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: jks@ksu.edu
Revised June 2016
(c) Copyright 2007-2016
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.
1-1
1-2
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/milli-Curie.
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
1-3
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
1-4
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 g/mol
12
=
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 post-1961 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
carbon-12 in its rest-energy 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. Pre-1961, 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
1-5
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
1-6
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 non-iron 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
1-7
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 atom-percent 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
1-8
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 atom-percent in 235 U with the remainder
being 238 U. What is the enrichment of 235U in weight-percent?
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
1-9
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)
!
From Eq. (2.5) in the text m = mo / 1 − v2 /c2 , which can be solved for v/c
to give
"
v
m2
1 m2o
mo
= 1 − o2 " 1 −
, if
<< 1.
(P2.2)
c
m
2 m2
m
(a) For an electron (mo = me ) with T = 108 eV = 100 MeV, Eq. (P2.1) gives
m=
100 MeV
+ me = 0.1074 u + 0.0005486 u = 0.1079 u.
931.5 MeV/u
Then m2e /m2 = (0.0005486/0.1079)2 = 2.59×10−5. Finally, from Eq. (P2.2)
above, we obtain
v
1 m2o
"1−
= 1 − 1.29 × 10−5 = 0.999987.
c
2 m2
(b) For an electron with T = 109 eV = 1000 MeV, we similarly obtain m =
1.0741 u and v/c = 0.99999987.
(c) For an electron with T = 1010 eV = 104 MeV, we similarly obtain m =
10.736 u and v/c = 0.9999999987.
Alternative solution: Use Eq. (P2.4) developed in Problem 2-3, namely
#
$
%2 &1/2
v
me c2
= 1−
.
c
T + me c2
2-1
2-2
Modern Physics Concepts
Chap. 2
2. Consider a fast moving particle whose relativistic mass m is 100! percent greater
than its rest mass mo , i.e., m = mo (1 + !). (a) Show that the particle’s speed
v, relative to that of light, is
!
v
1
= 1−
.
c
(1 + !)2
(b) For v/c << 1, show that this exact result reduces to v/c "
√
2!.
Solution:
(a) We are given
m − mo
mo ((1 + !) − 1)
=
= !.
mo
mo
But we also have
"
m − mo
1
=
mo
mo
Equating these two results yields
!=#
#
mo
1 − v2 /c2
$
− mo .
1
− 1.
1 − v2 /c2
Solving this result for v/c gives
v
=
c
!
1−
1
.
(1 + !)2
(P2.3)
(b) For ! << 1 we have (1 + !)−2 " 1 − 2! + · · · . Substitution of the approximation into Eq. (P2.3) above gives
√
v #
" 1 − (1 − 2!) = 2!.
c
3. In fission reactors one deals with neutrons having kinetic energies as high as
10 MeV. How much error is incurred in computing the speed of 10-MeV neutrons by using the classical expression rather than the relativistic expression
for kinetic energy?
Solution:
A neutron with rest mass mn = 1.6749288 × 10−27 kg has a kinetic energy
T = (107 eV)(1.602177 × 10−19 J/eV) = 1.602177 × 10−12 J. For the neutron
mn c2 = 939.56536 MeV.
Classically:
vc =
%
&1/2
#
2 × 1.602177 × 10−12
2T /mn =
= 4.373993 × 107 m/s.
1.6749288 × 10−27
Modern Physics Concepts
2-3
Chap. 2
Relativistically: From the text we have
T = mc2 − mo c2 = !
mo c2
1 − v2 /c2
− mo c2 .
Solving this equation for v yields the relativistic speed vr
"
#
mo c2
vr = c 1 −
T + mo c2
$2 %1/2
.
(P2.4)
Substitution then gives
"
#
939.56536
vr = c 1 −
10 + 939.56536
$2 %1/2
= 0.1447459c = 4.339373 × 107 m/s.
