) = 2 × 1.007825 = 2.01565 g/mol (b) A(H2 O) = 2 A(H) + A(O) = 2 × 1.007825 + 15.9994 = 18.0151 g/mol (c) A(HDO) = A(H) + A(D) + A(O) = 1.007825 + 2.014102 + 15.9994 = 19.0213 g/mol
14
Fundamental Concepts
Chap. 1
12. What is the mass in kg of a molecule of uranyl sulfate UO2 SO4 ? Solution: From Table A.3, A(U) = 238.0289 g/mol, A(O) = 15.9994 g/mol, and A(S) = 32.066 g/mol. The molecular weight of UO2 SO4 is thus A(UO2 SO4 ) = A(U) + 6A(O) + A(S) = 238.0289 + 6(15.994) + 32.066 = 366.091 g/mol = 0.336091 kg/mol.
Since one mol contains Na = 6.022 × 1023 molecules, the mass of one molecule of UO2 SO4 = A(UO2 SO4 )/Na = 0.366091/6.002 × 1023 = 6.079 × 10−25 kg/molecule.
13. Show by argument that the reciprocal of Avogadro’s constant is the gram equivalent of 1 atomic mass unit. Solution: By definition one gram atomic weight of one atom of 12 C is M (126 C) =
12
C is 12 g/mol. Thus the mass of
12 12 g/mol = g/atom. Na atoms/mol Na
But by definition, one atom of
12
C has a mass of 12 u. Therefore, 1u 12 1 12 1 u= g/( C atom) = g. Na 12 u/(12 C atom) Na
14. Prior to 1961 the physical standard for atomic masses was 1/16 the mass of the 16 12 8 O atom. The new standard is 1/12 the mass of the 6 C atom. The change led to advantages in mass spectrometry. Determine the conversion factor needed to convert from old to new atomic mass units. How did this change affect the value of the Avogadro constant? Solution From Table B.1, the 168 O atom has a mass of 15.9949146 amu. Thus, the pre1961 atomic mass unit was 15.9949146/16 post1961 units, and the conversion factor is thus 1 amu (16 O) = 0.99968216 amu (12 C). The Avogadro constant is defined as the number of atoms in 12 g of unbound carbon12 in its restenergy electronic state, i.e., the number of atomic mass units per gram. Using data from Table 1.5, one finds that Na is given by the reciprocal of the atomic mass unit, namely, [1.6605387×10−24]−1 = 6.0221420× 1023 mol−1. Pre1961, the Avogadro constant was more loosely defined as the number of atoms per mol of any element, and had the best value 6.02486×1023.
Fundamental Concepts
15
Chap. 1
15. How many atoms of
234
U are there in 1 kg of natural uranium?
Solution: From Table A.3, the natural abundance of 234 U in uranium is found to be f(234 U) = 0.0055 atom%. A mass m of uranium contains [m/A(U)]Na uranium atoms. Thus, the number of 234 U atoms in the mass m = 1000 g are N (234 U) = f(234 U)
mNa A(U)
= 0.000055
1000 × (6.022 × 1023) = 1.392 × 1020 atoms. 238.0289
16. A bucket contains 1 L of water at 4 ◦ C where water has a denisty of 1 g cm3 . (a) How many moles of H2 O are there in the bucket? (b) How many atoms of 1 2 1 H and 1 D are there in the bucket? Solution: (a) The relative atomic weight of water A(H2 O) = 2A(H)+A(O) = 2(1.00794)+ (15.9994) = 18.01528. Then the number of water molecules mols of H2 O =
mass(H2 O) 1000 g = = 55.5 mol. A(H2 O) 18.01258 g/mol
(b) Number of molecules of H2 O = 55.5 mol × Na mol−1 = 55.5 × 6.60221 × 1023 = 3.343 × 1025 molecules. Then the number of atoms of both 11 H and 21 D atoms = 2 × no. of H2 O molecules = 6.6856 × 1025 atoms. From Table A.4, the isoptopic abundances are found to be γ(11 H) = 0.999885 and γ(21 D) = 0.000115. Then N (11 H) = (0.999885)(6.6856 × 1025 ) = 6.69 × 1025 atoms and N (21 D) = (0.000115)(6.6856 × 1025 ) = 7.69 × 1021 atoms. 17. How many atoms of deuterium are there in 2 kg of water? Solution: Water is mostly H2 O, and so we first calculate the number of atoms of hydrogen N (H) in a mass m = 2000 g of H2 O is N (H) = 2N (H2 O) = 2 =2
mNa mNa '2 A(H2 O) A(H2 O)
2000 × (6.022 × 1023 ) = 1.34 × 1026 atoms of H. 18
16
Fundamental Concepts
Chap. 1
From Table A.4, the natural isotopic abundance of deuterium (D) is 0.015 atom% in elemental hydrogen. Thus, the number of deuterium atoms in 2 kg of water is N (D) = 0.00015 × N (H) = 2.01 × 1022 atoms. 18. Estimate the number of atoms in a 3000 pound automobile. State any assumptions you make. Solution: The car mass m = 3000/2.2 = 1365 kg. Assume most the this mass is iron. If the atoms in noniron materials (e.g., glass, plastic, rubber, etc.) were converted to iron, the car mass would increase to about mequiv = 1500 kg. Thus the number of atoms in the car is N =
mequiv Na (1.5 × 106 )(6.022 × 1023) = = 1.6 × 1028 atoms. A(Fe) 56
19. Calculate the relative atomic weight of oxygen. Solution From Table A.4, oxygen has three stable isotopes: 16 O, 17 O, and 18 O with percent abundances of 99.757, 0.038, and 0.205, respectively. Their atomic masses, in u, are found from Table B.1 and equal their relative atomic weights. Then from Eq. (1.2) A(O) = =
γ(16 O) 16 γ(17 O) 17 γ(18 O) 18 A( O) + A( O) + A( O) 100 100 100 99.757 0.038 0.205 15.994915 + 16.999132 + 17.999160 = 15.999405. 100 100 100
20. Natural uranium contains the isotopes 234U, relative atomic weight of natural uranium.
235
U and
238
U. Calculate the
Solution From Table A.4, the three isotopes 234 U, 235 U, and 238U have isotopic abundances of 0.0055%, 0.720%, and 99.2745%, respectively. Their atomic masses, in u, are found from Table B.1 and equal their relative atomic weights. Then from Eq. (1.2) A(O) = =
γ(234 U) 234 γ(235 U) 235 γ(238 U) 238 A( U) + A( U) + A( U) 100 100 100 0.0055 0.720 99.2745 234.040945 + 235.043923 + 238.050783 100 100 100
= 238.02891.
Fundamental Concepts
Chap. 1
17
21. Does a sample of carbon extracted from coal have the same relative atomic weight as a sample of carbon extracted from a plant? Explain. Solution The carbon extracted from coal has only two isotopes, namely 12 C and 13 C with with abundances of 98.93% and 1.07%, respectively. The relative atomic weight is thus is slightly larger than 12 that would result if there were no 13 C, namely 12.0107. Carbon extracted from plant material, however, also contains the radioactive isotope 14 C produced in the atmosphere by cosmic rays. Thus, the relative atomic weight is conceptually greater than that of carbon from coal in which all the 14 C has radioactively decayed away. However, as discussed in Section 5.8.1, the amount of 14 C in plant material is extremely small (1.23 × 10−12 atoms per atom of stable carbon). Thus, 14 C would increase the atomic weight only in the 12th significant figure! 22. Dry air at normal temperature and pressure has a mass density of 0.0012 g/cm3 with a mass fraction of oxygen of 0.23. What is the atom density (atom/cm3 ) of 18 O? Solution: From Eq. (1.5), the atom density of oxygen is N (O) =
wo ρNa 0.23 × 0.0012 × (6.022 × 1023) = = 1.04 × 1019 atoms/cm3 . A(O) 15.9994
From Table A.4 isotopic abundance of 18 O in elemental oxygen is f18 = 0.2 atom% of all oxygen atoms. Thus, the atom density of 18 O is N (18 O) = f18 N (O) = 0.002 × 1.04 × 1019 = 2.08 × 1016 atoms/cm3. 23. A reactor is fueled with 4 kg uranium enriched to 20 atompercent in 235U. The remainder of the fuel is 238 U. The fuel has a mass density of 19.2 g/cm3 . (a) What is the mass of 235 U in the reactor? (b) What are the atom densities of 235 U and 238U in the fuel? Solution: (a) Let m5 and m8 be the mass in kg of an atom of 235 U and 238U, and let n5 and n8 be the total number of atoms of 235U and 238U in the uranium mass MU = 4 kg. For 20% enrichment, n8 = 4n5 , so that m8 MU = n5 m5 + n8 m8 = n5 m5 + 4n5 m8 = n5 m5 1 + 4 . m5 Here n5 m5 = M5 is the mass of 235 U in the uranium mass MU . From this result we obtain using m5 /m8 ' 235/238 −1 −1 m8 238 M5 = MU 1 + 4 = 4 kg 1 + 4 = 0.7919 kg. m5 235
18
Fundamental Concepts
The mass of
238
Chap. 1
U M8 = MU − M5 = 3.208 kg.
(b) The volume V of the uranium is V = MU /ρU = (4000 g)/(19.2 g/cm3 ) = 208.3 cm3 . Hence the atom densities are N5 =
M5 Na (791.9 g)(6.022 × 1023 atoms/mol) = = 9.740×1021 cm−3 A5 V (235 g/mol)(208.3 cm3 )
N8 =
M8 Na (3208 g)(6.022 × 1023 atoms/mol) = = 3.896×1022 cm−3 A8 V (238 g/mol)(208.3 cm3 )
24. A sample of uranium is enriched to 3.2 atompercent in 235 U with the remainder being 238 U. What is the enrichment of 235U in weightpercent? Solution: Let the subscripts 5, 8 and U refer to 235U, 238 U, and uranium, respectively. For the given atom% enrichment, The number of atoms in a sample of the uranium are N5 = 0.0320NU The mass M5 and M8 of
235
U and
and 238
M5 = 0.0320NU m5
N8 = 0.9680NU .
U in the sample is
and
where m5 and m8 is the mass of an atom of
M8 = 0.9680NU m8 , 235
U and
238
U, respectively.
The enrichment in weight% is thus e(wt%) = 100 × =
M5 0.0320m5 = 100 × M5 + M8 0.0320m5 + 0.9680m8
100 × 0.0320 100 × 0.0320 ' 0.0320 + 0.9680(m8 /m5 ) 0.0320 + 0.9680(238/235)
= 3.16 wt%.
25. A crystal of NaCl has a density of 2.17 g/cm3 . What is the atom density of sodium in the crystal? Solution: Atomic weights for Na and Cl are obtained from Table A.3, so that A(NaCl) = A(Na) + A(Cl) = 22.990 + 35.453 = 58.443 g/mol. Thus the atom density of Na is N (Na) = N (NaCl) =
ρNaCl Na 2.17 × 6.022 × 1023 = = 2.24 × 1022 cm−3 . A(NaCl) 58.443
Fundamental Concepts
19
Chap. 1
26. A concrete with a density of 2.35 g/cm3 has a hydrogen content of 0.0085 weight fraction. What is the atom density of hydrogen in the concrete? Solution: From Eq. (1.5), the atom density of hydrogen is N (H) =
wH ρMa (0.0085)(2.35 g/cm3 )(6.022 × 1023 atoms/mol) = A(H) 1 g/mol
= 1.20 × 1022 atoms/cm3. 27. How much larger in diameter is a uranium nucleus compared to an iron nucleus? Solution: From Eq. (1.7) the nuclear diameter is D = 2Ro A1/3 so that DU = DF e
AU AF e
1/3
'
238 56
1/3
= 1.62.
Thus, DU ' 1.62 DFe. 28. By inspecting the chart of the nuclides, determine which element has the most stable isotopes? Solution: The element tin (Sn) has 10 stable isotopes. 29. Find an internet site where the isotopic abundances of mercury may be found. Solution: http://www.nndc.bnl.gov 30. The earth has a radius of about 6.35 × 106 m and a mass of 5.98 × 1024 kg. What would be the radius if the earth had the same mass density as matter in a nucleus? Solution: From the text, the density of matter in a nucleus is ρn ' 2.4 ×1014 g/cm3 . The mass of the earth M = ρ × V where the volume V = (4/3)πR3 . Combining these results and solving for the radius gives R=
3M 4πρ
1/3
=
3(5.98 × 1027 g) 4π(2.4 × 1014 g/cm3 )
1/3
= 1.81 × 104 cm = 181 m.
Chapter 2
Modern Physics Concepts
PROBLEMS 1. An accelerator increases the kinetic energy of electrons uniformly to 10 GeV over a 3000 m path. That means that at 30 m, 300 m, and 3000 m, the kinetic energy is 108 , 109 , and 1010 eV, respectively. At each of these distances, compute the velocity, relative to light (v/c), and the mass in atomic mass units. Solution: From Eq. (2.10) in the text T = mc2 − mo c2 we obtain m = T /c2 + mo . (P2.1) p From Eq. (2.5) in the text m = mo / 1 − v2 /c2 , which can be solved for v/c to give r v m2 1 m2o mo = 1 − 2o ' 1 − , if > mo , show that the de Broglie wavelength is given by r 17.35 × 10−8 mo √ cm. λ= m T Solution: (a) Rewrite the result of Problem 210 as −1/2 hc 1 m √ λ= √ 1+ . mo mo c2 T Substitute for the constants and use mo = me = 0.511 MeV/c2 to obtain λ=
(4.1357 × 10−15 eV s)(2.998 × 1010 cm/s) (1 + m/mo )−1/2 √ p 0.5110 × 106 eV T (eV)
−1/2 17.35 × 10−8 m cm. = p 1+ mo T (eV)
(P2.6)
p √ (b) For nonrelativistic electrons m ' mo , so that 1/ 1 + (m/mo ) ' 1/ 2, and the above result becomes λ=
12.27 × 10−8 p cm. T (eV)
p p (c) For very relativistic particles, m >> mo so that 1/ 1 + (m/mo ) ' mo /m. Eq. (2.4) above then becomes p 17.35 × mo /m p λ= × 10−8 cm. T (eV)
Modern Physics Concepts
211
Chap. 2
19. What are the wavelengths of electrons with kinetic energies of (a) 10 eV, (b) 1000 eV, and (c) 107 eV? √ Solution: From Eq. (2.17) p = (1/c) T 2 + 2T mo c2 and using the de Broglie relation λ = h/p we obtain the de Broglie wavelength as λ= √
T2
hc . + 2T mo c2
(P2.7)
Now apply this equation to the three electron energies. (a) Substitute mo c2 = me c2 = 0.5110 MeV and T = 10 eV into Eq. (P2.6) to obtain λ=
(4.135 × 10−15 eV s)(2.998 × 108 m/s) p = 3.88 × 10−10 m. 102 + 2(10)(0.5110 × 106 ) eV
(b) similarly, for T = 103 eV we find λ=
(4.135 × 10−15 eV s)(2.998 × 108 m/s) p = 3.87 × 10−11 m. 106 + 2(103 )(0.5110 × 106 ) eV
(c) similarly, for T = 107 eV we find λ=
(4.135 × 10−15 eV s)(2.998 × 108 m/s) p = 1.18 × 10−13 m. 1014 + 2(107 )(0.5110 × 106 ) eV
20. Low energy neutrons are often referred to by their de Broglie wavelength as measured in angstoms (˚ A) with 1 ˚ A= 1 × 10−10 m. (a) Derive a formula that gives the kinetic energy of such a neutron in terms of its de Broglie wavelength. (b) What is the energy of a neutron (in eV) of a 6˚ A neutron. Solution: (a) Equation (2.30) for a nonrelativistic particle reduces to p λ = h/ 2mo T , which, upon solving to T gives
T =
h2 . 2λ2 mo
(b) Here λ = 6 × 10−10 m and mo /c2 = 931.49 × 106 eV, so (4.135 × 10−15 eV s)2 2 ) (931.49 × 106 eV)/(2.998 × 108 m s−1 )2 (2)(6 × 10−10 m = 0.00229 eV.
T =
212
Modern Physics Concepts
Chap. 2
21. What is the de Broglie wavelength of a water molecule moving at a speed of 2400 m/s? What is the wavelength of a 3g bullet moving at 400 m/s? Solution: (a) A water molecule (H2 O) has a rest mass of about m = (18 u)(1.661 × 10−27 kg/u) = 2.989 × 10−26 kg. Its momentum when traveling at 2400 m/s is p = mv = (2.989×10−26 kg)× (2400 m/s) = 7.18 × 10−23 kg m s−1 = 7.18 × 10−23 J s m−1 . Thus the de Broglie wavelength of the water molecule is λ=
h 6.626 × 10−34 J s = = 9.23 × 10−12 m. p 7.18 × 10−23 J s m−1
(b) A 3g bullet moving at 400 m/s has a momentum p = mv = (0.003 kg) × (400 m/s) = 1.2 kg m s−1 = 1.2 J s m−1 . Its de Broglie wavelength is thus 6.626 × 10−34 J s h λ= = = 5.53 × 10−34 m. p 1.2 J s m−1
22. If a neutron is confined somewhere inside a nucleus of characteristic dimension ∆x ' 10−14 m, what is the uncertainty in its momentum ∆p? For a neutron with momentum equal to ∆p, what is its total energy and its kinetic energy in MeV? Verify that classical expressions for momentum and kinetic energy may be used. Solution: −14 From the uncertainty principle, ∆p∆x > m ∼ h/(2π) so that for ∆x ' 10
∆p =
6.626 × 10−34 J s h = = 1.05 × 10−20 J s m−1 . 2π∆x 2π × 10−14 m
A nonrelativistic (classical) particle has kinetic energy T = (1/2)mv2 = p2 /(2m). For a neutron with p ' ∆p = 1.05 × 10−20 J s m−1 T =
=
(∆p)2 (1.05 × 10−20 J s m−1 )2 = = 3.32 × 10−14 J 2mn 2(1.6749 × 10−27 kg) 3.32 × 10−14 J = 0.208 MeV. 1.602 × 10−13 J/MeV
This energy is well below the energy at which a neutron becomes relativistic, and hence justifies the use of classical mechanics. The neutron’s total energy is thus E = T + mn c2 = 0.207 MeV + 939 MeV ' mn c2 .
Modern Physics Concepts
213
Chap. 2
23. Repeat the previous problem for an electron trapped in the nucleus. HINT: relativistic expressions for momentum and kinetic energy must be used. Solution: −14 From the uncertainty principle, ∆p∆x > m ∼ h/(2π) so that for ∆x ' 10
∆p =
h 6.626 × 10−34 J s = = 1.05 × 10−20 J s m−1 . 2π∆x 2π × 10−14 m
For an electron with p ' ∆p = 1.05 × 10−20 J s m−1 p2 c2 = (1.05 × 10−20 J s m−1 )2 (3.00 × 108 m/s)2 = (3.15 × 10−12 J)2 = (19.7 MeV)2 . From the equation above Eq. (2.16) in the text, we see that p2 c2 = (mc2 )2 − (mo c2 )2 = E 2 −(mo c2 )2 . We use this relation to find the electron’s total energy E as p p E = p2 c2 + (me c2 )2 = 19.72 + 0.5112 MeV ' 20 MeV.
Since the electron’s total energy E is related to the kinetic energy T by E = T + me c2 = T + 0.511 MeV, in this problem the total energy is essentially the electron’s kinetic energy, i.e., E ' T . 24. The wavefunction for the electron in a hydrogen atom in its ground state (the 1s state for which n = 0, ` = 0, and m = 0 is spherically symmetric as shown in Fig. 2.14. For this state the wavefuntion is real and is given by ψ0 (r) = p
1 πa30
exp[−r/a0 ],
where ao = h2 o /(4π 2 me e2 ) ' 5.29 × 10−11 m. This quantity is the radius of the first Bohr orbit for hydrogen (see next chapter). Because of the spherical symmetry of ψo , dV in Eq. (2.40) is dV = 4πr 2 dr and the integral in Eq. (2.40) can be written as Z ∞ Z ∞ 4 ψ0 (r)ψ0∗ (r)4πdr = 3 r 2 e−αr dr, a0 0 0 where α ≡ 2/a0 . (a) Verify that the required normalization required by Eq. (2.40) is satisfied, i.e., the electron is somewhere in the space around the proton. (b) What is the probability the electron is found a radial distance r < a0 from the proton? Solution: (a) Integration by parts twice gives Z ∞ 4 4 2 4 a30 2 −αr r e dr = = = 1. a30 0 a30 α3 a30 4
214
Modern Physics Concepts
Chap. 2
(b) Replace upper limit in the above itegral by a0 . Then integration by parts twice gives Z a0 4 Prob{electron is inside r ≤ a0 } = 3 r 2 e−αr dr a0 0 2 4 a0 2a0 2 = 1 − 3 e−αa0 + 2 + 3 a0 α α α 3 3 a0 2a 2a3 4 + 0+ 0 = 1 − 3 e−2 a0 2 4 8 = 1 − 5e−2 = 0.323.