Thus the percent error in the classical speed is = 100(vc − vr )/vr = 0.798%.
4. What speed (m s−1 ) and kinetic energy (MeV) would a neutron have if its
relativistic mass were 10% greater than its rest mass?
Solution:
We are given (m − mo )/mo ≡ ! = 0.1. From Problem 2-2
&
'
v
1
1
= 1−
= 1−
= 0.4167.
2
c
(1 + !)
1.12
Thus the neutron’s speed is v = 0.4167c = 1.25 × 108 m/s.
The kinetic energy can be calculated from
(
2
2
T = mc − mo c = mo c
2
)
1
!
−1 .
1 − v2 /c2
For mo c2 = 939.6 MeV and v/c = 0.4167 we obtain
#
$
1
T = 939.6 √
− 1 = 94.0 MeV.
1 − 0.41672
5. Show that for a relativistic particle the kinetic energy is given in terms of the
particl’s momentum by
!
T = p2 c2 + m2o c4 − mc c2 .
Solution:
Squaring Eq. (2.17) and rearranging the terms one obtains
T 2 + 2T mo c2 − p2 c2 = 0
2-4
Modern Physics Concepts
Chap. 2
The solution of this quadratic equation gives
#
"
1!
T =
−2mo c2 ± 4m2o c4 + 4p2 c2
2
Only the + sign gives a physically meaningful result. Rearrangement gives the
desired realtion.
6. For a relativistic particle show that Eq. (2.21) is valid.
Solution:
From the definition of η one has
η 2 +1 =
P2
p 2 c2
(mc2 )2 − (mo c2 )2
+1 =
+1 =
+1 = (W 2 −1)+1 = W 2 .
2
2
2
(mo c)
(mo c )
(mo c2 )2
7. Prove the relationships given in (a) Eq. (2.19), (b) Eq. (2.20), and (c) Eq. (2.21).
Solution:
(a) From the definition of η and W one immediately has
β=
v
p
η
=
=
.
c
mc
W
(b) Because W 2 = 1 + η 2 , then
β2 =
$ v %2
c
=
η2
η2
=
.
W2
1 + η2
(c) Because β = η/W and W 2 = 1 + η 2 , one has
β2
η 2 /W 2
η 2 /(1 + η 2 )
η2
=
=
=
= η2 .
2
2
2
2
2
1−β
1 − η /W
1 − η /(1 + η )
(1 + η 2 ) − η 2
From this result we see
β2
p2
c2 p 2
= 2 2 =
,
2
1−β
mo c
(mo c2 )2
but we know p2 c2 = T 2 + 2T mo c2 , so
β2
T 2 + 2T mo c2
=
=
2
1−β
(mo c2 )2
&
T
mo c2
'2
+
2T
=
mo c2
&
T
mo c2
'2 &
'
2mo c2
1+
.
T
Modern Physics Concepts
Chap. 2
2-5
8. In the Relativistic Heavy Ion Collider, nuclei of gold are accelerated to speeds
of 99.95% the speed of light. These nuclei are almost spherical when at rest;
however, as they move past the experimenters they appear considerably flattened in the direction of motion because of relativistic effects. Calculate the
apparent diameter of such a gold nucleus in its direction of motion relative to
that perpendicular to the motion.
Solution: The relativistically contracted diameter D to the uncontracted diameter Do when v/c = 0.9995 is
!
!
!
D/Do = 1 − v2 /c2 = 1 − 0.99952 = 1 − (1 − 0.0005)2
!
√
" 1 − (1 − 2 × 0.0005) = 0.001 = 0.031.
Hence the gold nucleus appears to flatten to 3.1% of its at-rest width.
9. Muons are subatomic particles that have the negative charge of an electron
but are 206.77 times more massive. They are produced high in the atmosphere
by cosmic rays colliding with nuclei of oxygen or nitrogen, and muons are
the dominant cosmic-ray contribution to background radiation at the earth’s
surface. A muon, however, rapidly decays into an energetic electron, existing,
from its point of view, for only 2.20 µs, on the average. Cosmic-ray generated
muons typically have speeds of about 0.998c and thus should travel only a
few hundred meters in air before decaying. Yet muons travel through several
kilometers of air to reach the earth’s surface. Using the results of special
relativity explain how this is possible. HINT: consider the atmospheric travel
distance as it appears to a muon, and the muon lifetime as it appears to an
observer on the earth’s surface.
Solution:
Muon’s Point of View: A muon, with a lifetime to = 2.20 × 10−6 s and
traveling with a speed v = 0.998c, travels on the average a distance d = vto =
0.998(3.00 × 108 m/s)(2.29 × 10−6 s) = 660 m.
If the muon is created at an altitude Lo , from the muon’s point of view the
distance to the surface (approaching with speed v = 0.998c) is relativistically
narrowed or contracted to a distance
!
!
L = Lo 1 − v2 /c2 = Lo 1 − 0.9982 = 0.063Lo .
For example, if Lo = 10 km, L = 630 m, so that, on the average, almost half
of the muons will reach the surface.
Surface Observer’s Point of View: An observer on the earth’s surface
observes the muon approaching at a speed v = 0.998c and the muon’s lifetime
appears to expand (the muon’s internal clock appears to slow) as
to
to
t= !
= √
= 15.9to = 3.49 × 10−5 s.
1 − 0.9982
1 − v2 /c2
In such a lifetime, the muon can travel d = 0.998c × t = 10, 500 m so that it
can reach the surface from an altitude of 10 km before decaying.
2-6
Modern Physics Concepts
Chap. 2
10. A 1-MeV gamma ray loses 200 keV in a Compton scatter. Calculate the scattering angle.
Solution:
From Eq. (2.26) in the text we find
1 − cos θs = me c2
or
cos θs = 1 − me c2
!
!
1
1
−
E!
E
"
"
1
1
−
.
E!
E
Here me c2 = 0.511 MeV, E ! = 0.8 MeV, and E = 1 MeV so that
!
"
1
1
cos θs = 1 − 0.511
−
= 0.87225.
0.8 1
Thus the scattering angle θs = cos−1 (0.87225) = 29.3o
11. At what energy (in MeV) can a photon lose at most one-half of its energy in
Compton scattering?
Solution:
Eq. (2.26) in the text gives the basic Compton scattering relation:
1
1
1
−
=
(1 − cos θs ).
E!
E
me c2
By inspection, the maximum energy loss (the smallest E ! ) occurs when θs = π.
Here we are told E ! = E/2
2
1
1
2
2
−
=
=
=
.
E
E
E
me c2
0.511 MeV
From this result, we find E = 0.255 MeV. Above this incident photon energy,
the minimum scattered photon energy is less than one-half of the initial energy.
12. Derive for the Compton scattering process the recoil electron energy T as a
function of the incident photon energy E and the electron angle of scattering
φe . Show that φe is never greater than π/2 radians.
Solution:
Application of the law of cosines to the triangle in text Fig. 2.5 leads to
pλ! 2 = pλ 2 + pe 2 − 2pλ pe cos φe .
√
Substitute E/c for pλ , (E − T )/c for pλ! , and (1/c) T 2 + 2T me c2 for pe . Then
solve for T , with the result
T =
2me c2 E 2 cos2 φe
.
(E + me c2 )2 − E 2 cos2 φe
Modern Physics Concepts
2-7
Chap. 2
Examination of the triangle in Fig. 2.5 reveals that, since pλ! ≤ pλ , 0 ≤ φe ≤
π/2, confirming the commonsense observation that the target electron, initially
at rest, can recoil only in the forward hemisphere.
13. A 1 MeV photon is Compton scattered at an angle of 55 degrees. Calculate
(a) the energy of the scattered photon, (b) the change in wavelength, and (c)
the recoil energy of the electron.