Thus the electron has a 32.3% of being at a radial distance less that a0 .
Chapter 3
Atomic/Nuclear Models
PROBLEMS 1. Estimate the wavelengths of the first three spectral lines in the Lyman spectral series for hydrogen. What energies (eV) do photons with these wavelengths have? Solution: From text Eq. (3.1) we have 1 λn→no
= 1.0967758 × 107
1 1 − 2 n2o n
m−1 .
(P3.1)
For the Lyman series, no = 1 < n. The first three wavelengths are found from Eq. (P3.1) as λ2→1 = 1.2156844 × 10−7 m
λ3→1 = 1.0257337 × 10−7 m λ4→1 = 0.9725476 × 10−7 m
The photon energy is found from E = hν = hc/λ. The quantity hc = (2.99792458 × 108 m/s)(4.1356673 × 10−15 eV s) = 1.2398418 × 10−6 eV m. Then E2→1 = E3→1 = E4→1 =
hc = 10.198714 eV λ2→1 hc λ3→1
= 12.087364 eV
hc = 12.748392 eV λ4→1 31
32
Atomic/Nuclear Models
Chap. 3
2. Consider an electron in the first Bohr orbit of a hydrogen atom. (a) What is the radius (in meters) of this orbit? (b) What is the total energy (in eV) of the electron in this orbit? (c) How much energy is required to ionize a hydrogen atom when the electron is in the ground state? Solution: (a) The orbital radius can be calculated from text Eq. (3.4). Here n = 1, Z = 1, h = 6.626×10−34 J s, me = 9.1094×10−31 kg, e = 1.6022×10−19 C, and o = 8.8541 × 10−12 F m−1 (= C2 J−1 m−1 ). Substitution into Eq. (3.4) yields r1 =
h2 o = 5.292 × 10−11 m. πme e2
(b) The energy of the n = 1 orbital electron is given by Eq. (3.5) as E1 = −
me e4 = −2.180 × 10−18 J = −13.60 eV. 82o h2
(c) To free this groundstate electron, we must provide +13.60 eV of energy to produce an electron with zero net energy. 3. What photon energy (eV) is required to excite the hydrogen electron in the innermost (ground state) Bohr orbit to the first excited orbit? Solution: For hydrogen (Z = 1), the energy of the electron in the orbit with quantum number n is given by Eq. (3.5) En = −
me e4 . 82o h2 n2
Data: h = 6.626 × 10−34 J s, me = 9.1094 × 10−31 kg, e = 1.6022 × 10−19 C, and o = 8.88541 × 10−12 F m−1 (= C2 J−1 m−1 ).
The energy needed to excite an electron from the ground state (n = 1) to the first excited state (n = 2) is thus me e4 1 1 Eexcite = E2 − E1 = − 2 2 − 2 = 1.635 × 10−18 J = 10.20 eV. 8o h 22 1
4. What is the de Broglie wavelength of the electron in the first Bohr orbit? Compare this wavelength to the circumference of the first Bohr orbit. What does this comparison reveal the standing wave in the first Bohr orbit? Solution: For a nonrelativistic electron, the wavelength is λ=
h h 6.626 × 10−34 J s = = = 3.326−10 m. p me v1 (9.109 × 10−31 kg)(2.187 × 106 m/s)
Atomic/Nuclear Models
33
Chap. 3
The length ` of the circumference of the first Bohr orbit is ` = 2πr1 = 2π(5.293 × 10−11 m) = 3.326 × 10−10 m. Hence we see the wavelgth equals the circumference. This means as the electron makes one orbit its wave is exactly in phase with the phase it started with. 5. Calculate the limiting (smallest) wavelength of the Lyman, Balmer, and Paschen series for the Bohr model of the hydrogen atom. Solution: From Eq. (3.17), with R∞ replaced by RH , the smallest wavelength is obtained as n → ∞. Then the limiting wavelengths are given by λmin =
n2o n2o = m. RH 10 967 758
Thus for the Lyman series (no = 1), λmin = 9.1176 × 10−8 m. For the Balmer series (no = 2) and λmin = 3.6471 × 10−7 m, and for the Paschen series (no = 3) and λmin = 8.20589 × 10−7 m, 6. Based on the nucleon distribution of Eq. (3.11), by what fraction does the density of the nucleus decrease between r = R − 2a and r = R + 2a? Solution:
From Eq. (3.11) we find ρo = 0.8808ρo 1 + exp(−2) ρo ρ(R + 2a) = = 0.1192ρo 1 + exp(+2) ρ(R − 2a) =
Thus the nuclear density drops at R + 2a to 100 × 0.1182/0.8808 = 13.5% of its value at R − 2a. 7. Using the liquid drop model, tabulate the nuclear binding energy and the various contributions to the binding energy for the nuclei 40 Ca and 208Pb. Solution: A BASIC program is used to evaluate the terms in text Eq. (3.16). A program listing and results are given below. ’ program to calculate BE(A)/A from the liquid drop model CLS fmt$ = " ### ### ### ####.### ####.### ####.### ####.### #.### #####.### ##.### " OPEN "35.out" FOR OUTPUT AS #1
34
Atomic/Nuclear Models
PRINT PRINT PRINT PRINT
"LIQUID DROP PREDICTION of BE vs A" " A Z N BEv BEs BEc #1, "LIQUID DROP PREDICTION of BE vs A" #1, " A Z N BEv BEs BEc
BEa BEa
Chap. 3
BEp
BE
BE/A "
BEp
BE
BE/A "
’ Case of Ca40 A = 40: Z = 20 BEv = 15.835 * A BEs = 18.33 * A ^ (2 / 3) BEc = .714 * Z * Z / A ^ (1 / 3) BEa = 23.2 * (A  2 * Z) ^ 2 / A BEp = 11.2 / SQR(A) BE = BEv  BEs  BEc  BEa + BEp PRINT USING fmt$; A; Z; A  Z; BEv; BEs; BEc; BEa; BEp; BE; BE / A PRINT #1, USING fmt$; A; Z; A  Z; BEv; BEs; BEc; BEa; BEp; BE; BE / A ’ Case of Pb208 A = 208: Z = 82 BEv = 15.835 * A BEs = 18.33 * A ^ (2 / 3) BEc = .714 * Z * Z / A ^ (1 / 3) BEa = 23.2 * (A  2 * Z) ^ 2 / A BEp = 11.2 / SQR(A) BE = BEv  BEs  BEc  BEa + BEp PRINT USING fmt$; A; Z; A  Z; BEv; BEs; BEc; BEa; BEp; BE; BE / A PRINT #1, USING fmt$; A; Z; A  Z; BEv; BEs; BEc; BEa; BEp; BE; BE / A CLOSE END  RESULTS LIQUID DROP PREDICTION of BE vs A A Z N BEv BEs BEc BEa 40 20 20 633.400 214.389 83.510 0.000 208 82 126 3293.680 643.484 810.286 215.938
BEp 1.771 0.777
BE 337.272 1624.748
BE/A 8.432 7.811
8. From the difference in mass of a hydrogen atom (Appendix B) to the mass of a proton and an electron (Table 1.5), estimate the binding energy of the electron in the hydrogen atom. Compare this to the ionization energy of the ground state electron as calculated by the Bohr model. What fraction of the total mass is lost as the electron binds to the proton? Solution: From the mass data of Appendix B, we find ∆mass = mp + me − M (11 H) = 1.0072764669 + 0.0005485799 − 1.0078250321 = 1.47 × 10−8 u. Notice that using all the significant figures available in the mass data only yields three significant figures for the mass deficit. This mass deficit corresponds to the binding energy of the electron to the proton and equals BEe = (1.47 × 10−8 u)(931.5 × 106 eV/u) = 13.7 eV.
Atomic/Nuclear Models
35
Chap. 3
From text Eq. (3.5) for hydrogen (Z = 1), the energy of the electron in the ground state (n = 1) is E1 = −
me e4 = −13.606 eV 82o h2
which is the negative of the electron binding energy and in good agreement with the result from the mass deficit. The percent of mass lost in binding the electron to the proton in a hydrogen atom is % mass lost = 100 ×
∆mass 1.47 × 10−8 = 100 × = 1.46 × 10−6 %. 1 1.007 M (1 H)
9. Using the liquid drop model, plot on the same graph, as a function of A, in units of MeV/nucleon (a) the bulk or volume binding energy per nucleon, (b) the negative of the surface binding energy per nucleon, (c) the negative of the asymmetry contribution per nucleon, (d) the negative of the Coulombic contribution per nucleon, and (e) the total binding energy per nucleon ignoring the pairing term. For a given A value, use Z determined from Eq. (3.18) for the most stable member of the isobar. Solution: Program for the liquid drop model calculations and graph of results are given below. ’ Program to calculate BE(A)/A from the liquid drop model. Problem 37 CLS fmt$ = " ### ###.# ###.# ##.### ##.### ##.### ##.### ###.###" OPEN "37.out" FOR OUTPUT AS #1 PRINT "LIQUID DROP PREDICTION of BE/A vs A" PRINT " A Z N BEv/A BEs/A BEc/A BEa/A BE/A" PRINT #1, "LIQUID DROP PREDICTION of BE/A vs A" PRINT #1, " A Z N BEv/A BEs/A BEc/A BEa/A BE/A" FOR A = 2 TO 250 STEP 2 denom = 1 + .00769397# * (A ^ (2 / 3)) Z = (A / 2) * 1.013958 / denom BEvA = 15.835 BEsA = 18.33 / A ^ (1 / 3) BEcA = .714 * Z * Z / A ^ (4 / 3) BEaA = 23.2 * (A  2 * Z) ^ 2 / A ^ 2 BEA = BEvA  BEsA  BEcA  BEaA PRINT USING fmt$; A; Z; A  Z; BEvA; BEsA; BEcA; BEaA; BEA PRINT #1, USING fmt$; A; Z; A  Z; BEvA; BEsA; BEcA; BEaA; BEA NEXT CLOSE END  RESULTS LIQUID DROP PREDICTION of BE/A vs A A Z N BEv/A BEs/A BEc/A 2 1.0 1.0 15.835 14.549 0.284
BEa/A 0.000
BE/A 1.002
36
Atomic/Nuclear Models
4 6 8 10 12 14 16 18 20 22 240 242 244 246 248 250
2.0 3.0 3.9 4.9 5.8 6.8 7.7 8.7 9.6 10.5 . . 93.8 94.5 95.1 95.8 96.4 97.1
Chap. 3
2.0 15.835 11.547 0.445 0.001 3.842 3.0 15.835 10.087 0.576 0.003 5.168 4.1 15.835 9.165 0.691 0.006 5.973 5.1 15.835 8.508 0.794 0.010 6.523 6.2 15.835 8.006 0.889 0.015 6.925 7.2 15.835 7.605 0.977 0.020 7.233 8.3 15.835 7.274 1.059 0.026 7.476 9.3 15.835 6.994 1.137 0.032 7.672 10.4 15.835 6.753 1.211 0.038 7.833 11.5 15.835 6.542 1.281 0.045 7.967 . . . . . . . . . . . . . . . . . . . . . . 146.2 15.835 2.950 4.212 1.106 7.567 147.5 15.835 2.941 4.225 1.116 7.553 148.9 15.835 2.933 4.237 1.126 7.538 150.2 15.835 2.925 4.250 1.136 7.524 151.6 15.835 2.918 4.262 1.146 7.509 152.9 15.835 2.910 4.274 1.156 7.495
10. In radioactive beta decay, the number of nucleons A remains constant although the individual number of neutrons and protons change. Members of a such betadecay chain are isobars with nearly equal masses. Using the atomic mass data in Appendix B, plot the mass difference [70 − A Z X] (in u) of the nuclei 70 70 70 70 70 70 70 versus Z for the isobar chain 70 36 Kr, 35 Br, 34 Se, 33 As, 32 Ge, 31 Ga, 30 Zn, 29 Cu, 70 70 28 Ni, and 27 Co. Compare the position of maximum nuclear stability with that predicted by Eq. (3.18). Solution: The liquid drop model gives the most stable proton number for a given A from Eq. (3.18), namely, A 1 + (mn − mp )c2 /(4aa ) . Z(A) = 2 1 + ac A2/3 /(4aa )
From the text we find 4aa = 92.80 MeV, ac = 0.714 MeV, and from Table 1.5 (mn − mp )c2 = 939.56533 − 938.27200 = 1.2933 MeV. Thus for A = 70 we find 70 1 + 1.2933/92.8 Z(70) = = 31.39. 2 1 + 0.714(70)2/3/92.8
Atomic/Nuclear Models
37
Chap. 3
The figure below shows Ap. B masses of the isotoopes that are members of the isobar A = 70 as well as masses calculated by the liquid drop model. In this 2 model the mass of an atom is calculated as A Z X = Zmp +N mn +Zme −BEld /c . Here BEld is the nuclear binding energy as calculated by the term in braces in Eq. (3.16). Nuclear mass from liquid drop A Z N BE(MeV) mn(MeV) 70 27 43 584.634 65150.020 70 28 42 597.668 65135.691 70 29 41 602.350 65129.715 70 30 40 609.389 65121.383 70 31 39 608.075 65121.402 70 32 38 609.117 65119.066 70 33 37 601.808 65125.082 70 34 36 596.855 65128.742 70 35 35 583.549 65140.758 70 36 34 572.600 65150.410
model for ISOBAR A=70 mn(u) Matom(u) Matom70 69.9414 69.9562 0.04376 69.9260 69.9414 0.05860 69.9196 69.9355 0.06447 69.9107 69.9271 0.07286 69.9107 69.9277 0.07229 69.9082 69.9258 0.07425 69.9147 69.9328 0.06724 69.9186 69.9372 0.06276 69.9315 69.9507 0.04932 69.9418 69.9616 0.03841
11. Equation (3.18) can not be solved analytically to give A(Z) that produces the isotope (fixed Z) with the smallest mass. But it can be put into form that can be solved iteratively, i.e., Ai = f(Z, Ai−1 ),
i = 1, 2, 2, . . .
with A0 = 0.
Find A when Z = 20. Solution: Rearrange Eq. (3.18) to give A = 2Z
1 + (mn − mp )c2 /(4aa ) . 1 + ac /(4ac)
Because ac = 0.714 MeV, aa = 23.20 MeV, and (mn − mp )c2 = 1.293 MeV, the iteration scheme for the above result becomes 2/3
Ai = 2Z
1 + 0.00769Ai−1 . 1.01393
For calcium (Z = 20) and with A0 = 0, one obtains the following: A1 = 39.450, A2 = 42.968, A3 = 43.174, A4 = 43.186, A5 = 43.187, and A6 = 43.187.
38
Atomic/Nuclear Models
Chap. 3
12. From the data in Appendix B, plot the mass parabolas for nuclei in the isobar with A = 184. Show on the plot the neutron number N for maximum stability as calculated from Eq. (3.18). Solution: From Appendix B the following data is extracted. N 113 112 111 110 109 108 107 106 105 104 103 102
Z 71 72 73 74 75 76 77 78 79 80 81 82
M183.9 (mu) 61.170 55.450 54.009 50.9326 52.524 52.491 57.390 59.900 67.470 71.900 81.760 88.200
A=184 Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb
stable stable
Finally, from Eq. (3.18) it is found that Z(114) = 74.691 so that the most stable nuclide is one with N = A − Z = 109.309. The resulting plot is shown below.
Chapter 4
Nuclear Energetics
PROBLEMS 1. Complete the following nuclear reactions based on the conservation of nucleons: (a) (b) (c) (d)
238 1 92 U + 0 n −→ (?) 14 1 1 7 N + 0 n −→ (?) + 1 H 226 4 88 Ra −→ (?) + 2 He 4 (?) −→ 230 90 Th + 2 He
Solution: (a)
238 92 U
+ 10 n −→ 239 92 U
(b)
14 7N
(c)
226 88 Ra
(d)
234 92 U
1 + 10 n −→ 14 6 C + 1H
4 −→ 222 86 Rn + 2 He
4 −→ 230 90 Th + 2 He
2. What is the rest mass energy equivalent in MeV of 1 atomic mass unit as calculated directly from E = mc2 ? Solution: From Table 1.5 1 u= 1.660538 × 10−27 kg. Then E = mc2 = (1.660538 × 10−27 kg)(2.997925 × 108 m/s)2 = 1.492418 × 10−10 J × (1.602176 × 10−19J/eV)−1
= 9.31494 × 108 eV = 931.494 MeV
(P4.1)
This conversion factor of 931.5 MeV/u makes it very easy to calculation nuclear reaction energies when the masses of atoms are given in atomic mass units as they are in Appendix B.
41
42
Nuclear Energetics
Chap. 4
3. Determine the binding energy (in MeV) per nucleon for the nuclides: (a) 235 (b) 178 O, (c) 56 26 Fe, and (d) 92 U.
16 8 O,
Solution: We use text Eq. (4.12) with masses expressed in atomic mass units so that, per nucleon, 1 A BE(A Z X)/A = [ZM (1 H) + (A − Z)mn − M (Z X)]/A (u) × 931.5 (MeV/u).
Use the atomic mass tables of Ap. B to find masses. The results are (a) BE(168 O)/16 = [8 × 1.007825 + 8 × 1.008664 − 15.994914]/16 × 931.5 = 7.976 MeV. (b) Similarly, BE(178 O)/17 = 7.751 MeV. (c) Similarly, BE(56 26 Fe)/56 = 8.790 MeV. (d) Similarly, BE(235 92 U)/235 = 7.951 MeV.
4. Calculate the binding energy per nucleon and the neutron separation energy for 168 O and 178 O. Solution: (a) Binding Energies: From Eq. (4.12) and the data in Ap. B we have BE(168 O) = [8M (11 H) + 8mn − M (168 O)]/16 (u) × 931.49 (MeV/u) = 7.976 MeV. BE(178 O) = [8M (11 H) + 9mn − M (178 O)]/17 (u) × 931.49 (MeV/u) = 7.751 MeV.
(b) From Eq. (4.13) for the neutron separation energy and the data in Ap. B we have Sn (168 O) = [M (158 O) + mn − M (168 O)] (u) × 931.49 (MeV/u) = 15.66 MeV.
Sn (178 O)
= [M (168 O) + mn − M (178 O)] (u) × 931.49 (MeV/u) = 4.143 MeV.
Nuclear Energetics
43
Chap. 4
5. What is the net energy released (in MeV) for each of the following fusion reactions? (a) 21 H + 21 H −→ 32 He + 10 n and (b) 21 H + 31 H −→ 42 He + 10 n Solution:
(a) The energy released is the Qvalue of the reactions. Q = 2M (21 H) − M (32 He) − mn c2 = {2 × 2.0141018 − 3.0160293 − 1.0086649} (u) × 931.5 (MeV/u) = 3.27 MeV
(b) Similarly Q = M (21 H) + M (31 H) − M (42 He) − mn c2 = {2.0141018 + 3.0160493 − 4.0026032 − 1.0086649} × 931.5 = 17.59 MeV
6. Calculate the binding energy and the binding energy per nucleon for for 235 92 U. What is the significance of these results? Solution: From Eq. (4.12) the BE for
56 26 Fe
56 26 Fe
and
is found as
1 56 BE(56 26 Fe) = [26M (1 H) + (56 − 26)mn − M (26 Fe)] (u) × 931.5 (MeV/u) = [26(1.007825030 + 30(1.00866492) − 55.9349421] × 931.5 = 492.26 MeV.
From this result BE/A = 492.26/56 = 8.790 MeV. Similarly for
235 92 U
1 235 BE(235 92 U) = [92M (1 H) + (235 − 92)mn − M (92 U)] (u) × 931.5 (MeV/u)
= [92(1.007825030 + 143(1.00866492) − 235.0439231] × 931.5 = 1783.9 MeV.
From this result BE/A = 1783.9/235 = 7.592 MeV. This simple calculation shows that the nucleons in 56 26 Fe are more tightly boound thaat tthose in B235 U. Thus by splitting (or fissioning) a heavy atom like 92B 235 U the nucleons become more tightly bound. The differences in the binding 92 energies is emitted exoergically.
44
Nuclear Energetics
Chap. 4
7. Generally, energies of chemical reactions can not be calculated by finding the difference between the masses of the reactants and the products because the mass must be known to 10 or more significant figures. However, the mass of the proton and hydrogen atom are known to 10 significant figures. Estimate the binding energy of the electron BEe in the 11 H atom and compare this result to what the Bohr model predicts. Discuss this comparison. Solution: The binding energy reaction can be written as 1 1p
+
0 −1 e
−→ 11 H + BEe .
Use mass values found in Table A.1 and Appendix B. Then the binding energy is estimated as BEe = [mp + me − M (11 H)] (u) × 931.5 (MeV/u) = 1.0072764669 + 5.48579909 × 10−4 − 1.0078250321 × 931.5 = 1.3887 × 10−5 MeV = 13.887 eV.
(P4.2)
From the Bohr model BEe is found to be 13.606 eV. Although the estimate of 13.887 is correct to 2 significant figures, to get a better estimate the proton and hydrogen atomic masses must be known to at least 13 significant figures, an accuracy beyond present day technology.
8. What is (a) the BE of 3 He, and (b) the neutron separation energy? Solution: (a) The BE(32 He) is BE(32 He) = [211 H + mn − 32 He] (u) × 931.5 (MeV/u) = [2(1.00982503) + 1.00866492 − 3.01602931]931.5 = 7.718 MeV. (b) The netron separation energy from Eq. (4.13) is Sn (32 He) = M (22 He) + mn − M (32 He). But 22 He doesn’t exist. By removing the neutron, two 11 H atoms are created and the atom is totally dismantled. So the neutron separation energy for this isotope is the same as the BE!
Nuclear Energetics
45
Chap. 4
9. Verify Eq. (4.15) on the basis of the definition of the binding energy. Solution: The proton separation reaction may be written as A−1 Z−1 X
+ 11 p −→ A Z Y.
The energy released in this reaction Sp is the energy required to separate a proton from the nucleus A Z Y. This energy is given in terms of the change in nuclear masses, i.e., 2 A Sp = m(A−1 Z−1 X) + mp − m(Z Y) c Use Eq. (4.8) to express the nuclear masses in terms of atomic masses as e e 2 2 1 Sp = [M (A−1 Z−1 X) − (Z − 1)me + BEZ−1 /c ] + [M (1 H) − me + BE1 /c ] e 2 2 −[M (A Z Y) − Zme + BEZ /c ] c e e e 2 1 A = [M (A−1 Z−1 X) + M (1 H) − M (Z Y)]c + [BEZ−1 + BE1 − BEZ ]
The electron binding energies BEe in the last term tend to cancel and any small nonzero value is negligible compared to the nuclear binding energies. Finally, A using Eq. (4.12) to express M (A−1 Z−1 X) and M (Z Y) in terms of the hydrogen atom’s mass, i.e., 2 2 2 1 Sp = [(Z − 1)M (11 H) + (A − Z)mn − BE(A−1 Z−1 X)/c ]c + [M (1 H)]c 2 2 −[ZM (11 H) + (A − Z)mn − BE(A Z Y)/c ]c
A−1 = BE(A Z Y) − BE(Z−1 X).
which agrees with Eq. (4.15). 10. A nuclear scientist attempts to perform experiments on the stable nuclide 56 26 Fe. Determine the energy (in MeV) the scientist will need to 1. remove a single neutron. 2. remove a single proton. 3. completely dismantle the nucleus into its individual nucleons. 4. fission it symmetrically into two identical lighter nuclides
28 13 Al.
Solution: Atomic masses from Appendix B are used in the solution. (a) The energy needed is the neutron separation energy. From Eq. (4.13) we obtain 2 56 Sn = M (55 26 Fe) + mn − M (26 Fe) c = 11.20 MeV.
(b) The energy needed is the proton separation energy. From Eq. (4.15) we obtain 2 1 56 Sp = M (55 25 Mn) + M (1 H) − M (26 Fe) c = 10.18 MeV.
46
Nuclear Energetics
Chap. 4
(c) The energy needed is the binding energy of 56 26 Fe. From Eq. (4.12) we obtain 2 1 56 BE(56 26 Fe) = 26M (1 H) + 30mn − M (26 Fe) c = 492.25 MeV.
28 (d) Symmetric fission would give 56 26 Fe → 2 [13Al]. The energy required is 2 28 Ef = M (56 26 Fe) − 2M (13 Al) c = 26.90 MeV.
11. Write formulas for the Qvalues of the reactions shown in Section 4.4. With these formulas, evaluate the Qvalues. Solution: For the binary reaction x + X → Y + y, the Q value is given by Q = [(Mx + MX ) − (My + MY )]c2 . Results are summarized below. Reaction
Mx
MX
MY
My
Q (MeV)
(α,p) (α,n) (γ,n) (p,γ) (γ,αn) (n,p)
4.002603 4.002603 0.000000 1.007825 0.000000 1.008665
14.003074 9.012182 2.014102 7.016004 16.999132 15.994915
16.999132 12.000000 1.007825 8.005305 12.000000 16.006101
1.007825 1.008665 1.008665 0.000000 5.011268 1.007825
−1.19 5.70 −2.22 17.26 −11.31 −9.64
12. What is the Qvalue (in MeV) for each of the following possible nuclear reactions? Which are exothermic and which are endothermic? 10 5B + γ 9 1 5B + 0n 9 Be + 1 p 4 1 1 9 1 p + 4 Be −→ 8 2 Be + 4 1H 7 3 4 Be + 1 H 6 4 3 Li + 2 He Solution:
Consider the 9 Be(p,γ)10 B reaction. The Qvalue is 2 Q = M (11 H) + M (94 Be) − M (10 5 B) c
= {1.00782503 + 9.0121821 − 10.0129370} (u) × 931.5 (MeV/u) = 6.586 MeV
Nuclear Energetics
47
Chap. 4
The other reactions are treated similarly. The results are tabulated below. Reaction 9
10
Be(p,γ) B Be(p,n)9 B 9 Be(p,p)9 Be 9 Be(p,d)8 Be 9 Be(p,t)7 Be 9 Be(p,α)6 Li 9
Qvalue (MeV)
type
6.586 −1.850 0.0 0.559 −12.083 2.125
exoergic endoergic exoergic endoergic exoergic
13. Neutron irradiation of 6 Li can produce the following reactions. 7 3 Li + γ 6 1 3 Li + 0 n 1 6 6 1 0 n + 3 Li −→ 2 He + 1 p 5 2 2 He + 1 H 3 4 1 H + 2 He What is the Qvalue (in MeV) for each reaction?
Solution: Results are summarized in the table below. Reaction 6
Li(n,γ)7 Li Li(n,n)6 Li 6 Li(n,p)6 He 6 Li(n,d)5 He 6 Li(n,t)4 He 6
Qvalue (MeV)
type
7.250 0.0 −2.725 −2.361 4.783
exoergic endoergic endoergic exoergic
14. Calculate the Qvalues for the following two beta radioactive decays. 22 0 38 38 0 (a) 22 11 Na −→ 10 Ne + +1 e + ν and (b) 17 Cl −→ 18 Ar + −1 e + ν. Solution:
Because in beta and positron decay the number of protons in the parent and daughter are different from the number of electrons in neutral atoms of the parent and daughter are different. This change in electron number must be accounted for in the decay reactions. (a) This positron reaction, in terms of neutral atoms, is written as 0 22 22 0 11 Na −→ 10 Ne + −1 e + +1 e + ν. The Qvalue is then 2 22 Q = M (22 11 Na) − M (10 Ne) − 2me c
= {21.994437 − 21.991386 − 2 × 0.0005486} (u) × 931.5 (MeV/u)
= 1.820 MeV
48
Nuclear Energetics
Chap. 4
(b) This beta minus reaction, in terms of neutral atoms, is written as 38 0 38 0 17 Cl + −1 e −→ 18 Ar + −1 e + ν. The Qvalue is then 2 38 Q = M (38 17 Cl) − M (18 Ar) c = {37.968011 − 37.962732} (u) × 931.5 (MeV/u) = 4.917 MeV 15. Reactions employed in cyclotron production of radionuclides for PET scanning are listed in Table 14.3. Select at least one reaction and compute the Qvalue for the reaction. Solution The reaction 188 O(11 p,10 n)189 F is imbalanced in charge, as discussed in Sec. 4.6. There are eight electrons in the reactants and nine in the products. An electron from the target milieu must enter into the reaction. The Qvalue may thus be computed as Q = [M (11 H) + M (188 O) − mn − M (189 F)]c2 = [1.0078250321 + 17.9991604 −1.0086649233 − 18.0009377] × 931.5 = −2.438MeV The reaction is endothermic, so the necessary reaction energy must be supplied by kinetic energy of the proton, generally 11 MeV or greater. 60 − 16. The radioactive isotope 60 27 Co decays to an excited state of 27 Ni by β emission. The reaction is 60 60 + 0 ¯, 27 Co −→ [28 Ni*] −1 e + ν
where the superscript + indicates that the Ni atom is produced as a positive ion. The energy of excitation E ∗ = 2.205 MeV. Calculate the Qvalue of this reaction. Solution: First add an electron to both sides of the reaction to neutralize the ion. Thus the reaction becomes 60 27 Co
+
0 −1 e
−→ 60 28 Ni* +
0 −1 e +
ν¯,
or
60 27 Co
−→ 60 ¯. 28 Ni* + ν
Then the Qvalue of this last reaction, after ignoring the negligible mass of the antineutrino and using Eq. (4.28), is found as 60 Q = [M (60 27 Co) − M (28 Ni*)] (u) × 931.5 (MeV/u)
60 = [M (60 27 Co) − M (28 Ni)] (u) × 931.5 (MeV/u) − 2.505 MeV
= [59.933822 − 59.930791]931.5 − 2.505 = 0.3184 MeV.
Chapter 5
Radioactivity
PROBLEMS 1. (a) Identitify the type of radioactive decay and the unknown in each of the following radioactive decay reactions, and (b) calculate the total kinetic energy of all the decay products. 1. 2. 3. 4. 5. 6. 7.
210 4 84 Po −→ (?) + 2 He 38 0 16 S −→ (?) + −1 e + ν e 0 (?) −→ 27 13 Al + +1 e + νe 145 145 62 Sm −→ 61 Pm + (?) ∗ 137 1 (E ∗ = 6.71) MeV 54 Xe −→ (?) + 0 n ∗ 108 107 (E ∗ = 3.4 MeV) 52 Te −→ 51 Sb + (?) ∗ 60 0 (E ∗ = 0.125 MeV; 28 Ni −→ (?) + −1 e
BEeK = 8.33 keV)
Solution: (a)
α decay
210 84 Po
4 −→ 206 82 Pb + 2 He
β − decay
38 16 S
β + decay
27 14 Si
electron capture
145 62 Sm
neutron decay
∗ 137 54 Xe
proton decay
108 52 Te
∗ 1 −→ 107 51 Sb + 1 p
internal conversion
60 ∗ 28 Ni
−→ 60 28 Ni +
−→ 38 17 Cl + −→ 27 13 Al +
0 −1 e
+ νe
0 +1 e
+ νe
−→ 145 61 Pm + νe 1 −→ 136 54 Xe + 0 n
0 −1 e
(b) The total kinetic energy released is the Qvalue of the decay reaction. 1. From Eq. (5.7) and the relative masses of atoms in Appendix B, it is found that Qα = 5.407 MeV. 2. From Eq. 5.14 and the masses in Appendix B, it is found that Qβ− = 2.937 MeV. 3. From Eq. (5.18) and the masses in Appendix B, it is found that Qβ+ = 5.835 MeV. 51
52
Radioactivity
Chap. 5
4. Similarly, from Eq. (5.22), it is found QEC = 6.194 MeV. 5. Unlike Eq. (5.24) for neutron decay that leaves the daughter in an excited state, here the parent is in an excited state and decays to the ground state of the daughter. Thus, Eq. (5.24) is modified as follows. Qn /c2 = M (137 Xe∗ ) − M (136 Xe) = [M (137Xe) + (2.4 MeV)/c2 ] − M (136 Xe) From this result it is found that Qn = 2.684 MeV. 6. From Eq. (5.25), one finds Qp = 1.086 MeV. 7. From Eq. (5.29), QIC = (E ∗ − BEeK ) = 125 − 8.3 = 124.2 keV. 2. Consider a stationary nucleus of mass mn in an excited state with energy E ∗ above the ground state. When this nucleus decays to the ground state by gamma decay, the emitted photon has an energy Eγ . (a) By considering the conservation of both energy and momentum of the decay reaction explain why Eγ < E ∗ . (b) Show that the two energies are related by Eγ = mn c
2
(r
2E ∗ 1+ −1 mn c2
)
'E
∗
E∗ 1− . 2mn c2
(c) Use an explicit example to verify that the difference between E ∗ and Eγ is for all practical purposes negligible. Solution: (a) If the excited nucleus is initially at rest, the products of the decay must have zero net linear momentum, i.e., the photon and the groundstate daughter nucleus must travel in opposite directions each with the same amount of linear momentum. The Qvalue of the decay reaction is E ∗ and must be equal to the sums of the kinetic energies of the photon and groundstate nucleus. Hence it follows that Eγ < E ∗ , the difference equal to the kinetic energy of the recoil groundstate nucleus. (b) Conservation of total energy requires mn c2 + E ∗ = mn c2 + En + Eγ ,
(P5.1)
and conservation of linear momentum (treating the recoil nucleus as a classical particle) requires 0 = Eγ /c −
p 2mn En .
Substitute Eq. (P5.1) into Eq. (P5.2) to eliminate En yields Eγ2 + 2mn c2 Eγ − 2mn c2 E ∗ = 0.
(P5.2)
Radioactivity
53
Chap. 5
Solving this quadratic equation for Eγ produces ( ) r 2E ∗ 2 Eγ = mn c −1 ± 1 + . mn c2 Only the + sign yields a positive real value for Eγ so that r ∗ 2E −1 Eγ = mn c 1+ mn c2 2
Since E ∗ is typically much less than mn√ c2 ' thousands of MeV, then ∗ 2 defining ≡ (2E )/(mn c ) 3.179 Ex > 2.755 none
7.250 0 0.454 0.394 4.783
2. For each of the following possible reactions, all of which create the compound nucleus 10 B, 10 B+γ 6 Li + α 1 p + 9 Be −→ 10 B∗ −→ 8 Be + d 9 B+n 5 Li + α + n
calculate (a) the Qvalue, (b) the kinematic threshold energy of the proton, (c) the threshold energy of the proton for the reaction, and (d) the minimum kinetic energy of the products. Summarize your calculations in a table. Solution: For a binary, twoproduct reaction x +X −→ y +Y we use the following results from Ch. 6. (a) The Qvalue is obtained from Eq. (6.6), Q = (Mx + MX − My − MY )c2 , and from the atomic masses in Ap. B. NOTE: for the last reaction, a third product mass must be subtracted from the righthand side of the Qvalue equation. (b) The kinematic and Coulombic threshold energies are obtained from Eq. (6.15) (needed when Q < 0) and from Eq. (6.19) (needed when x and X are both charged particles). mx Zx ZX Exth ' − 1 + Q and ExC ' 1.2 1/3 . 1/3 mX Ax + AX (c) The minimum KE of the products is (Ey + EY )min = Q + (Exth )min . Results are tabulated below. Reaction 9
10
Be(p,γ) B Be(p,α)6 Li 9 Be(p,d)8 Be 9 Be(p,n)9 B 9 Be(p,αn)5 Li 9
Q (MeV)
ExC
Exth
6.586 2.125 0.559 −1.850 −3.541
1.558 1.558 1.558 1.558 1.558
0 0 0 2.056 3.934
condition Ex Ex Ex Ex Ex
> 1.558 > 1.558 > 1.558 > 2.056 > 3.934
(Ey + E Y )min 8.144 3.683 2.117 0.206 0.393
Binary Nuclear Reactions
63
Chap. 6
3. Consider the following reactions caused by tritons, nuclei of 3 H, interacting with 16 O to produce the compound nucleus 19 F 18 F+n 17 O+d 3 H + 16 O −→ 19 F∗ −→ 18 O+p 16 N + 3 He
For each of these reactions calculate (a) the Qvalue, (b) the kinematic threshold energy of the triton, (c) the threshold energy of the triton for the reaction, and (d) the minimum kinetic energy of the products. Summarize your calculations in a table. Solution: For a binary, twoproduct reaction x +X −→ y +Y we use the following results from Ch. 6. (a) The Qvalue is obtained from Eq. (6.6), Q = (Mx + MX − My − MY )c2 , and the atomic masses in Ap. B. (b) The kinematic and Coulombic threshold energies are obtained from Eq. (6.15) (needed when Q < 0) and from Eq. (6.19) (needed when x and X are both charged particles). mx Zx ZX Exth ' − 1 + Q and ExC ' 1.2 1/3 . 1/3 mX Ax + AX (c) The minimum KE of the products is (Ey + EY )min = Q + (Exth )min .
Results are tabulated below. Reaction 16
Q (MeV)
18
O(t,n) F 1.268 O(t,d)17 O −2.114 16 O(t,p)18 O 3.706 16 O(t,3 He)16 N −10.402 16
ExC
Exth
condition
(Ey + E Y )min
2.423 2.423 2.423 2.423
0 2.510 0 12.352
Ex > 2.423 Ex > 2.510 Ex > 2.423 Ex > 12.352
3.691 0.396 6.192 1.950
4. The compound nucleus 15 N∗ can be formed in many ways. Shown below are a few of these production paths and some of the ways 15 N∗ decays. Calculate the thresshold condition for each reaction and the minimum kinetic energy of the products. 14 N + 1n 4 11 15 ∗ He + B −→ N −→ 14 C + 1H 3
H + 12 C −→ 15 N∗ −→ 14 C + 1 H 11 B + 4 He 2 13 15 ∗ H + C −→ N −→ 12 C + 3H
2
H + 13 C −→ 15 N∗ −→ 14 N + 1 n 1 n + 14 N −→ 15 N∗ −→ 11 B + 4 He
64
Binary Nuclear Reactions
Chap. 6
Solution: Use the same techniques as describe in Example 6.1 to obtain the following results. Reaction Path 11
B(α,n)14 N B(α,p)14 C 12 C(t,p)14 C 13 C(d,α)11 B 13 C(d,t)12 C 14 C(p,n)14 N 14 N(n,α)11 B 11
Qvalue (MeV)
ExC (MeV)
Exth (MeV)
Reaction Condition
min(Ey + EY ) (MeV)
0.157 0.7839 4.64 5.17 1.31 0.626 0.157
3.15 3.15 1.93 1.99 1.99 2.11 0.0
0 0 0 0 0 0.671 0.168
Ex > ExC Ex > ExC Ex > ExC Ex > ExC Ex > ExC Ex > ExC Ex > Exth
3.307 3.933 6.57 7.16 3.30 1.484 0.011
5. Start with Eq. (6.17) and derive Eq. (6.15). Solution: Substitution of Eq. (6.15) into Eq. (6.17) gives WC −
e2 Zx ZX . 1/3 4πo Ro A1/3 x +A X
−19
From Table 1.5, e = 1.602 × 10 C and o = 8.854 × 10−12 F m−1 . From −15 Eq. (1.7), Ro ' 1.2 × 10 m. Thus,
or
e2 (1.602 × 10−19)2 = = 1.922 × 10−13 J, 4πo Ro 4π(8.854 × 10−12)(1.2 × 10−15 ) e2 = (1.922 × 10−13 J)/(1.602 × 10−13 J/eV) = 1.20 MeV. 4πo Ro
6. Verify the values reported in the table of Example 6.1. Solution: (a) For 2 H + 13 C −→ 3 H + 12 C we have: Q/c2 = M (2 H) + M (13 C) − M (3 H) − M (12 C) = 2.01410178 + 13.00335484 − 3.01606927 − 12.00000000 = 0.0014074 u Q = 0.0014074 u × 931.5MeV/u = 1.311 MeV. ExC = 1.2 Exth
Zx ZX 1/3 AX
+
1/3 AX
= 1.2
1×6 = 1.992 MeV 21/3 + 131/3
= 0. min(Ey + EY ) = Q + (Exth )min = 1.311 + 1.992 = 3.305 MeV.