Solution:
(a) From Eq. (2.26)
1
1
1 − cos θs
1
1 − cos 55
=
+
=
+
= 1.835 MeV−1 .
E!
E
me c2
1 MeV 0.511 MeV
Thus the scattered photon energy is E ! = 1/1.835 = 0.545 MeV.
(b) From Eq. (2.25) we have
∆λ = λ! − λ =
=
h
hc
(1 − cos θs ) =
(1 − cos θs )
me c
me c2
(4.135 × 10−21 MeV s)(3.00 × 108 m/s)
(1 − cos 55)
0.511 MeV
= 1.04 × 10−12 m.
(c) The kinetic energy of the recoil electron is Er = E − E ! = 1 − 0.545 =
0.455 MeV.
14. When light with wavelengths > 475 nm = λmax impinges on of a certain metalic
surface photoelectrons are observed to be emitted. What is the work functiion
of this metal in eV?
Solution:
The frequency of light corresponding the the maximum wavelgth is νmin =
c/λmax = (2.998 × 108 m s−1 /(475 × 10−9 m) = 6.31 × 1014 s−1 . From Example
2.3, the work function is A = hνmin = (4.136 × 10−15 eV s)(6.31 × 1014 s−1 ) =
2.61 eV.
2-8
Modern Physics Concepts
Chap. 2
15. Consider the experimental arrangement shown in Fig. 2.3. The surface of a
sodium sample was illuminated by monochromatic light of various wavelengths,
and the retarding potentials required to stop the collection of the photoelectrons
were observed. The results are shown below.
wavelemgth (nm)
retarding potential (V)
253.6
2.60
283.0
2.11
303.9
1.81
330.2
1.47
366.3
1.10
435.8
0.57
Present these data graphically to verify the photoelectric equation eVo = hν −
A. From the graph estimate the value of Planck’s constant h and the work
function A for sodium.
Solution:
The frequency of the light is related to the wavelength by
[ν =
c
2.997 × 1017 −1
=
s .
λ
λ (nm)
Then plot the following data:
eVo (eV)
ν × 10−14
2.60
11.82
2.11
10.59
1.81
9.682
1.47
9.076
1.10
8.182
0.57
6.877
Fit a straight line to the plotted data as shown below.
From the least-squares fit it is found that h = 4.142 × 10−15 eV s and that
the work function for sodium is A = 2.271 eV.
Modern Physics Concepts
2-9
Chap. 2
16. Consider the electron scattering experiment of Davisson and Germer described
in Section 2.2.4. For the nickel crystal they used the interatomic spacing was
d = 2.15 Å = 2.15 × 10−10 m. (a) For an incident electrons with an arbitrary
energy of T eV, show that the constructive interference peaks occur at angles
! "
!
"
nλ
5.705n
−1
−1
√
θ = sin
= sin
,
n = 1, 2, 3, . . ..
d
T eV
(b) What are the angles of the peaks when T = 54 eV (as used by Davisson
and Germer) and when T = 300 eV?
Solution:
√
(a) From Eq. (2.30) for non-relativistic electrons λ = h/ 2me T . Recall the
rest mass of the electron is me /c2 = 5.11 × 106 eV. Substitution of of these
values gives
!
"
nhc
√
θ = sin−1
d 2me T
#
%
n(4.136 × 10−15 eV s)(2.998 × 108 m s−1 )
−1
$
= sin
(2.15 × 10−10 m) (2 × 0.555 × 106 eV)(T eV)
= sin−1
!
5.705n
√
T eV
"
.
(P2.5)
(b) For T = 54 eV the only angle is θ = 50.9◦ (n = 1). For T = 300 eV the
angles are θ = 19.2◦ (n = 1), 41.2◦ (n = 2), and81.2◦ (n = 3).