Binary Nuclear Reactions
65
Chap. 6
(b) For 1 H + 14 C −→ 1 n + 14 N we have: Q/c2 = M (1 H) + M (14 C) − M (1 n) − M (14 N) = 1.00782503 + 14.00324199 − 1.00866492 − 14.00307400 = −0.006719 u Q = 0.006719 u × 931.5MeV/u = −0.6159 MeV.
1×6 = 2.111 MeV 11/3 + 141/3 mx 1 = − 1+0 = 1+ = 0.6706 MeV. mX 14
ExC = 1.2 Exth
Zx ZX
1/3 AX
1/3 + AX
= 1.2
min(Ey + EY ) = Q + Exth = −0.6715 + 2.111 = 1.485 MeV. (c) For 1 n + 14 N −→ 4 He + 11 B we have: Q/c2 = M (1 n) + M (14 N) − M (4 He) − M (11 B)
= 1.00866492 + 14.00307400 − 4.00260325 − 11.00930550 = −0.00016983 u Q = 0.00016983 u × 931.5MeV/u = −0.1582 MeV. ExC = 1.2 Exth
Zx ZX
= 0. 1/3 + AX mx 1 = − 1+0 = 1+ = 0.1695 MeV. mX 14 1/3 AX
min(Ey + EY ) = Q + Exth = −0.1582 + 0.1695 = 0.0113 MeV. 7. Derive Eq. (6.21) from Eq. (6.11). Solution: Begin with the general result of Eq. (6.11), namely p Ey = ±
s
mx my Ex cos θy (my + mY )2
s
mx my Ex mY − mx mY Q 2θ + cos E + . (P6.1) y x (my + mY )2 (my + mY ) (my + mY )
For a heavy particle scattering elastically from an electron at rest, we identify particles X and y in Eq. (P6.1) as the electron, so mX = my = me , Ey = Ee (the recoil electron energy), θs = θe , mx = mY = M (the mass of the heavy particle), and Ex = EM (the kinetic energy of the incident heavy particle). For this scattering process, there is no change in the rest masses of the reactants,
66
Binary Nuclear Reactions
Chap. 6
i.e., Q = 0, and only the + sign of the ± choice in the above equation has physical significance. Substitution of these variables into Eq. (P6.1) gives s s p M me EM M me EM Ee = cos θe + cos2 θe + 0 2 (me + M ) (me + M )2 =
p 2 M me EM cos θe . (M + me )
Squaring this result and using M >> me , we find that the recoil energy of the electron is me Ee = 4 EM cos2 θe . M 8. A 2MeV neutron is scattered elastically by 12 C through an angle of 45 degrees. What is the scattered neutron’s energy? Solution: From Eq. (6.25) n√ o2 p 1 2 − 1 + cos2 θ ) E cos θ + E(A s s (A + 1)2
E0 =
o2 p 1 n√ 2 (45)) 2 cos(45) + 2(144 − 1 + cos = 1.90 MeV. 132
=
9. The first nuclear excited state of 126 C is 4.439 MeV above the ground state. (a) What is the Qvalue for neutron inelastic scattering that leaves 12 C in this excited state? (b) What is the threshold energy for this scattering reaction? (c) What is the kinetic energy of an 8MeV neutron scattered inelastically from this level at 45 degrees? Solution: (a) The Qvalue is computed as Q = {M (126 C) − M (126 C∗ )}c2 = {M (126 C) − [M (126 C) + E ∗ /c2 ]}c2 = −E ∗ = −4.439 MeV. (b) From Eq. (6.15) 1 mn Q =− 1+ (−4.439) = 4.81 MeV. Enth = − 1 + MC 12 (c) From Eq. (6.25) E0 =
n√ o2 p 1 2 − 1 + cos2 θ ) + A(A + 1)Q E cos θ + E(A s s (A + 1)2
Binary Nuclear Reactions
=
67
Chap. 6
o2 p 1 n√ 2 (45)) + 12(13)(−4.439) 8 cos(45) + 8(144 − 1 + cos 132
= 3.22 MeV. 10. An important radionuclide produced in watercooled nuclear reactors is 16 N which has a halflife of 7.13 s and emits very energetic gamma rays of 6.1 and 7.1 MeV. This nuclide is produced by the endoergic reaction 16 O(n, p)16 N. What is the minimum energy of the neutron needed to produce 16 N? Solution: From Eq. (6.6), and the atomic masses in Ap. B we find for the reaction
16
O(n, p)16 N
Q = [M (168 O) + mn − M (167 N) − M (11 H)]c2 = −9.638 MeV The neutron threshold energy for this reaction is found from Eq. (6.15) 1 mx th Q=− 1+ (−9.638) = 10.24 MeV. Ex ' − 1 + mX 16 11. The isotope 18 F is a radionuclide used in medical diagnoses of tumors and, although usually produced by the 18 O(p,n)18 F reaction, it can also be produced by irradiating lithium carbonate (Li2 CO3 ) with neutrons. The neutrons interact with 6 Li to produce tritons (nuclei of 3 H) which in turn interact with the oxygen to produce 18 F. (a) What are the two nuclear reactions? (b) Calculate the Qvalue for each reaction. (c) Calculate the threshold energy for each reaction. (d) Can thermal neutrons (average energy 0.00253 eV) be used to create 18 F? Solution: (a) The first reaction is 6 Li(n,t)4 He followed by
16
O(t,n)18 F.
(b) Using the masses in Ap. B, the Qvalues for these reactions are 6
Li(n,t)4 He:
Q = [M (63 Li) + mn − M (42 He) − M (31 H)]c2 = 4.78 MeV,
and for the second reaction 16
O(t,n)18 F: Q = [M (168 O) + M (31 H) − mn − M (189 F)]c2 = 1.27 MeV.
(c) Because both reactions are exoergic, there are no kinematic thresholds. However, the 16 O(t,n)18 F reaction has a Coulombic threshold. From Eq. (6.19), the minimum energy of the triton EtC needed to cause this reaction is EtC ' 1.2
Zx ZX 1/3 Ax
+
1/3 AX
. = 1.2
(1)(8) = 2.42 MeV. + 161/3
31/3
68
Binary Nuclear Reactions
Chap. 6
(d) The energy of the triton produced by a thermal neutron induced 6 Li(n,t)4 He reaction is given by Eq. (6.12), namely MHe 4 Et = Q = 4.78 = 2.73 MeV. MHe + Mt 4+3 This is greater than EtC , and hence thermal neutrons can be used. 12. Consider the reaction 188 O(p,n)189 F mentioned in problem 11. This is one of the reactions employed in cyclotron production of radionuclides for PET scanning. Calculate (a) the Qvalue, (b) the kinematic threshold energy of the proton, (c) the threshold energy of the proton for the reaction, and (d) the minimum kinetic energy of the products. Solution The Qvalue is computed as Q = [M (11 H) + M (188 O) − mn − M (189 F)]c2 = [1.0078250321 + 17.9991604 −1.0086649233 − 18.0009377] × 931.5 = −2.438 MeV
The kinematic threshold is given by Eq. 6.15 as Exth ' −Q(1 + mx /mX ) = 2.438 × (1 + 1/18) = 2.573 MeV. The Coulombic barrier is given by Eq. (6.19) as ExC = 1.2
Zx ZX 1/3 Ax
+
1/3 AX
=
1.2 × 8 = 2.651MeV. 1 + 181.3
The proton threshold energy, from Eq. 6.20 is thus the Coulombic barrier, 2.651 MeV and the minimum kinetic energy of the products is 2.438 + 2.651 = 0.213 MeV. 13. Consider the following two neutronproducing reactions caused by incident 5.5MeV α particles. α + 7 Li −→ 10 B + n α + 9 Be −→ 12 C + n (a) What is the Qvalue of each reaction? (b) What are the kinetic energies of neutrons emitted at angles of 0, 30, 45, 90 and 180 degrees.
Binary Nuclear Reactions
69
Chap. 6
Solution: (a) Using the masses in Appendix B, the Qvalues for these reactions are 7
Li(α,n)10B: Q = [M(73 Li)+M(42 He)−mn −M(105 B)]c2 = −2.790 MeV,
and 9
Be(α,n)12 C: Q = [M(94 Be) + M(42 He) − mn − M(126 C)]c2 = 5.701 MeV.
(b) For the 7 Li(α,n)10B reaction, the neutron energy can be calculated from Eq. (6.11). Here mx = M (42 He) ' 4, mX = M (73 Li) ' 7, mY = M (105 B) ' 10, and my = mn ' 1. With these masses, Eq. (6.11) reduces to p p En = 0.4264 cos θn + 0.1818 cos2 θn + 0.4636. The energy of the neutrons emitted at various angles is tabulated below. θn (deg)
0◦
30◦
45◦
90◦
180◦
En MeV
1.512
1.308
1.009
0.464
0.142
(c) For the 9 Be(α,n)12 C reaction, the neutron energy can be calculated from Eq. (6.11). Here mx = M (42 He) ' 4, mX = M (94 Be) ' 9, mY = M (126 C) ' 12, and my = mn ' 1. With these masses, Eq. (6.11) reduces to p p En = 0.3608 cos θn + 0.1302 cos2 θn + 8.647. The energy of the neutrons emitted at various angles is tabulated below. θn (deg)
0◦
30◦
45◦
90◦
180◦
En MeV
11.045
10.690
10.283
8.647
6.770
14. Show that for inelastic neutron scattering, the minimum and maximum energies of the scattered neutron are √ 2 √ 2 A 1+∆−1 A 1+∆+1 0 0 Emin =E and Emax =E A+1 A+1 where ∆ ≡ (A + 1)E/(EA).
Solution:
The energetics of neutron scattering are given by Eq. (6.25) which can be written ( )2 r E0 1 A(A + 1)Q 2 2 = cos θs ± (A − 1 + cos θs ) + , E (A + 1)2 E
610
Binary Nuclear Reactions
Chap. 6
The maximum scattered neutron energy occurs for θs = 0. The above equation becomes ( )2 r √ 2 0 Emax 1 (A + 1)Q 1±A 1+∆ = 1±A 1+ = , (P6.2) E (A + 1)2 AE A+1 where ∆ ≡ (A + 1)Q/(EA) .
At the threshold of this reaction, Eq. (6.15) gives E th = −(A + 1)Q/A so that A + 1 . th A+1 . A+1 ∆th = Q E = Q − Q = −1. A A A th As √ the incident neutron energy increases above E , ∆ approaches zero and 1 + ∆ is positive and approaches unity. Thus only the + sign in Eq. (P6.2) is meaningful, and we obtain
0 Emax
√ 2 1+∆+1 =E . A+1
The minimum energy of the scattered neutron occurs for θs = π and Eq. (6.25) becomes ( )2 r 0 (A + 1)Q 1 Emin = −1 ± A 1 + . E (A + 1)2 AE Again the + sign gives the only meaningful result, so that 0 Emin
=E
√
1+∆−1 A+1
2
.
15. How many elastic scatters, on the average, are required to slow a 2MeV neutron to below 1 eV in (a) 16 O and in (b) 56 Fe? Solution: The average logarithmic energy loss per elastic scatter, ξ, is given by Eq. (6.29) E α ln =1+ ln α ≡ ξ, E0 1−α where α ≡ (A − 1)2 /(A + 1)2 . For α = 0.9311 and ξ = 0.03529.
16
O, α = 0.7785 and ξ = 0.1199. For
56
Fe,
The average number of scatters to decrease a neutron’s kinetic energy from E1 = 2 × 106 eV to E2 = 1 eV is given by Eq. (6.30) 1 E1 1 n = ln = ln(2 × 106 ). ξ E2 ξ For
16
O, we find n = 121 and
56
Fe, n = 411.
Binary Nuclear Reactions
611
Chap. 6
16. How many neutrons per second are emitted spontaneously from a 1 mg sample of 252 Cf? Solution: The number of neutrons produced by spontaneous fission per second, Sn , from a mass m of a spontaneously fissioning radionuclide is mNa ln 2 Sn = (N atoms)(λsf s−1 )(ν n/fiss.) = Pf fiss./decay ν A T1/2 From Table 6.3, we obtain Pf = 0.0309 and other needed data to find Sn =
(10−3 )(6.022 × 1023 ) 252
ln 2 × 0.0309 (2.638 y)(3.15 × 107 s/y)
3.73
= 2.30 × 109 n/s. Alternative: From Table 6.2 we are told Sˆn = 2.3 × 1012 n/(s g). Thus, 1 mg of 252 Cf has a source strength Sn = 10−3 Sˆn = 2.3 × 109 n/s. 17. In a particular neutroninduced fission of 235 U, 4 prompt neutrons are produced and one fission fragment is 121Ag. (a) What is the other fission fragment? (b) How much energy is liberated promptly (i.e., before the fission fragments begin to decay)? (c) If the total initial kinetic energy of the fission fragments is 150 MeV, what is the initial kinetic energy of each? (d) What is the total kinetic energy shared by the four prompt neutrons. Solution: (a) The fission fragments are 235 92 U
111 + 10 n −→ 4(10 n) + 121 47 Ag + 45 Rh.
(b) The prompt energy emission is the Qvalue for the above reaction, namely 121 111 2 Q = [235 92 U + mn − 4mn − 47 Ag − 45 Rh]c = 173.6 MeV.
(c) From Eq. (6.41) for a total fission fragment kinetic energy of Eff = 150 MeV EH = EL
mL mL mL = (Eff − EH ) = Eff = 71.8 MeV. mH mH mL + mH
The kinetic energy of the light fission fragment is then EL = Eff − EH = 78.20 MeV. (d) The kinetic energy of the fission neutrons is En = Q − EH − EL = 23.6 MeV.
612
Binary Nuclear Reactions
Chap. 6
18. Consider the following fission reaction 1 0n
90 142 1 + 235 92 U −→ 36 Kr + 56 Ba + 4(0 n) + 6γ
where 90 Kr and 142Ba are the initial fission fragments. (a) What is the fission product chain created by each of these fission fragments? (b) What is the equivalent fission reaction taken to the stable end fission products? (c) How much energy is liberated promptly? (d) What is the total energy eventually emitted? Solution: (a) From the Chart of the Nuclides we find β−
90 36 Kr
−→
142 56 Ba
−→
32 s β−
10.7 m
β − 90 β − 90 β − 90 90 37 Rb −→ 38 Sr −→ 39 Y −→ 40 Zr 2.4 m 29.1 y 2.67 d 142 57 La
β−
−→
1.54 h
142 58 Ce
(stable)
(stable)
(b) The fission reaction taken to its stable endpoint fission products is 1 0n
90 142 1 + 235 92 U −→ 40 Zr + 58 Ce + 4(0 n).
The photons produced in the fission event and in the decay of the fission products are not shown. (c) The prompt energy release is the Qvalue of the initial fission reaction, i.e., 2 90 142 Efp = Q = [M(235 92 U)+mn −M(36 Kr)−M( 56 Ba)−4mn ]c = 169.5 MeV.
(d) The energy released over the time required for the fission fragments to decay to their stable endpoints is 2 90 142 Efend = Q = [M(235 92 U)+mn −M(40 Zr)−M( 58 Ce)−4mn ]c = 190.0 MeV.
19. A 10 g sample of 235 U is placed in a nuclear reactor where it generates 100 W of thermal fission energy. (a) What is the fission rate (fission/s) in the sample? (b) After one year in the core, estimate the number of atoms of 99 43 Tc in the sample produced through the decay chain shown in Eq. (6.37). Notice that all fission products above 99 43 Tc in the decay chain have halflives much shorter than 1 year; hence all of these fission products can be assumed to decay to 99 43 Tc immediately. Solution: (a) From Section 6.6.3 of the text, it is shown that a fission power of 1 W corresponds to 3.1 × 1010 fission/second. Thus, in our sample generating 100 W the fission rate is Rf = (100 W)(3.1 × 1010 fiss/(s W)) = 3.1 × 1012 fiss/s.
Binary Nuclear Reactions
Chap. 6
613
(b) After one year, the number of fissions that have occurred in the sample is Nf = (Rf fiss/s)(3.16 × 107 s/y) = 9.78 × 1019 fissions/year. From the Chart of the Nuclides (or Fig. 6.6), we see the fission chain yield for A = 99 is y(99) = 6.1%. This is the yield per fission of one of the nuclides in the fission product chain of Eq. (6.37). If we assume there is negligible direct production of 99 Ru as a fission product, and because of all the halflives of the chain members above 99 Tc are much less than one year, we can assume that for every fission in the sample, about 0.061 atoms of 99 Tc are produced. Hence, the number of 99 Tc atoms in the sample after one year is estimated as N (99 Tc) = 0.061 × Nf = 6.0 × 1018 atoms. 20. (a) How much 235 U is consumed per year (in g/y) to produce enough electricity to continuously run a 100 W light bulb? (b) How much coal (in g/y) would be needed (coal has a heat content of about 12 GJ/ton)? Assume a conversion efficiency of 33% of thermal energy into electrical energy. Solution: With a 33% electrical conversion process, the amount of thermal energy required to light the 100 W bulb for one year is th Ebulb = 100(Wy/y)/0.33 = 300(Wy/y) = 1.10 × 105 (Wd/y) = 9.46 × 109(J/y).
(a) Since 1 MWd of thermal energy comes from the consumption of 1.24 g of 235 U, the amount of 235 U that is consumed to produce the thermal energy for the electricity to light the bulb is M235 = 0.110(MWd/y) × 1.24(g/MWd) = 0.136 g/y. (b) The mass of coal that must be burned is Mcoal =
9.46 × 109 (J/y) = 0.786 tons/y = 0.715 × 106 g/y. 12 × 109 (J/ton)
21. An accident in a fuel reprocessing plant, caused by improper mixing of 235U, produced a burst of fission energy liberating energy equivalent to the detonation of 7 kg of TNT (4.2 GJ/ton = 4.6 kJ/g). About 80% of the fission products were retained in the building. (a) How many fissions occurred? (b) Three months after the accident, what is the rate (W) at which energy is released by all the fission products left in the building? Solution: (a) The number of fissions, assuming 200 MeV release per fission, is Nf = Ereleased(MeV)/200(MeV/fiss) g J 1 1 = 7kg(TNT) × 103 × 4600 × × kg g 1.602×10−13J/MeV 200MeV/fiss = 1.00 × 1018 fissions.
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Chap. 6
(b) From Fig. 6.9, we obtain for t = 3 months = 7.78 × 106 s Fγ+β (t) = 1.2 × 10−8 × Nf = 1.2 × 1010 MeV/s = 1.9 mW. Alternatively, from Eqs. (6.44) and (6.45) we obtain for t = 7.78 × 106 s Fγ+β (t) = 2.66t−1.2 × Nf = 1.14 × 1010 MeV/s = 1.8 mW. 22. Estimate the available DD fusion energy in an 8 ounce glass of water. For how long could this energy provide the energy needs of a house with an average power consumption of 10 kW? Solution: (a) First find the number of deuterium atoms in the glass of water. From Table 1.3, 1 oz = 29.57 cm3 . Thus in 8 oz of water we have 8 × 29.57 = 236.6 cm3 of water, or with a density of 1 g/cm3 , a mass m of 236.6 g. The number of atoms of hydrogen NH is NH = 2NH2 O = 2
mNa (236.6)(6.022 × 1023) =2 = 1.583 × 1025 atoms. AH2 O 18
From Table A.4, the atomic abundance of deuterium in elemental hydrogen is 0.015%. Thus, the glass of water contains ND = (0.00015)(1.583 × 1025 ) = 2.375 × 1021 atoms of deuterium. From Example 6.6, we see that each atom of deuterium, when fully fussioned to produce 4 He, has a fusion energy potential of ED = 23.82/2 = 11.9 MeV. Thus the DD fusion energy potential in an 8oz glass of water is Etot = ED × ND = 2.826 × 1022 MeV = 4.53 × 109 J.
(b) The annual energy consumption in the house is PH = (104 J/s)(365.25 × 24 × 3600) s/y = 3.16 × 1011 J/y. Thus, the deuterium in the 8oz glass of water could provide the house’s energy needs for Etot /PH = 0.014 y = 5.24 d. 23. The sun currently is still in its hydrogen burning phase, converting hydrogen into helium through the net reaction of Eq. (6.46). It produces power at the rate of about 3.8×1026 W. (a) What is the rate (in kg/s) at which mass is being converted to energy? (b) How many 4 He nuclei are produced per second? (c) What is the radiant energy flux (W cm−2 ) incident on the earth? The average distance of the earth from the sun is 1.5 × 1011 m. Solution:
(a) Recall that E(J) = M (kg)c2 (m2 /s2 ). Thus the rate of mass converted into energy is dM Ps J/s 3.8 × 1026 = = = 4.22 × 109 kg/s. dt (c m/s)2 (3 × 108 )2
Binary Nuclear Reactions
615
Chap. 6
(b) From Eq. (6.47), we see that the formation of one 4 He nuclei is accompanied by the release of 26.72 MeV of energy. The power emitted by the sun is Ps = 3.8 × 1026 J/s/1.602 × 10−13 Mev/J = 2.37 × 1039 MeV/s. Thus, the number of 4 He nuclei produced per second is dNHe Ps MeV/s = = 8.88 × 1037 4 He/s. dt 26.72 MeV/He (c) The earth is about Re = 93 × 106 mi = 1.50 × 1013 cm from the sun. A sphere with this radius has a surface area of A = 4πR2e = 2.81 × 1027 cm2 . The radiant power crossing this spherical surface is Ps or the energy flux per unit area is solar constant =
Ps 3.8 × 1026 W = = 0.135 W/cm2 . 2 4πRe 2.81 × 1027 cm2
24. The sun has a lifetime of about 1010 y. Assume that it releases energy at the same rate as it is doing now (see previous Problem). Compare the total energy the sun will produce in its lifetime and compare this energy to that released in a few seconds by a type Ia supernova. Solution: The total energy that will be emitted by the sun in its lifetime is Et = (3.8 × 1026 J/s)(3.154 × 107 s/y)(1010 y) = 1.20 × 104 4 J. From the text a type Ia supernova emits, in a few seconds about 1.5 × 1044 J. Thus such a super nova emits about as much energy as our sun will in its entire lifetime. 25. Explain how electricity generated from hydroelectric power, wind turbines, coalfired power plants, and nuclear power plants are all indirect manifestations of fusion energy generated in stars. Solution: Fusion reactions in stars (and our sun) result in radiant energy flowing into space. Solar energy absorbed and reflected by our atmosphere causes our weather. The heating of the atmosphere causes convective currents or winds from which we can extract energy with wind turbines. The solar energy also causes the water cycle in our atmosphere in which water is evaporated and, after condensing, falls as rain replenishing the rivers from which we obtain hydroelectric energy. Coal and all other fossil fuels are the fossilized remains of vegetation, which required sunlight to grow. Finally, when stars die, some become supernovas and, during their brief explosive end, all the heavy elements, including uranium, are formed. The debris
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Chap. 6
from such novas becomes part of interstellar gas clouds and new stars and their planets that are eventually formed from these dust clouds. The uranium used in nuclear reactors was thus formed in stars.
Chapter 7
Radiation Interactions with Matter
PROBLEMS 1. A broad beam of neutrons is normally incident on a homogeneous slab 6cm thick. The intensity of neutrons transmitted through the slab without interactions is found to be 40% of the incident intensity. (a) What is the total interaction coefficient µt for the slab material? (b) What is the average distance a neutron travels in this material before undergoing an interaction? Solution: From Eq. (7.4), neutrons are exponentially attenuated as they pass through a material, i.e., I o (x) = exp[−µt x]. I o (0) (a) Here we are given that I(6 cm)/I(0) = exp[−µt 6] . Solving for µt gives o 1 I (t) 1 µt = − ln o =− ln(0.4) = 0.153 cm−1 . t I (0) 6 cm (b) The average distance a neutrons travels before interacting with the medium is given by Eq. (7.8), namely x = 1/µt = 6.55 cm. 2. With the data of Appendix C, calculate the halfthickness for 1MeV photons in (a) water, (b) iron, and (c) lead. Solution: The halfthickness is given by Eq. (7.9). Thus, from the data of Ap. C we obtain 71
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Radiation Interactions
Chap. 7
(a) Water: (µ/ρ) = 0.07066 cm2 /g and ρ = 1.00 g cm−3 . Thus, .µ x1/2 = ln 2 ρ = 9.81 cm. ρ (b) Iron: (µ/ρ) = 0.05951 cm2 /g and ρ = 7.874 g cm−3 . Thus, .µ x1/2 = ln 2 ρ = 1.48 cm. ρ (c) Lead: (µ/ρ) = 0.06803 cm2 /g and ρ = 11.35 g cm−3 . Thus, .µ x1/2 = ln 2 ρ = 0.898 cm. ρ 3. Based on the interaction coefficients tabulated in Appendix C, plot the tenththickness (in centimeters) versus photon energy from 0.1 to 10 MeV for water, concrete, iron and lead. Solution: By analogy to Eq. (7.9), the tenth thickness is x1/10 = ln 10/µ. With the data in Table C.3 the following plot is obtained.
Radiation Interactions
73
Chap. 7
4. A material is found to have a tenththickness of 2.8 cm for 1.25MeV gamma rays. (a) What is the linear attenuation coefficient for this material? (b) What is the halfthickness? (c) What is the meanfreepath length for 1.25MeV photons in this material? Solution: (a) By analogy to Eq. (7.9), the tenth thickness is x1/10 = ln 10/µ. Thus we have ln 10 µ= = 0.822 cm−1 . x1/10 (b) From Eq. (7.9) x1/2 =
ln 2 = 0.843 cm. µ
(c) The average distance traveled before interaction is, from Eq. (7.8), x=
1 = 1.22 cm. µ
5. Consider two adjacent infinite homogeneous slabs numbered, from left ot right, 1 and 2. The slab thickness and total interaction coefficients are ti and µi (i = 1, 2). Normally incident on the left slab is a beam of gamma rays. (a) What is the probability a gamma rays has its first interaction slab 1? (b) What is the probability a gamma ray has its first interaction in slab 2? (c) What is the probability a gamma ray pentetrates both slabs without interacting? Solution: (a) From Eq. (7.5) P (t1 ) = 1 − e−µ1 t1 .
(b) The probability the photon penetrates slab 1 without interaction is e−µ1 t1 . Then from the result of part (a) the probability it first interacts in slab 2 is P (t2 ) = e−µ1 t1 (1 − e−µ2 t2 ).
(c) The penetrates both slabs without interacting is simply the product of the probabilities the photon escapes both slabs without interacting, i.e., P (t1 + t2 ) = e−µ1 t1 e−µ2 t2 . 6. Consider a generaliztion of the previous problem in which there are N adjacent homogeneous slabs with thickness ti and interaction coefficient µi (i = 1 . . . N ). (a) What is the probability an incident photon has its first interaction in slab i? (b) What is the probability a photon penetrates all N slabs without interacting? Solution: (a) The probability the photon first interacts in slab i is Pi = {photon penetrate slabs 1 . . . (i − 1) w/o interacting} × { photon interacts in slab i}
74
Radiation Interactions
=
i Y
j=1
Chap. 7
exp(−µj tj ) × [1 − exp(−µi ti )].
(b) The probability a photon doen not interact in any of the slabs is the product of the probabilities it escape interaction in each slab, i.e., P (t1 + . . . + tN ) =
N Y
exp(−µj tj ).
j=1
7. Show the details to obtain Eq. (7.8) Solution: Integrate by parts to obtain Z ∞ ∞ Z ∞ x = µt xe−µt x dx = −xe−µt x + e−µt x 0 0 0 Z ∞ ∞ 1 =0+ e−µt x = − e−µt x µt 0 0 1 1 = − [0 − 1] = . µt µt 8. In natural uranium, 0.720% of the atoms are the isotope 235 U, 0.0055% are 234 U, and the remainder 238U. From the data in Table C.1, what is the total linear interaction coefficient (macroscopic cross section) for a thermal neutron in natural uranium? What is the total macroscopic fission cross section for thermal neutrons? Solution: From the data in Table A.3, the atom density of natural uranium is NU =
ρU Na (18.95)(6.022 × 1023) = = 4.794 × 1022 atoms/cm3 . AU 238.0289
(a) The atomic abundances fi of the three isotopes of uranium are found in Table A.4 and the thermal neutron microscopic cross sections are given in Table C.1. The total macroscopic cross section for natural uranium is X Σt = Ni σti = NU f234 σt234 + f235 σt235 + f238 σt238 i
= 4.794 × 1022 {(0.000055)(116) + (0.0072)(700) +(0.99275)(12.2)} × 10−24 = 0.823 cm−1 .
(b) Similarly, the macroscopic fission cross section is X Σf = Ni σfi = NU f234 σf234 + f235 σf235 + f238 σf238 i
= 4.794 × 1022 {(0.000055)(0.465) + (0.0072)(587) + (0.99275)(11.8 × 10−6 ) × 10−24 = 0.203 cm−1 .
Radiation Interactions
75
Chap. 7
9. Calculate the linear interaction coefficients in pure air at 20◦ C and 1 atm pressure for a 1MeV photon and a thermal neutron (2200 m s−1 ). Assume that air has the composition 75.3% nitrogen, 23.2% oxygen, and 1.4% argon by mass. Use the following data: Photon Element
2
µ/ρ (cm g
Nitrogen Oxygen Argon
Neutron −1
)
0.0636 0.0636 0.0574
σtot (b) 11.9 4.2 2.2
Solution: The density of dry air at standard conditions is ρair = 0.0012 g/cm3 . The atom density of the elements comprising air are Ni = wi
ρair Na Ai
(P7.1)
where wi is the mass fraction of the ith element (i = N, O, Ar). Photons: Use the gammaray data in Ap. C to obtain X µ X X µ X µ ρi air µγ = µi = ρi = ρair = ρair wi ρi ρ i ρair ρ i i i i i = 0.0012[(0.753)(0.0636) + (0.232)(0.0636) + (0.014)(0.0574)] = 7.61 × 10−5 cm−1 . Neutrons: Use the neutrons cross section data in Ap. C and Eq. (P7.1) to obtain X X fi σ i t i µair = N σ = ρ N i air a n t Ai i i (0.753)(11.9) (0.232)(4.2) (0.014)(2.2) = (0.0012)(0.6022) + + 14.007 15.9994 39.948 = 5.07 × 10−4 cm−1 . 10. From the data in Appendices A.3 and C.1, calculate the meanfreepath length of a thermal neutron in graphite. Solution: First find the atom density of graphite. N (C) =
ρNa (2.37 g/cm3 )(6.022 × 1023 atoms/mol) = A 12.0107 g/mol
= 1.188 × 1023 atoms/cm3 .
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Chap. 7
Multiply by the abundances of the the two isotopes in graphite to obtain the atom densities of each isotope, i.e., N (12 C) = 0.9889 × N (C) = 1.175 × 1023 atoms/cm3 N (13 C) = 0.0111 × N (C) = 1.319 × 1021 atoms/cm3 .
(P7.2)
Now calculate the total interaction coefficient (macroscopic cross section) for each isotope. µt (12 C) = σt (12 C)N (12 C) = (4.74 × 10−24 cm2 )(1.175 × 1023 atoms/cm3 = 0.5570 cm−1 (P7.3) .
Similarly it is found µt (12 C) = 0.00553 cm−1 .
The total macroscopic cross section for graphite is thus µt (C) = µt (12 C) + µt (13 C) = 0.5625 cm−1 . Thus from Eq. (7.8), the meanfreepath length is 1/µt (C) = 1.78 cm. 11. At a particular position the flux density of particles is 4.5 × 1012 cm−2 s−1 . (a) If the particles are photons, what is the density of photons at that position? (b) If the particles are thermal neutrons (2200 m/s), what is the density of neutrons? Solution: Since the flux density is defined as φ = nv, the particle density is n = φ/v. Thus, (a) Photons: Here v = c and n=
φ 4.5 × 1012 cm−2 s−1 = = 150 cm−3 . c 3.00 × 1010 cm s−1
(b) Thermal Neutrons: n=
φ 4.5 × 1012 cm−2 s−1 = 2.05 × 107 cm−3 . = v 2.2 × 105 cm s−1
12. A beam of 3MeV photons with intensity 108 cm−2 s−1 irradiates a small sample of water. (a) How many photonwater interactions occur in one second in one cm3 of the water? (b) How many positrons are produced per second in one cm3 of the water? Solution: Here φ = 108 cm−2 s−1 and we obtain (µ/ρ) data for water from Table C.3. (a) The reaction rate density for all types of interactions is given by Eq. (7.15), namely, µ b Rt = µt φ = ρ φ = (0.03968 cm2 /g)(1.00 g/cm3 )(108 cm−2 s−1 ) ρ t = 3.97 × 106 interactions cm−3 s−1 .
Radiation Interactions
77
Chap. 7
(b) The positron production rate equals the rate of pair production interactions in the water sample. Thus bβ+ = µpp φ = µ ρ φ R ρ pp = (1.131 × 10−3 cm2 /g)(1.00 g/cm3 )(108 cm−2 s−1 )
= 1.13 × 105 positrons cm−3 s−1 . 13. A small homogeneous sample of mass m (g) with atomic mass A is irradiated uniformly by a constant flux density φ (cm−2 s−1 ). If the total atomic cross section for the sample material with the irradiating particles is denoted by σt (cm2 ), derive an expression for the fraction of the atoms in the sample that interact during a 1h irradiation. State any assumptions made. Solution: The number of interactions that occur, during a time interval ∆t, in the sample with atom density N , a volume ∆V , a mass density ρ, and an atomic weight A is bt ∆V ∆t = µt φ∆V ∆t = (N σt )φ(m/ρ)∆t Nint = R ρNa m mNa σt φ = ∆t = σt φ∆t. A ρ A
The number of atoms in the sample, Natoms = (mNa )/A. Hence the fraction of atoms in the sample that experience an interaction is Fraction reacting =
no. interactions Nint = = σt φ∆t. no. atoms Natoms
14. A 2mCi source of 60 Co is placed in the center of a cylindrical waterfilled tank with an inside diameter of 20 cm and depth of 100 cm. The tank is make of iron with a wall thickness of 1 cm. What is the uncollided flux density at the outer surface of the tank nearest the source? Solution: From Ap. D or Fig. 5.12, we see that 60 Co emits a 1.17 MeV and a 1.33 MeV gamma ray, each with a frequency of almost 100% per decay. For simplicity, we will assume that each decay of 60 Co emits 2 photons each with an average energy of 1.25 MeV. Thus, the source emits Sp = 2 × (2 × 10−3 Ci)(3.7 × 1010 decays/Ci) = 1.48 × 108 γ/s each with an energy of 1.25 MeV.
To reach the outside of the tank nearest the source, an uncollided photon must pass through t1 = 10 cm of water and t2 = 1 cm of iron. From Eq. (7.27) the maximum uncollided flux density outside the tank is φo =
Sp exp[−µH2 O t1 − µF e t2 ]. 4π(t1 + t2 )2
(P7.4)
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Radiation Interactions
Chap. 7
From Ap. C we find for 1.25MeV photons, µH2 O = 0.0632 cm−1 and µF e = 0.4191 cm−1 . Then substitution into Eq. (P7.2) gives φo =
1.48 × 108 exp[−(0.0632)(10) − (0.4191)(1)] = 3.40 × 104 cm−2 s−1 . 4π(10 + 1)2
15. What is the maximum possible kinetic energy (keV) of a Compton electron and the corresponding minimum energy of a scattered photon resulting from scattering of (a) a 100keV photon, (b) a 1MeV photon, and (c) a 10MeV photon? Estimate for each case the range the electron would have in air of 1.2 mg/cm3 density and in water of 1 g/cm3 density. Solution: Let E and E 0 be the photon energy before and after scattering. By conservation of energy, the loss in photon energy must equal the recoil kinetic energy T of the electron, i.e., T = E − E 0 . From Eq. (7.33) we then have 1 T = E − E0 = E 1 − . 1 + (me c2 /E)(1 − cos θs ) By inspection, T is greatest when θs = π, so that 1 me c2 =E 1− . Tmax = E 1 − 1 + (2me c2 /E) 2E + me c2 The range of the recoil electron can be estimated from Eq. (7.47) and the data in Table 7.2. Results for the three specified photon energies are tabulated below. E (MeV)
Tmax (MeV)
Rair (cm)
RH2 O (cm)
0.1 1.0 10.0
0.0281 0.797 9.751
1.55 288 4,530
1.64 × 10−3 0.309 5.17
16. From Fig. 7.7, the total microscopic cross section in iron for neutrons with energy of 27 keV is about 0.4 b, and for a neutron with an energy of 28 keV about 90 b. (a) Estimate the fraction of 27keV neutrons that pass through a 10cm thick slab without interaction. (b) What is this fraction for 28keV neutrons? Solution: The fraction of the neutrons that are transmitted through a slab of iron with a thickness x = 10 cm without interaction is, from Eq. (7.4), I o (x) = e−µx ≡ e−Σt x . I o (0)
(P7.5)
Radiation Interactions
79
Chap. 7
The macroscopic total cross section is Σt (E) = N F e σtF e (E). The atom density of iron is NFe =
ρF e Na (7.875 g/cm3 )(6.022 × 1023 atoms/mol) = AF e 55.845 g/mol
= 8.49 × 1022 atoms/cm3 . (a) 27keV neutrons: The total interaction coefficient or macroscopic cross section is Σt (27 keV) = N F e σtF e (27 keV) = 0.03397 cm−1 . Then from Eq. (P7.3), the fraction of 27keV neutrons transmitted through the slab is Fraction transmitted = exp[−Σt x] = 0.712. (b) 28keV neutrons: The total interaction coefficient or macroscopic cross section is Σt (27 keV) = N F e σtF e (28 keV) = 7.644 cm−1 . From Eq. (P7.3), the fraction of 28keV neutrons transmitted through the slab is Fraction transmitted = exp[−Σt x] = 6.34 × 10−34 . 17. When an electron moving through air has 5 MeV of energy, what is the ratio of the rates of energy loss by bremsstrahlung to that by collision? What is this ratio for lead? Solution: (a) The Z number for air is Z air ' 0.8Z N +0.2Z O = 7.2. Then from Eq. (7.43) we find for M = me 2 (−dE/ds)rad EZ air me 2 (5)(7.2) 1 ' = = 0.05. (−dE/ds)coll 700 M 700 1
(b) The Z number for lead is 82. Hence, from Eq. (7.43) we find 2 EZ P b me 2 (5)(82) 1 (−dE/ds)rad ' = = 0.59. (−dE/ds)coll 700 M 700 1
18. About what thickness of aluminum is needed to stop a beam of (a) 2.5MeV electrons, (b) 2.5MeV protons, and (c) 10MeV alpha particles? Hint: For parts (a) and (b), use Table 7.2 and compare your values to ranges shown in Fig. 7.16. For part (c) use the range interpolation rules on page 196. Solution: (a) The range in aluminum of a 2.5MeV electron is obtained with the empirical formula of Eq. (7.47) and the data in Table 7.2. With x = log10 2.5 = 0.3979 2
ρRe (2.5 MeV) = 10−0.27957+1.2492x−0.18247x = 1.543 g/cm2 . Since ρ = 2.7 g/cm3 for aluminum, Re (2.5 MeV) = 0.571 cm. This value agrees with data of Fig. 7.16.
710
Radiation Interactions
Chap. 7
(b) The range in aluminum of a 2.5MeV proton is obtained with the empirical formula of Eq. (7.47) and the data in Table 7.2. With x = log10 2.5 = 0.3979 2
ρRp (2.5 MeV) = 10−2.3829+1.3494x+0.1967x = 0.01532 g/cm2 . Since ρ = 2.7 g/cm3 for aluminum, Rp(2.5 MeV) = 0.00567 cm. This value agrees with data of Fig. 7.16. (c) First find the kinetic energy of a proton with the same speed as a 10MeV alpha particle. Classical mechanics, appropriate for these energies, gives (1/2)mp vp2 Ep mp 1 = = = . Eα (1/2)mα vα2 mα 4 The proton energy with the same speed as a 10MeV alpha particle is, thus, Ep = Eα /4 = 2.5 MeV. Then from rule 3 on page 196 for particles of the same speed in the same medium , we have Rα (10 MeV) mα zp2 41 = = = 1. 2 Rp (2.5 MeV) mp zα 14 Thus the range of the alpha particle is Rα (10 MeV) = Rp (2.5 MeV) = 0.00567 cm (from part (b)). 19. Estimate the range of a 10MeV tritium nucleus in air. Solution: First find the kinetic energy of a proton with the same speed as a 10MeV triton. Classical mechanics, appropriate for these energies, gives (1/2)mp vp2 mp 1 Ep = = = . Et (1/2)mt vt2 mt 3 The proton energy with the same speed as a 10MeV triton is, thus, Ep = Et /3 = 3.333 MeV. Now find the range in air of a 3.333MeV proton using the empirical formula of Eq. (7.47) and the data in Table 7.2. With x = log10 3.333 = 0.5229 2
ρRp (3.333 MeV) = 10−2.5207+1.3729x+0.21045x = 0.01798 g/cm2 . For air at STP ρ = 0.0012 g/cm3 , and hence Rp (3.333 MeV) = 14.98 cm. Finally, we use rule 3 on page 196 for the range of different charged particles of the same speed in the same medium: Rt (10 MeV) mt zp2 31 = = = 3. Rp (3.333 MeV) mp zt2 11 The range of the triton is then Rt (10 MeV) = 3Rp(3.333 MeV) = 44.9 cm.
Radiation Interactions
711
Chap. 7
20. The CSDA range for protons in aluminum (in units of mass thickness g/cm2 ) are 2.953×10−4 , 4.020×10−3 , and 0.1692 for protons with energies of 0.1 MeV, 1 Mev, and 10 MeV, respectively [Janni 1982]. Use the empircal Eq. (7.47) to estimate these ranges and compute the percent difference. Solution: Use of a simple program or a spread sheet gives the following results: E (MeV) (MeV) 0.1 1.0 10.0
CSDA range mg/cm2 2.953 × 10−4 4.020 × 10−3 1.692 × 10−1
Eq. (7.47) mg/cm2 2.913 × 10−4 4.141 × 10−3 1.456 × 10−1
Difference % 1.34 3.00 13.9
21. Compare the ranges of the median heavy fission fragment with energy E = 67.9 MeV in air and gold as determined from Eq. (7.48) to those from Fig. 7.19. Solution: Air: density ρ = 1.205 mg/cm3 From Eq. (7.48): ρR = 0.14E 2/3 = 2.33 mg/cm2 . Hence R = 1.93 cm. From Fig. 7.19: ρR ' 2.5 mg/cm2 or a range R ' 2.2 cm
Gold: density ρ = 1.928 × 104 mg/cm3 From Eq. (7.48): ρR = 0.5E 2/3 = 8.32 mg/cm2. Hence R = 4.32 µm. From Fig. 7.19: ρR ' 9.1 mg/cm2 or a range R ' 4.7 µm
712
Radiation Interactions
Chap. 7
Chapter 8
Detection and Measurement of Radiation
PROBLEMS 1. Should the quenching gas in a GM tube have a higher or lower ionization potential than the major tube gas? Why? Solution: The quench gas should have a lower ionization potential than that of the major tube gas (hereafter, simply called the tube gas). As the positive tubegas ions drift toward the cathode wall, they collide with quench gas molecules and, because of their lower ionization potential, there is a tendency to transfer an electron from the quench gas molecule to the tubegas ion thereby neutralizing it. The resulting positively charged quenchgas molecule then drifts to the cathode where the energy liberated in the neutralization of the molecule causes the molecule to dissociate rather than to free an electron from the cathode surface (and thus start another avalanche). 2. What effect does each of the following changes have on the performance of a proportional counter? (a) The diameter of the anode wire is increased. (b) The pressure of the fill gas is increased. (c) The atomic number of the wall material is increased. Solution: (a) By increasing the diameter of the central anode wire, the electric field is decreased near the wire surface. This reduces the multiplication factor of the avalanche mechanism and thus reduces the magnitude of the output pulses from the detector. (b) Increasing the gas pressure increases the density of gas atoms in the chamber. Consequently, more atoms are near the anode so that there is a slight increase in the avalanche multiplication effect. Also increased gas pressure 81
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increases the detector efficiency for charged particle detection since there is a higher chance a charged particle interacts in the tube gas. (c) Increasing the Z number of the wall material decreases the probability a lowenergy photon can reach the interior of the detector through the wall. However, for a “thin” wall through which photons can pass, increasing Z increases the production of photoelectrons into the tube’s interior and, thus, the detector’s efficiency is increased. 3. For 1MeV photons, sketch how the detector efficiency of a proportional or GM chamber varies with the thickness of the cylindrical wall. Solution: For very thin walls, the photons will pass through the detector with little chance of interaction and hence of being detected. As the wall increases in thickness, more interactions occur in the wall that liberate electrons into the gaseous interior where an event is registered. Thus, the detector efficiency initially increases as the wall thickness increases. However, for very large wall thicknesses, the wall begins to act as a shield to the tube interior, and fewer incident photons will reach the inner portions of the wall surface, where electron production and subsequent multiplication via the avalanche process leads to a detector response. Thus, fewer photons are detected as the wall thickness increases to very large values. η 6

æ
wall thickness
4. A given GM tube has a dead time of 0.25 ms. If the measured count rate is 900 counts per second, what would be the count rate if there were no dead time? Solution: From Eq. (8.4) we obtain the count rate n from a detector with no dead time as m 900 n= = = 1160 cps. 1 − mτ 1 − (900)(0.25 × 10−3 )
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5. The decay constant for NaI(Tl) fluorescence radiation is about 230 ns. How long must one wait to collect 90% of the scintillation photons? Solution: The decay constant for excited atoms that emit the scintillation photons is λ = 1/t = 1/(230 × 10−9 s) = 4.35 × 106 s−1 . The number of excited atoms N (t) at time t after their creation is N (t) = N (0) exp[−λt]. Here N (0) is the number of photon emitters created by a radiation event at t = 0 and also equals the total number of scintillation photons that eventually are emitted. The number of photons emitted in the time interval (0, to) is n(to ) =
Z
to
λN (t) dt = λN (0)
0
Z
0
to
e−λt dt = N (0)[1 − e−λto ].
We seek the time to such that n(to )/N (0) = 0.9. From the above result we obtain 1 n(to ) 1 ln(1 − 0.9) = 53.0µs. to = − ln 1 − =− λ N (0) 4.35 × 106 s−1 6. In anthracene, scintillation photons have a wavelength of 447 nm. If 1MeV of energy is deposited in an anthracene crystal and 20,000 scintillation photons are produced, what is the scintillation efficiency? Solution: The energy of a scintillation photon is E = hν =
hc (4.1357 × 10−15 eV s)(2.998 × 108 m/s) = = 2.774 eV. λ 447 × 10−9 m
The energy of all 20,000 scintillation photons is thus Es = 20, 000 × 2.774 = 0.05548 MeV. Thus the scintillation efficiency is η=
Es 0.05548 = = 5.55%. Eγ 1.0
7. Why is air often used as a tube fill gas in an ionization chamber, but not in a proportional counter? Solution: Air would be the choice for an ion chamber used in measurement of exposure, which is defined in terms of ionization in air. The difficulty with air is that the oxygen component has a large affinity for capturing a free electron to produce a negatively charged molecule of oxygen. The significance of this depends on how the gas chamber is operated. An ionization chamber usually operates in the “current mode” measuring the current of flowing charges in the gas chamber. Whether or not the electrons
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produced by radiation interactions in the gas are left as free electrons or attached to an oxygen molecule does not affect the charge flow (i.e., current) in the tube. By contrast, a proportional counter usually is operated in the “pulse mode” in which the rapid collection of the free electrons at the anode (before the positive ions are collected at the cathode) produces a rapid voltage pulse whose magnitude is proportional to the number of ionizations produced in the gas by the incident radiation particle. If air were included in the chamber gas, many of the initially free electrons would become attached to oxygen molecules before reaching the anode. The mobility of negative oxygen molecules is very much lower than free electrons, and the collection of the negative charge on the anode would occur over a much longer time interval resulting in a very slow voltage change that would be nullified by a simultaneously slow voltage change produced by the collection of positive ions at the cathode. Thus, no rapid electron voltage pulse would occur. 8. If the energy resolution of a NaI(Tl) detector is 8%, what is the FWHM of the fullenergy peak for a 137Cs source? Solution: NOTE: Chapter 8 does not define resolution although it is discussed frequently. Consequently, this question is somewhat beyond the text. The resolution of a spectroscopic energy peak is define as Res(%) = 100
FWHM Emax
where Emax is the energy at the which the peak is a maximum and FWHM is the width of the peak at onehalf the maximum height. From Ap. D, 137 Cs emits a 0.662 MeV gamma ray. This the FWHM of the 0.662MeV peak in a NaI spectrum is FWHM = Emax
8 Res(%) = 0.662 = 0.00662 MeV. 100 100
9. Which detector has the greatest fullenergy peak efficiency: NaI(Tl), HPGe, Si(Li), or CZT for (a) 100keV gamma rays, and (b) for 1MeV gamma rays? Justify your answer. Solution: The fullenergy peak efficiency is the probability an incident gamma ray will interact in the crystal and also deposit all its energy in the crystal (i.e., no secondary radiation escapes). Thus, the efficiency is the product of (1) the probability Pint the gamma ray interacts at all and (2) the probability Pdep that the interaction leads to total energy deposition in the crystal. The probability of interaction is Pint ' 1 − e−µx where x is the crystal thickness and µ the total interaction coefficient.
Radiation Detection
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85
Photoelectric interactions almost always produce a fullenergy event, while Compton scattering and pair production interactions may or may not depending on whether the secondary photons are also absorbed in the scintillator crystal. For small crystals, such as those of typical HPGe, Si(Li) and CZT spectrometers, only the photoelectric interactions produce counts in the fullenergy peak. By contrast NaI crystals can be very large and many Compton and pairproduction interactions also lead to complete energy deposition. For small crystals Pe ' µph /µ, while, for large crystals, Pdep is also a function of the crystal size and material. (a) 100keV Photons: For 100keV photons, the total interaction coefficient µ ' µph in all four detector materials. Thus the crystal material with the largest photoelectric cross section (or the largest average Z number) has the largest fullenergy peak efficiency. Here is it is the CZT detector. (b) 1MeV Photons: At this energy Compton scattering dominates and the size of the detector becomes very important. Although CZT detectors still have the largest photoelectric coefficient, most interactions produce scattered photons which generally escape typically small CZT crystals. The same is true for typically small HPGe and Si(Li) crystals. However, NaI crystals can be very large and Compton scattering events can lead to fullenergy deposition, giving them the largest fullenergy efficiency.
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Chapter 9
Radiation Doses and Hazard Assessment
PROBLEMS 1. In an infinite homogeneous medium containing a uniformly distributed radionuclide source emitting radiation energy at a rate of E MeV cm−3 s−1 , energy must be absorbed uniformly by the medium at the same rate. Consider an infinite air medium with a density of 0.0012 g/cm3 containing tritium (halflife 12.33 y and emitting beta particles with an average energy of 5.37 keV/decay) at a concentration of 2.3 pCi/L. What is the airkerma rate (Gy/h)? Solution: The kerma rate, by definition, is the rate at which kinetic energy of secondary charged particles is released per unit mass of the medium. Here we have ˙ K(Gy/h) = KE(J) of beta particles released per h per kg of air = [decays per hour per L][av. beta KE(J) per decay]/[kg of air/L] = [(2.3 × 10−12 Ci/L)(3.7 × 1010 decays per s/Ci)(3600 s/h)]
×[(5.37 MeV/decay)(1.602 × 10−13 J/MeV)]/[0.0012 kg/L]
= 2.196 × 10−7 J kg−1 h−1 = 2.196 × 10−7 Gy/h = 22.0 µGy/h. Note that for this infinite homogeneous medium, the rate of energy emitted by all radioactive sources in a unit mass must equal the radiation energy absorbed in a unit mass. Thus the kerma rate equals the absorbed dose rate.
91
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2. A 137 Cs source has an activity of 900 µCi. A gamma photon from 137mBa with energy 0.662 MeV is emitted with a frequency of 0.845 per decay of 137 Cs. At a distance of 2.5 meters from the source, what is (a) the exposure rate, (b) the kerma rate in air, and (c) the dose equivalent rate? Solution: First we find the uncollided flux density 2 meters from the source. From Eq. (7.23) φo =
Sp (900 × 10−6 Ci)(3.7 × 1010 decays/Ci)(0.845 photons/decay) = 4πr 2 4π(250 cm)2
= 35.83 cm−2 s−1 . (a) Using the interaction coefficient data for air in Ap. C and linearly interpolating between tabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (µen /ρ)air = 0.02931. Then from Eq. (9.9) we find µen X˙ = 1.835 × 10−8 E φo ρ air = (1.835 × 10−8 )(0.662)(0.02931)(43.54) = 1.276 × 10−8 R/s = 45.9 µR/h. (b) Using the interaction coefficient data for air in Ap. C and linearly interpolating between tabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (µtr /ρ)air = 0.02937. Then from Eq. (9.5) we find µtr K˙ = 1.602 × 10−10 E φo ρ air = 1 × (1.602 × 10−10 )(0.662)(0.02937)(35.83) = 1.116 × 10−10 Gy/s = 0.402 µGy/h.
(c) Assume charged particle equilibrium so that kerma rate K˙ equals the ab˙ The dose equivalent equals the quality factor for sorbed dose rate D. photons (QF = 1) times the absorbed dose in tissue. We approximate tissue by water. Using the interaction coefficient data for water in Ap. C and linearly interpolating between tabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (µen /ρ)H2 O = 0.0.03260. Then from Eqs. (9.5) and (9.10) we find µen H˙ = QF × D˙ = 1.602 × 10−10 E φo ρ H2 O = 1 × (1.602 × 10−10)(0.662)(0.03260)(35.83) = 1.239 × 10−10 Gy/s = 0.446 µSv/h.
Radiation Doses and Hazards
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93
3. Suppose the source in the previous problem is placed in a large tank of water. Considering only the uncollided photons, at 0.4 m from the source, what is (a) the exposure rate, (b) the kerma rate in air, and (c) the dose equivalent rate? Solution: First find the uncollided flux density 50 cm from the source. The total interaction coefficient for water at 0.662 MeV, from linear interpolation of water data in Ap. C, is µ = 0.08604 cm−1 . Then from Eq. (7.25) φo = Sp e−µr /(4πr 2 ) or (900 × 10−6 Ci)(3.7 × 1010 decays/Ci)(0.845 γ/decay)e−0.08604×40 4π(50 cm)2 −2 −1 = 44.80 cm s .
φo =
(a) Using the interaction coefficient data for air in Ap. C and linearly interpolating between tabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (µen /ρ)air = 0.02931. Then from Eq. (9.9) we find µen φo X˙ = 1.835 × 10−8 E ρ air = (1.835 × 10−8 )(0.662)(0.02931)(44.80) = 1.595 × 10−8 R/s = 57.4 µR/h. (b) Using the interaction coefficient data for air in Ap. C and linearly interpolating between tabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (µtr /ρ)air = 0.02937. Then from Eq. (9.5) we find µtr −10 ˙ K = 1.602 × 10 E φo ρ air = 1 × (1.602 × 10−10 )(0.662)(0.02937)(44.80) = 1.395 × 10−10 Gy/s = 0.502 µGy/h.
(c) Assume charged particle equilibrium so that kerma rate K˙ equals the ab˙ The dose equivalent equals the quality factor for sorbed dose rate D. photons (QF = 1) times the absorbed dose in tissue. We approximate tissue by water. Using the interaction coefficient data for water in Ap. C and linearly interpolating between tabulated values at E = 0.6 MeV and E = 0.8 MeV, we find for E = 0.662 MeV that (µen /ρ)H2 O = 0.0.02940. Then from Eqs. (9.5) and (9.10) we find µen H˙ = QF × D˙ = 1.602 × 10−10 E φo ρ H2 O = 1 × (1.602 × 10−10)(0.662)(0.03260)(40.80) = 1.549 × 10−10 Gy/s = 0.558 µSv/h.
94
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4. What is the gammaray absorbed dose rate (Gy/h) in an infinite air medium at a distance of 10 cm from a 1mCi point source of (a) 16 N and (b) 43 K? Energies and frequencies of the emitted gamma rays from these radionuclides are given in Appendix D. Solution: At distances near the source, attenuation of radiation by the air can be neglected. The uncollided flux density of photons of energy Ei at distance r from the point source is from Eq. (7.25) φoi (cm−2 h−1 ) =
S(Bq)fi × 3600(s/h) 4π[r(cm)]2
where fi is the frequency of the ith photon, i.e., the probability of emission per radionuclide decay. The absorbed dose rate in air is then obtained by using Eq. (9.6) for each photon energy and then summing over the N discrete photon energies, i.e., air N (1.602 × 10−10)S(Bq) X µen ˙ D(Gy/h) = fi Ei (MeV) (cm2 /g). 4πr 2 ρ i i=1 (a) Using the data in Ap. D and with linear interpolation of µ/ρ values in Ap. C, we generate the following table for 16 N. Ei (MeV)
fi
(µen /ρ)air i
fi Ei (µen/ρ)air i
6.129 7.115
0.690 0.050
0.01639 0.01579
0.08931 0.00562 0.09493
For a 16 N source with 1 mCi activity (3.7 × 107 Bq) we then calculate the dose rate in air as (1.602 × 10−10 )(3.7 × 107 ) D˙ = × 3600 × 0.09493 = 1.611 mGy/h. 4π102 (b) Similarly, for
43
K we generate the following table:
Ei (MeV) 0.2206 0.3728 0.3969 0.5934 0.6175 1.0218
fi 0.0411 0.8727 0.1143 0.1103 0.8051 0.0188
(µen /ρ)air i
fi Ei (µen/ρ)air i
0.02713 0.02928 0.02947 0.02954 0.02880 0.02778
0.00025 0.00953 0.00134 0.00193 0.01432 0.00053 0.02789
Thus for a 43 K source with 1 mCi activity (3.7 × 107 Bq) we then calculate the dose rate in air as (1.602 × 10−10 )(3.7 × 107 ) D˙ = × 3600 × 0.02789 = 0.4736 mGy/h. 4π102
Radiation Doses and Hazards
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Chap. 9
5. What are the uncollided absorbed dose rates (Gy/h) in air at a distance of 10 cm from the point sources of the previous problem if the sources are placed in an infinite iron medium? Solution: In this case the attenuation by the iron must be explicitly included in the dose calculation. The uncollided flux density of photons of energy Ei at distance r from the point source is from Eq. (7.26) φoi (cm−2 h−1 ) =
S(Bq)fi e × 3600(s/h) × exp[−µF i r] 4π[r(cm)]2
where fi is the frequency of the ith photon, i.e., the probability of emission per radionuclide decay. The absorbed dose rate in air is then obtained by using Eq. (9.6) for each photon energy and then summing over the N discrete photon energies, i.e., air N µen (1.602 × 10−10 )S(Bq) X e ˙ (cm2 /g) exp[−µF fi Ei (MeV) D(Gy/h) = i r]. 4πr 2 ρ i i=1
(a) Using the data in Ap. D and with linear interpolation of µ/ρ values in Ap. C, we generate the following table for 16 N. Ei (MeV) 6.129 7.115
fi 0.690 0.050
(µen /ρ)air i
e cm−1 µF i
0.01639 0.01579
0.2403 0.2378
Fe
−µi fi Ei (µen/ρ)air i e
r
0.006269 0.000521 0.006790
For a 16 N source with 1 mCi activity (3.7 × 107 Bq) we then calculate the uncollided dose rate in air as (1.602 × 10−10)(3.7 × 107 ) D˙ = × 3600 × 0.006790 = 0.1153 mGy/h. 4π102 (b) Using the data in Ap. D and with linear interpolation of µ/ρ values in Ap. C, we generate the following table for 43 K. Ei (MeV)
fi
(µen /ρ)air i
e cm−1 µF i
0.2206 0.3728 0.3969 0.5934 0.6175 1.0218
0.0411 0.8727 0.1143 0.1103 0.8051 0.0188
0.02713 0.02928 0.02947 0.02954 0.02880 0.02778
1.0189 0.7485 0.7223 0.6005 0.5174 0.4627
F er
−µi 105fi Ei (µen /ρ)air i e
0.0009 0.5348 0.0975 0.4769 8.1067 0.5221 9.7389
For a 43 K source in iron with 1 mCi activity (3.7×107 Bq) we then calculate the uncollided air dose rate as (1.602 × 10−10 )(3.7 × 107 ) D˙ = × 3600 × 9.7389 × 10−5 = 1.654 µGy/h. 4π102
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6. A rule of thumb for exposure from point sources of photons in air at distances over which exponential attenuation is negligible is as follows: 6CEN X˙ = , r2 where C is the source strength (Ci), E is the photon energy (MeV), N is the number of photons per disintegration, r is the distance in feet from the source, and X˙ is the exposure rate (R h−1 ). 1. Reexpress this rule in units of Bq for the source strength and meters for the distance. 2. Over what ranges of energies is this rule accurate within 20%? Solution: (a) The approximate formula can be transformed thusly: 6C(Ci)E(MeV)N X˙ approx (R/h) = [r(ft)]2 . 2 C(Bq) r(m) =6 3.7 × 1010(Bq/Ci) 0.3048(m/ft) =
1.507 × 10−11 C(Bq)E(MeV)N . [r(m)]2
(b) Now calculate the exact exposure rate. The uncollided flux density of photons a distance r(ft) from a source of activity C(Ci) emitting N photons/decay of energy E(MeV) is φo (cm−2 h−1 ) =
3.7 × 1010(Bq/Ci) C(Ci) N × 3600(s/h). 4π[30.48(cm/ft)]2 [r(ft)]2
The exact exposure rate is given by Eq. (9.9) air µen −8 ˙ Xexact (R/h) = 1.835 × 10 E(MeV) (cm2 /g) φo (cm−2 h−1 ). ρ Substitute for φo in this expression and combine the constants to obtain C(Ci) E(MeV)N X˙ exact (R/h) = 209.36(µen/ρ)air (cm2 /g) . [r(ft)]2
The ratio of the approximate to exact exposure rates is thus −1 X˙ approx /X˙ exact = 6 209.36(µen/ρ)air (cm2 /g)
Evaluating and plotting this ratio versus energy gives the results and plot on the next page. From this plot we find that X˙ approx is within ±25% of X˙ exact for photon energies between 0.05 and 2 MeV.
Radiation Doses and Hazards
Chap. 9
Problem 96: Data for plot E air Xap/Xex (MeV) mu_en/rho (%) 0.01 4.74200 0.60 0.02 0.53890 5.32 0.03 0.15370 18.65 0.04 0.06833 41.94 0.06 0.03041 94.19 0.08 0.02407 119.06 0.10 0.02325 123.26 0.20 0.02672 107.26 0.40 0.02942 97.41 0.60 0.02953 97.05 1.00 0.02789 102.76 2.00 0.02345 122.21 4.00 0.01870 153.26 6.00 0.01647 173.98 10.00 0.01450 197.65 
97
98
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7. A medium sized banana (˜ 120 g) contains 422 mg of potassium. If Reference Man eats one such banana, what committed effective dose equivalent does he receive. If he ate one banana each morning for a year what is his committeed effective dose equivalent? Solution: The dose comes from ingesting 40 K. From Table 5.2 the percent isotopic abundance is f = 0.0117% and the halflife is T1/2 = 1.27 × 109 y. Thus, each banana contain a mass M (40 K)(0.000117)(422) = 49.4 µg of 40 K. The number of 40 K atoms is N (40 K) = M (40 K)Na /40 = 7.44 × 1016 atoms. The decay constant of 40 K is ln 2 ln 2 λ(40 K) = = = 4.15 × 10−16 s−1 . 9 T1/2 (1.27 × 10 y)(365.24 d/y)(3600 s/d)
The activity in each banana is A(40 K) = λ(40 K)N (40 K) = 30.9 Bq. From Table b E = 5.1 × 10−9 Sv/Bq. 9.4, the specific committed effective dose equivalent is H Thus his committed dose is b E (40 K) × A(40 K) = (5.1 × 10−9)(30.9) = 1.57 × 10−7 Sv = 15.7 µrem. HE = H
Multiply this dose per banana by 365 days, then eating one banana a day for a year results in a committed effective dose equivalent of 5.7 mrem.
8. The maximum permissible body burden for 226Ra has long been established as 0.1 µCi. In Reference Man, about 78 percent of the radium in the body resides in the 7 kg skeleton. Suppose the 226Ra is in equilibrium with its daughters to, but not including 210Pb. What is the average alphaparticle dose rate (rad y−1 ) in Reference Man if the body burden is the maximum permitted? Solution: From Fig. 5.20 the alpha producing nuclides are seen to be 226Ra, 222Rn, 218 Po, and 214 Po, which, from Appendix D, are seen to release alpha particles with energies 4.774 MeV, 6.288 MeV, 6.001 MeV, and 7.687 MeV, respectively. Because these radionuclides are in secular equilibrium, they all decay at the same rate, and, per decay, a total of 24.75 MeV of alpha particle kinetic energy is released. The maximum body burden thus represents (10−7 Ci) × (3.7 × 1010 Bq/Ci) = 3.7×103 Bq. Thus, each second (24.75 MeV)(3.7×103 Bq) = 9.157×104 MeV/s of alpha particle energy is released in the body. But only 78% of this energy is absorbed in the skeleton, so the dose rate to the skeleton is D˙ = (0.78)(9.157 × 104 )/7 = 1.020 × 104 MeV kg−1 s−1 = (1.020 × 104 MeV kg−1 s−1 )(3600 s/h)(24 h/d)(365.24 d/y = 3.220 × 1011 MeV kg−1 s−1
= (3.220 × 1011 MeV kg−1 s−1 )(1.602 × 10−13 J/MeV) = 0.0515 Gy/y = 5.15 rad/y.
Radiation Doses and Hazards
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99
9. A male worker at a nuclear facility receives an accidental wholebody exposure of 2.3 Gy. Describe what physical symptoms the individual is likely to have and when they occur. Solution: The deterministic effects for a 2.3 Gy dose are found from Table 9.7 and from the discussion in Section 9.5.2. Specifically, • sperm will be suppressed after about 74 days for months thereafter
• nausea, vomiting, fatigue, and weakness is likely to start after 2 weeks
• after 3 weeks, there will be about a 30% drop in blood cell production 10. A population of 500,000 around a nuclear facility receives an average wholebody dose of 0.005 Gy (0.5 rad) as a result of an accidental release of radionuclides. (a) Estimate how many children subsequently born in the first two generations would experience significant hereditary defects as a result of this exposure to their parents and grandparents. (b) Estimate how many such defects would occur naturally in the absence of this accidental exposure. (c) Estimate the cancer mortality risk imposed on this population, both absolutely and relative to natural cancer mortality. Solution: (a) From Table 9.11, per million progeny and per Gy exposure, there is an expectation of, for example in the case of autosomal disorders, 1300 to 2500 cases in the first two generations. Note that column 3 in the table combines first and second generation cases. For the conditions of this problem, for a static population there would be an expectation of: • For autosomal disorders: (1300 to 2500) × 500, 000/1, 000, 000 × 0.005 = 3 to 6 cases. • For all disorders: (3930 to 6700) × 500, 000/1, 000, 000 × 0.005 = 10 to 17 cases. (b) From Table 9.11, per million progeny, there is an expectation of, for example in the case of autosomal disorders disorders, 24,000 cases per million progeny. Thus, in two generations of 500,000 births each, one would expect 24,000 cases; for all disorders, including chronic multifactorial, one would expect 738,000 cases. (c) We assume equal numbers of males and females in the exposed population. Then, from the data in Table 9.14 (Case 1), the radiogenic cancer mortality risk is (1/2)(480 + 660) per 100,000 personGy collective exposure of low LET radiation. Here the collective population dose is 500, 000×0.005 Gy = 2.5 × 103 personGy. The expected number of radiogenic cancer deaths resulting from the accidental exposure is, therefore, 1 2.5 × 103 personGy (480 + 660) = 143 deaths. 2 104 personGy
910
Radiation Doses and Hazards
Chap. 9
Similarly, the expected number of naturally occurring cancer deaths, using data from Table 9.14, is 1 5 × 105 persons (22, 810 + 18, 030) = 102, 100 deaths. 2 105 persons Comparing these two results, we see the number of radiogenic cancer deaths is 0.14% that of the natural incidence. 11. A static population of 900,000 people lives in an area with a background radiation level that gives each person an excess 125 mrem (wholebody) per year of low LET exposure over that received by people in other parts of the country. (a) Estimate the annual collective gonad dose to the reproductive population. (b) Estimate the number of radiationinduced cases of hereditary illness that occurs in one generation as a result of this excess exposure to the parental and grandparental generations. (c) Compare this number to the natural incidence of genetic illness in this population. Solution: (a) We assume a steadystate (static) population with a 50:50 male to female ratio and that all persons, early in their lives, are part of the reproductive population and receive the excess 125 mrem/y. Thus an average person in this population receives the excess annual gonad dose of 125 mrem/y. If we assume a mean reproductive age of 30, each person receives a cumulative gonad dose to age 30 of 30 y×125 mrem/y = 3.75 rem = 0.0375 Gy. Then the collective gonad dose each generation receives is Dg = 0.0375 Gy × 900, 000 people = 3.38 × 104 personGy. (b) From the estimates in Table 9.11, there are 3930 to 6700 cases in one generation resulting from exposure of parental and grandparental generations to one million personGy. The number of cases of genetic illness in one generation caused by the excess background radiation received by two previous generations is then Nrad =
3930 to 6700 cases × 3.38 × 104 personGy = 133 to 226 cases. 106 personGy
(c) From Table 9.11, the expectation of the natural frequency of genetic illness is 738,000 cases per million liveborn, including 650,000 cases of chronic multifactorial disorders. In a static population of 900,000 with an average lifespan of 75 y and mean age to reproduction of 30 y, there are (30/75) × 900, 000 = 360, 000 liveborn each generation. Thus, the number of naturally occurring genetic illness cases, per generation, is Nnat = 738, 300
360, 000 = 266, 000 cases. 1, 000, 000
Radiation Doses and Hazards
Chap. 9
911
12. On September 13, 1987, a 137Cs source for radiotherapy was stolen from an abandoned treatment facility in Goiˆ aia, Brazil. The activity of the source was estimated to be about 5 × 1013 Bq. The decay of an atom of 137 Cs leads to emission of a gamma ray with an energy of 0.662 MeV 94.4% of the time. (a) The unsuspecting thief had no idea what it was and eventually got it out of its shielding. If the thief spent one hour approximately 1 meter from the unshielded source, what is his total absorbed dose? (b) What are the expected health implications of that dose? (c) Although four people ultimately died soon after the event at Goiˆ aia, Brazil, another 1000 people were exposed to doses estimated to be as large as 200 mSv. How many of these people are expected to get cancer due to that dose? How many are expected to die from cancer? (d) How many of the 1000 people in part (c) would be expected to die from cancer of natural causes? Solution: (a) Neglect air attenuation so the gammaray flux 1 meter from the source is φ=
(0.947)(5 × 1013 ) = 3.77 × 108 cm−2 s−1 . 4π × 1002
Because a person is wellapproximated by water, the mass energy absorption coefficient for water at E = 0.662 MeV is found by linearly interpolating the tabulated values in in Table C.3 at E1 = 0.6 and E2 = 0.8 MeV, i.e., µen (E) µen (E2 ) E − E2 µen (E1 ) µen (E2 ) = + − ρ ρ E1 − E2 ρ ρ 0.662 − 0.8 = 0.03206 + [0.03284 − 0.03206] = 0.03260 cm2 g−1 0.6 − 0.8 From Eq. (9.6), the dose rate one meter from the source is µen −10 ˙ D = 1.602 × 10 E φ ρ
= 1.602 × 10−10(0.662)(0.03206)(3.77 × 108 ) = 0.00128 Gy/s.
Hence, the total absorbed dose over 1 hour is (0.00128 Gy/s)(3600 s/h) = 4.61 Gy. (b) Because the computed dose is greater than the LD50/60 dose of 3–3.5 Gy, one would expect the thief to die. The thief was actually estimated to have received a somewhat larger dose, but he ultimately survived. Unfortunately, four others perished because of his actions, and several more may end up suffering longterm effects.
912
Radiation Doses and Hazards
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(c) Assume the exposed people are half males and half females and had mix of ages similar to that of the US population. Then from Table 9.14, the risk of developing cancer is risk =
0.5(900 + 1370) cases = 0.1135 cases/personGy. 105 people × 0.1 Gy
Because the gamma radiation from 137 Cs is low LET radiation, the dose of 0.2 Sv is equivalent to 0.2 Gy. Thus the expected number of cancer cases is (0.1135 cases/personrad) × (1000 people) × (0.2 Gy) = 22.7 ' 23 cases. The risk of dying from the cancer is found similarly, i.e., risk =
0.5(480 + 660) deaths = 0.0570 deaths/personGy. 105 people × 0.1 Gy
Thus, the expected number of cancer deaths is (0.0570 cases/personrad)× (1000 people) × (0.2 Gy) = 11.4 ' 11 deaths. (d) Again using data from Table 9.14 natural risk =
0.5(22810 + 18030) deaths = 0.204 deaths/person. 105 people
Thus the expected number natural cancer deaths is (1000 people)(0.204deaths/person) = 204 deaths. 13. A 40year old female worker receives an xray exposure of 5.2 rad (wholebody) while carrying out emergency procedures in a nuclear accident. Discuss the health risks assumed by this worker as a result of the radiation exposure. Solution: From Table 9.13, the radiogenic cancer mortality risk for a 30year old female is (455 + 52)/105 per 10 rad of gammaray exposure. Thus for a 5.2 rad exposure, the female’s risk is (455 + 52) 5.2 risk = = 2.64 × 10−3 . 105 10 This is a risk of about one chance in 380, or 0.26%. 14. How many people in the U.S. might be expected to die each year as a result of cancer caused by natural background radiation (excluding radon lung exposures)? Assume an average wholebody exposure of 200 mrem and a population of 330 million. Compare this to the natural total death rate by cancer. Solution: (a) Use the data in Table 9.14 (Case 2 for continuous lowlevel exposure). If we assume equal numbers of males and females, the average radiogenic
Radiation Doses and Hazards
Chap. 9
913
cancer mortality risk is (1/2)(480 + 660) per 105 people per 1 mGy annual exposure. Thus, in the U.S. population of 330 million, the number of people expected to eventually die of radiogenic cancer is 480 + 660 330 × 106 people 2 mGy Nrad = = 3.76 × 106 . 2 105 people 1 mGy If we now assume a static population with an average lifetime of 73.1 y (see Table 9.15), the death rate from radiogenic cancer is 3.76 × 106 deaths N˙ rad = = 51, 464 per year. 73.1 y (b) Similarly, from Table 9.14 (Case 2) we see the number of cancer deaths from natural causes is Nnat =
330 × 106 people 1 (22, 810 + 18, 030) = 67.4 × 106 deaths. 2 105 people
For a static population with an average life span of 73.1 years, the annual natural cancer death rate is then 67.4 × 106 deaths = 922, 000 per year. N˙ nat = 73.1 y
15. A radiology technician receives an average occupational dose of 3 mGy (300 mrad) per year over her professional lifetime. What cancer risk does she assume as a result of this activity? Solution: Use the risk estimates for Case 3 of Table 9.14 for this problem. From this table we see a female occupation worker has a risk of 2389/105 for an exposure of 10 mGy per year for each year of her working lifetime. Thus her probability of dying from radiogenic cancer for an annual dose of 3 mGy is 2389 3 mGy risk = = 7.2 × 10−3 . 105 10 mGy Thus her risk is about one chance in 140 or about 0.7%.
16. A male reactor operator receives a whole body dose equivalent of 0.95 rem over a period of one week while working with gammaray sources. For gammaray exposure, the dose is thus 9.5 mGy. What is the probability he will (a) die of cancer, (b) die of cancer caused by the radiation exposure, (c) have a child with an hereditary illness due to all causes, and (d) have a child with an hereditary illness due to the radiation exposure?
914
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Solution: (a) For males, the risk of natural occurrence of cancer mortality is seen from data in Table 9.14 to be 22, 810/105 = 0.22 = 22%. (b) From Table 9.14 for a single exposure of 100 mGy (10 rad), the risk of radiation induced cancer mortality for males is 480/105. For an exposure of 9.5 mGy, the risk is (480/105 )(9.5/100) = 4.6 × 10−4 = 0.046%. Alternatively, the radiogenic cancer risk, averaged over all population ages, is estimated in the second paragraph of Section 9.7.2 as about 5 ×10−4 per rem. Thus for an exposure of 0.95 rem, the radiogenic cancer mortality risk to the individual is about (5 × 10−4 )(0.95) ' 4.7 × 10−4 ' 0.046%. (c) Total risk of natural hereditary illness is obtained from the first column of Table 9.11. Thus natural risk is • risk (autosomal) = 28, 000/106 = 2.87%. • risk (multifactorial) = 710, 000/106 = 71%. (d) From Table 9.11, the radiation induced risk to the first generation, per Gy gonad dose to the parents is seen to be 750 to 1500 per million for autosomal disorders and 2250 to 3200 per million for multifactorial disorders. Thus for a parental gonad dose of = 0.0095 Gy the radiogenic hereditary risk for the male operator is • risk (autosomal) = 0.0095 × (750 to 1500)/106 = 7 to 14 per million • risk (multifactorial) = 0.0095 × (2250 to 3200)/106 = 21 to 30 per million.
17. An individual is exposed 75% of the time to radon with a physical concentration of 4.6 pCi/L and an equilibrium factor of F = 0.6. The remaining 25% of the time, the individual is exposed to radon at a concentration of 1.3 pCi/L and with an equilibrium factor of F = 0.8. What is the annual radon exposure (on an EEC basis) in MBq h m−3 ? Solution:
P The annual radon exposure is calculated as R = i Fi Ci hi , where Ci is the radon concentration at location i, Fi is the equilibrium factor at that location, and hi is the number of hours per year spent in location i. Thus, for this problem we have R = (4.6 pCi/L)(0.6)(0.75 × 365.25 d/y × 24 h/d)
+(1.3 pCi/L)(0.8)(0.25 × 365.25 d/y × 24 h/d) = 2.04 × 104 pCi h/L.
From Section 9.7.1, we find that an EEC of 4 pCi/L is equivalent to 150 Bq m−3 . With this conversion factor, the annual Rn EEC exposure is 150 Bq/m3 4 −1 R = (2.04 × 10 pCi h L ) = 0.766 MBq h m−3 . 4 pCi/L
Radiation Doses and Hazards
915
Chap. 9
18. If the individual in the previous problem is a nonsmoking male and receives the same annual radon exposure for his entire life, what is the probability he will die from lung cancer as a result his radon exposure? Solution: From Table 9.15, we find that for a nonsmoking male receiving an annual exposure of 1 MBq h m−3 for his entire life the probability he dies from radon induced lung cancer is 0.016. Thus, for a lifetime annual exposure of 0.766 MBq h m−3 the probability of death from radon induced lung cancer is 0.016 (MBq h m−3 )−1 × 0.766 (MBq h m−3 ) = 0.012 = 1.2%, or about 1 chance in 83. 19. Consider a female exposed 75% of the time to radon with an 222 Rn EEC of 25 Bq m−3 and 25% of the time to a concentration of 5 Bq m−3 for her entire life. (a) What is the probability that she will die from radoninduced lung cancer if she is a nonsmoker? (b) What is the probability if she smokes? Solution: The annual radon exposure is calculated as X R= F i C i hi i
where Ci is the radon concentration at location i, Fi is the equilibrium factor at that location, and hi is the number of hours per year spent in location i. Here we are given the EEC concentrations, i.e., Ci Fi . Thus, for this problem the female receives an annual exposure of R = (25 Bq m−1 )(0.75 × 365.25 d/y × 24 h/d) + (5 Bq m−1 )(0.25 × 365.25 d/y × 24 h/d) = 0.175 MBq h m−3 . (a) From Table 9.15, we find a nonsmoking female has a radoninduced cancer mortality risk of 0.0088 per MBq h m−3 annual exposure. Thus her probability of death from radoninduced lung cancer is 0.0088 (MBq h m−3 )−1 × 0.175 (MBq h m−3 ) = 0.0015 = 0.15%, or 1 chance in 670. (b) From Table 9.15, we find a smoking female has a radoninduced cancer mortality risk of 0.081 per MBq h m−3 annual exposure. Thus her probability of death from radoninduced lung cancer is 0.081 (MBq h m−3 )−1 × 0.175 (MBq h m−3 ) = 0.014 = 1.4%, or 1 chance in 71.
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20. Consider people who receive an annual radon exposure (EEC basis) of 0.11 MBq h m−3 for the first 30 years of their lives. At age 30, through radon reduction remediation actions, they decrease their annual exposure to 0.02 MBq h m−3 for the remainder of their lives. (a) What is their mortality risk for radoninduced lung cancer? (b) If they had not undertaken remedial measures, what would be their radon mortality risk? Solution: (a) From Table 9.16, we find people receiving an annual radon exposure of 1 MBq h m−3 for the first thirty years of life have a radiogenic cancer mortality risk of 0.0054. Thus for an annual exposure of 0.11 MBq h m−3 for the first thirty years, the probability they eventually die from radoninduced lung cancer is P0−30 = 0.0054 (MBq h m−3 )−1 × 0.11 (MBq h m−3 ) = 0.000594. Similarly, from Table 9.16, we find, people receiving an annual radon exposure of 1 MBq h m−3 from age 30 for the rest of their lives have a radiogenic cancer mortality risk of 0.012. Thus for a continuous annual exposure of 0.02 MBq h m−3 beginning at age 30, the probability they die from radoninduced lung cancer is P30−∞ = 0.012 (MBq h m−3 )−1 × 0.02 (MBq h m−3 ) = 0.000240. Thus, their total lifetime risk of dying from radoninduced lung cancer is 0.000594 + 0.000240 = 0.00083 = 0.083% or 1 chance in 1200. (b) From Table 9.16, we find people receiving an annual radon exposure of 1 MBq h m−3 for their entire lives have a radiogenic cancer mortality risk of 0.014. Thus, for an annual exposure of 0.11 MBq h m−3 for their entire lives, the probability they die from radoninduced lung cancer is P = 0.014 (MBq h m−3 )−1 × 0.11 (MBq h m−3 ) = 0.00154 = 0.15%, or 1 chance in 650. 21. Estimate how many radoncaused lung cancer deaths there are every year in the U.S. if everyone were exposed to the U.S. residential average radon concentration of 1.25 pCi/L 75% of the time. Assume an equilibrium factor of F = 0.5. Assume a population of 250 million. State and justify any other assumptions you make. Solution: First assume the other 25% of time the radon exposure is negligible. From page 253, we find that an EEC of 4 pCi/L is equivalent to 150 Bq m−3 . Then the annual radon exposure received by people in this population is 150 Bq/m3 R = (1.25 pCi/L)(0.5)(0.75 × 365.25 d/y × 24 h/d) 4 pCi/L = 0.154 MBq h m−3 .
Radiation Doses and Hazards
Chap. 9
917
We assume this annual radon exposure does not change over time. Now we assume a steady population in which the death rate equals the birth rate. From Table 9.15, the average lifetime in a mixed population is 73.1 y. Thus, the death rate in our steady population of 250 million is 250×106 /73.1 = 3.42 × 106 y−1 .
From Table 9.15 we see that the excess radoninduced cancer mortality risk is 0.039 per annual MBq h m−3 . Thus the radon mortality risk for an annual exposure of 0.154 MBq h m−3 is PRn = 0.039 × 0.154 = 0.0060. The number of annual radon cancer deaths in the population is then estimated as radon deaths/y = deaths/y×PRn = (3.42×106 )(0.0060) = 20, 500 deaths/y. 22. Why do the skin and extremities of the body have a higher annual dose limit than for other organs and tissues? Solution: Compared to other regions of the body, the skin and extremities (hands and feet) produce far fewer vital chemicals and cells needed to sustain the human body. Consequently, the skin and extremities are far less radiosensitive compared to the other organs and tissues. 23. Describe in your own words the rationale for the NCRP [1987] limit of 5 rem a year wholebody exposure for occupational workers. Give arguments why you do or do not believe this limit to be reasonable. Solution: In Section 10.8.1 it is stated that the annual average occupational death rate from accidents is about 100 per million workers, or 10−4 y−1 . Also on in Section 10.8.1, the probability of radiationinduced cancer mortality is said to be about 10−4 per rem of whole body exposure. The average dose equivalent received by a radiation worker is, from p 257, 0.23 rem. An average radiation worker therefore has an annual probability of dying from radiogenic cancer (usually far in the future) of (0.23)(10−4) = 0.23 × 10−4, which is considerably less than the occupational accidental death rate in nonradiation industries. To set an limit on the annual dose a radiation worker can receive, one could determine the annual dose that a radiation worker would have to receive so that his probability of dying from radiogenic cancer equals the accidental death rate accepted by nonradiation workers, namely Annual dose limit =
=
deaths/y for nonradiation workers risk of death from radiation/rem 10−4 y−1 = 4.3 rem/y ' 5 rem/y. 0.23 × 10−4 rem−1
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Chapter 10
Principles of Nuclear Reactors
PROBLEMS 1. In a liquid metal fast breeder reactor, no neutron moderation is desired and sodium is used as a coolant to minimize fissionneutron thermalization. How many scatters with sodium, on the average, would it take for 2MeV neutrons to reach an average thermal energy of 0.025 eV? HINT: review Section 6.5.1. Solution: From Eq. (6.30) the average number of scatters n needed to reduce a neutron’s kinetic energy from E1 to E2 is 1 E1 n = ln ξ E2 where ξ is the mean logarithmic energy lost by a neutron in an elastic scatter with a nucleus of atomic mass number A, namely ξ =1+
α ln α 1−α
with
α≡
(A − 1)2 . (A + 1)2
For sodium (A = 23), we find α = 0.8403 and ξ = 0.08448. Then the number of elastic scatters with carbon required to reduce a neutron’s energy from 2 MeV to 0.025 eV is 1 E1 1 2 × 106 n = ln = ln = 215. ξ E2 0.08448 0.025
101
102
Principles of Nuclear Reactors
Chap. 10
2. Discuss the relative merits of water and graphite for use in a thermal reactor. Solution: Water: • much less expensive • can also be used as the coolant for a power reactor • boils at lower temperatures and thus require a pressure vessel • σa is relatively large (0.664 b) • larger ξ and slows neutrons with fewer scatters • much more inert chemically • cannot be used as a structural element of the core • can produce radioactive 3 H and 16 N by activation • can dissociate and produce explosive hydrogenoxygen gas mixtures Graphite: • reactorgrade (high purity) graphite is expensive • cannot be used as the coolant • solid to very high temperatures; thus no need for a pressure vessel • σa is very small (34 mb) • smaller ξ and requires more scatters to slow neutrons • burns and thus hot graphite must be isolated from oxygen • can be used as structural elements of the core • does not activate and produce radionuclides • stores energy and must be periodically annealed to avoid a sudden energy release (Wigner effect). 3. List five desirable properties of a moderator for a thermal reactor. Explain the importance of each property. Solution: (a) It should have a small A number so that it can thermalize neutrons with relatively few scatters. (b) It should have a small absorption cross section and a large scattering cross section over all neutron energies to avoid nonfuel absorption (to maximize f) and to promote thermalization by scattering interactions. (c) It should be relatively inexpensive and easily incorporated into the core (d) It should not change phase over the anticipated range of core temperatures to avoid the need for expensive pressure vessels. (e) It should be chemically stable to reduce its longevity and prevent possible accident conditions (f) It should not absorb neutrons to produce unwanted radioactivity. 4.
234
U has a half life of 246,000 y so any primordial 234U has long ago decayed away. What is the origin of the 234U found in natural uranium today?
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Solution: Most
234
U comes from the alpha decay of 238 92 U
α
−→
4.47 Gy
234 90 Th
238
β−
−→ 24 d
U via
234 92 Pa
β−
−→
6.7 h
234 92 U
In spent fuel a minor source is from the alpha decay of
238
Pu.
5. From the data in Tables 10.1 and 10.2, what is the Wescott non1/v factor ga (To ) for 235 U at room temperature? Solution: Rearrangement of Eq. (10.5) yields 235
Σ ga (To ) = 235a Σa (Eo )
T To
1/2
σ 235 2 2 592.5 2 √ = 235a √ = √ = 0.9780. π σa (Eo ) π 683.68 π
6. In Eq. (10.6) the relation that VU /VW = 0.6974(N238/NW ) was used for natural uranium. Derive this relationship. Solution: The number of atoms
238
U and number of moleculars of water are given by
N238 = f238
ρU Na VU AU
and
NW =
ρW Na VW AW
Thus the volume ratio is VU ρW AU N238 = . VW f238 ρU AW NW Now calculate the ratio AU f234 A(234 U) + f235 A(235 U) + f238 A(238 U) = AW 2A(1 H) + A(16 O) From the atomic weights in Appendix B, this ratio is found to be AU /AW = (238.02985)/(18.000988) = 13.223155. Finally substitue into the second equation to obtain VU (1.000) N238 N238 = (13.223155) = 0.69737 . VW (0.992745)(19.1) NW NW
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7. What is the thermal fission factor η for natural uranium? Solution: Expansion of Eq. (10.12) and with the abundances and cross sections given in Example 10.1 gives η = ν 235
235 σ 235 f N 234 σ 234 a N
235 + σ 238N 238 + σ 235 a N a 235 U ν 235σ 235 N /N f = 234 234 U 235 /N U + σ 238 N 238/N U σ a N /N + σ 235 a N a 235 ν 235σ f f 235 = 234 234 235 + σ 238f 238 σa f + σ 235 a f a (2.437)(505.9)(0.007204) = = 1.3379. (89.49)(0.000055) + (592.6)(0.007204) + (2.382)(.997245)
8. What is the thermal averaged macroscopic absorption cross section for natural uranium at room temperature? Solution: 235 235 U Σa = (f 234 σ 234 σ a + f 239 σ 239 a +f a )N U 235 235 239 239 ρ Na + f σ + f σ ) = (f 234 σ 234 a a a AU = [(0.000055)(89.49) + (0.007204)(592.6) + (0.992745)(2.382)] × (19.1)(0.602214) = 0.3208. 238.0289
9. What is the thermal fission factor η for 5 atom% enriched uranium? Solution: Cross section data are taken from Table 10.1. From Eq. (10.12) we have F
η=ν
Σf
F
Σa =
235
= ν 235
Σf 235
Σa
238
+ Σa
=
ν 235σ 235 f 238 238 /N 235) σ 235 + σ (N a a
2.4367 × 505.9 = 1.933. 592.6 + 2.382(0.95/0.05)
10. What atom% enrichment of uranium is needed to produce a thermal fission factor of η = 1.85? Ignore any 234 U. Solution:
Principles of Nuclear Reactors
105
Chap. 10
All data are taken from Tables 10.1 and 10.2. The thermal fission factor η for uranium can be calculated as (see Problem 10.7)
η=
ν 235σ 235 f 238 238 /N 235) σ 235 a + σ a (N
.
Solving for r ≡ N 238 /N 235 we find r=
1 σ 238 a
"
# ν 235 × σ 235 2.4367 × 505.9 1 f 235 − σa − 592.6 = 30.96 = η 2.382 1.85
The enrichment e = N 235 /(N 235 + N 238 ) = 1/(1 + r) = 1/(1 + 30.96) = 0.0313 = 3.13%.
11. Plot the thermal fission factor for uranium as a function of its atom% enrichment in 235 U. Solution: By definition, the enrichment is e = N 235/N U = N 235/(N 235 + n238). From this definition N 235 e = . N 238 1−e Then from Eq. (10.12) we have
η(e) =
ν 235σ 235 f 238 238 /N 235) σ 235 a + σ a (N
=
ν 235σ 235 f 238 σ 235 a + σ a (1 − e)/e
.
From the data in Tables 10.1 and 10.2 the following results are obtained e (%) 0.7204 1.0 1.5 2.0 2.5
η(e) 1.3386 1.4880 1.6457 1.7379 1.7982
e (%) 3.0 4.0 5.0 6.0 7.0
η(e) 1.8409 1.8972 1.9326 1.9570 1.9747
e (%) 8.0 10.0 15.0 20.0 30.0
from which the following plot is obtained
η(e) 1.9883 2.0076 2.0339 2.0473 2.0609
e (%) 40.0 50.0 60.0 80.0 100.0
η(e) 2.0677 2.0719 2.0746 2.0781 2.0802
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Principles of Nuclear Reactors
Chap. 10
12. A soluble salt of fully enriched uranium is dissolved in water to make a solution containing 1.5 × 10−3 atoms of 235 U per molecule of water. (a) Explain why p ' 1 for this solution.
(b) What is k∞ for this solution? Neglect any neutron absorption by other elements in the uranium salt. (c) This solution is used to fill a bare spherical tank of radius R. Plot keff versus R and determine the radius of the tank needed to produce a critical reactor. Solution: Needed nuclear data are taken from Tables 10.1, 10.2 and 10.4. (a) Almost all of the neutron absorption during slowing down in a uranium fueled reactor is due to 238 U. Similarly, almost all fast fission is caused by fission neutrons interacting with 238 U. Thus, without any 238 U, p ' 1. (b) From Eq. (10.3) η = ν 235
σ 235 f σ 235 a
= 2.4367
505.9 = 2.080. 592.6
From Eq. (10.2) f=
σ 235 592.6 a = = 0.5723 235 H2 O H O 235 2 592.6 + 0.6644(1/1.5 × 10−3 ) σ a + σ a (N /N )
Hence we have k∞ = pηf ' ηf = (2.080)(0.5723) = 1.190.
Principles of Nuclear Reactors
107
Chap. 10
(c) The critical core buckling for a sphere is Bc2 = (π/R)2 where R is the radius needed to make the core critical, i.e., keff = 1. In terms of the f buckling, the PNL is found from Eq. (10.6) 2
2
f PNL = e−Bc τ = e−27Bc , th and the PNL is obtained from combining Eqs. (10.13) and (10.14) as th PNL =
1 1
+ L2H2 O (1
−
f)Bc2
=
1 1 = . 2 1 + 8.12(1 − 0.5723)Bc 1 + 4.511Bc2
To produce a critical configuration we must have f th keff ≡ k∞ PNL PNL =
1.1903 exp(−27Bc2 ) = 1. 1 + 4.511Bc2
Solve this equation by trial and error for Bc2 or, equivalently, for R. The critical buckling is found to be 5.5382 × 10−3 cm−2 which corresponds to a critical radius of 42.21 cm. 13. Consider a homogeneous, bare, spherical, sourcefree, critical, uraniumfueled reactor operating at a power Po . Explain how and why the power increases, decreases, or remains unchanged as a result of each of the separate changes to the reactor. (a) The reactor is deformed into the shape of a football (ellipsoid). (b) A person stands next to the core. (c) The temperature of the core is raised. (d) A neutron source is brought close to the core. (e) An energetic electron beam impacts the core. (f) The reactor is run at high power for a long time. (g) The core is launched into outer space. (h) A sheet of cadmium is wrapped around the core. (i) The enrichment of the fuel is increased. Solution: (a) k∞ remains unchanged (property of core material alone). But increasing the surfacetovolume ratio causes the neutron leakage to increase and f th hence PNL and PNL decrease and the reactor becomes subcritical and the power decreases. (b) k∞ remains unchanged, but the person acts as a neutron reflector causing the nonleakage probabilities to increase. Hence, the reactor becomes supercritical and the power increases.
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Principles of Nuclear Reactors
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(c) Reactors that are licensed to operate must have a “negative temperature coefficient” so that as the core temperature increases keff must decrease causing the reactor to become subcritical and hence the power to decrease. Increasing the temperature of our hypothetical spherical reactor causes the core to expand. k∞ changes little; but the expanded core has a larger f surface area which allows more neutrons to leak from the core. Hence PNL th and PNL decrease and keff becomes < 1 and the power decreases. (d) A neutron source has no effect on keff ! However, in each neutron cycle it adds extra neutrons that are faithfully maintained by the critical chain reaction. Thus, the neutron population increases in time and the power increases even though keff = 1. (e) Electrons produce bremsstrahlung as they slow down in the core material. These bremsstrahlung photons can have energies up to that of the kinetic energy of the electrons. Hence energetic photons will be produced. These photons can then cause (γ,n) reactions in the core. The photoneutrons then act as a neutron source. As in the previous change, keff is unchanged but the power increases. (f) At high power levels, fuel is consumed (η decreases) and fission products are produced (f decreases) so that keff decreases. Offsetting this effect is the breeding of new 239 Pu fissile fuel from 238 U. However, unless the reactor is a breeder reactor, the former effects dominate and k∞ decreases to less that one and the power will decrease. (g) When the reactor is in space there will be less neutron reflection than when it was on the launch pad where the ground and atmosphere act as f th a (poor) reflector. Thus, in space PNL and PNL decrease, and the reactor becomes subcritical and the power decreases. (h) Although cadmium is a strong thermal neutron absorber, it does have a small scattering cross section. Thus, the cadmium acts as a reflector (although a very poor one) and keff becomes > 1 and the power increases. (i) As the fuel enrichment increases, η increases causing k∞ to increase, thereby making the reactor supercritical. The power thus increases.
14. For a given amount of multiplying material with k∞ > 1, what is the shape of a bare core with the smallest mass of this material? Explain. Solution: For a given material, k∞ is fixed, and the geometry affects only the nonleakage f th probabilities PNL and PNL . These nonleakage probabilities both increase as the critical buckling Bc decreases. For a given volume, Bc2 is smallest for a sphere (see Table 10.6). Thus, a sphere has the smallest critical mass. Alternatiely, for a given volume, a sphere has the smallest surface area and hence the least amount of neutron leakage. Thus, if the sphere is critical, any other shape must have a greater volume (mass) of material to be critical.
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Chap. 10
15. If the uranium fuel enrichment in a reactor is increased, what is the effect on k∞ ? Explain. Solution: As the uranium enrichment increases, only the thermal fission factor η is affected. The thermal fission factor is
η=
ν 235σ 235 f 238 238 /N 235) σ 235 a + σ a (N
.
As the enrichment increases, the ratio N 238/N 235 decreases, η increase, and k∞ increases.
16. Consider a homogeneous mixture of fully enriched 235 U and graphite. Plot k∞ versus N 235/N C . What is the fueltomoderator ratio that yields the maximum value of k∞? Solution: For fully enriched uranium, there is no 238U in the fuel. As a consequence, p ' 1 since 238 U is responsible for almost all fast fission and for most of the absorption of neutrons as they slow down. For this fully enriched material
k∞ =
ν 235σ 235 f C σ 235 a + σ a (1/r)
,
235 C where r is the fueltomoderator ratio, √ √ i.e., r ≡ N /N . From the data in C C Table 10.4 σ a = ( π/2)σA (Eo ) = ( π/2)0.003861 = 0.003422 b.
k∞ =
2.4367 × 505.9 . 592.6 + 0.003422(1/r)
(P10.1)
By inspection of Eq. (P10.1), the maximum value of k∞ is obtained for a fully rich fueltomoderator mixture, i.e., r becomes very large. This value is (k∞ )max = 2.080. The plot below shows how keff varies with r.
1010
Principles of Nuclear Reactors
Chap. 10
NOTE: This is poorly conceived problem. The assumption p ' 1 is valid only for dilute mixtures, r