17. Show that the de Broglie wavelength of a particle with kinetic energy T can be
written as
&
'−1/2
h
1
m
√
λ= √
1+
mo T
mo
where mo is the particles’s rest mass and m is its relativistic mass.
Solution: From Eq. (2.17)
p=
√
1$ 2
T$
T + 2T mo c2 =
T + 2mo c2 .
c
c
But T = mc2 − mo c2 so the above result can be written as
√
√ √ $
T$ 2
p=
mc + mo c2 = T mo 1 + (m/mo ).
c
Finally, use of the de Broglie relation λ = h/p in the above result gives
&
'−1/2
h
1
m
√
λ= √
1+
.
mo T
mo
2-10
Modern Physics Concepts
Chap. 2
18. Apply the result of the previous problem to an electron. (a) Show that when the
electron’s kinetic energy is expressed in units of eV, its de Broglie wavelength
can be written as
!
"−1/2
17.35 × 10−8
m
√
λ=
1+
cm.
mo
T
(b) For non-relativistic electrons, i.e., m # mo , show that this result reduces
to
12.27 × 10−8
√
λ=
cm.
T
(c) For very relativistic electrons, i.e., m >> mo , show that the de Broglie
wavelength is given by
#
17.35 × 10−8 mo
√
λ=
cm.
m
T
Solution:
(a) Rewrite the result of Problem 2-10 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
√
$
0.5110 × 106 eV
T (eV)
!
"−1/2
17.35 × 10−8
m
$
1+
cm.
mo
T (eV)
(P2.6)
$
√
(b) For non-relativistic electrons m # mo , so that 1/ 1 + (m/mo ) # 1/ 2,
and the above result becomes
λ=
12.27 × 10−8
$
cm.
T (eV)
$
$
(c) For very relativistic particles, m >> mo so that 1/ 1 + (m/mo ) # mo /m.
Eq. (2.4) above then becomes
$
17.35 × mo /m
$
λ=
× 10−8 cm.
T (eV)
Modern Physics Concepts
2-11
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)
!
= 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)
!
= 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)
!
= 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 (Å) with 1 Å= 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-Å neutron.
Solution:
(a) Equation (2.30) for a non-relativistic particle reduces to
!
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 =
2-12
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 3-g 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 3-g 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
h
6.626 × 10−34 J s
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 =
h
6.626 × 10−34 J s
=
= 1.05 × 10−20 J s m−1 .
2π∆x
2π × 10−14 m
A non-relativistic (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
2-13
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
!
!
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) = !
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
" ∞
" ∞
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
" ∞
4
4 2
4 a3
r 2 e−αr dr = 3 3 = 3 0 = 1.
3
a0 0
a0 α
a0 4
2-14
Modern Physics Concepts
Chap. 2
(b) Replace upper limit in the above itegral by a0 . Then integration by parts
twice gives
! 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
"
#
4 −2 a30
2a30
2a30
= 1− 3e
+
+
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 =
hc
λ2→1
= 10.198714 eV
E3→1 =
hc
= 12.087364 eV
λ3→1
E4→1 =
hc
= 12.748392 eV
λ4→1
3-1
3-2
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.
8!2o h2
(c) To free this ground-state 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
.
8!2o 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
−
= 1.635 × 10−18 J = 10.20 eV.
8!o h 22
12
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
3-3
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 "3-5.out" FOR OUTPUT AS #1
3-4
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 Ca-40
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 Pb-208
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
3-5
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
8!2o 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 3-7
CLS
fmt$ = " ### ###.# ###.# ##.### ##.### ##.### ##.### ###.###"
OPEN "3-7.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
3-6
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
beta-decay 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
Kr,
Br,
Se,
As,
36
35
34
33
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
3-7
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) Matom-70
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.
3-8
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
M-183.9 (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
230
(?) −→ 90 Th + 42 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.
4-1
4-2
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
4-3
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 Q-value 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
92B U. Thus by splitting (or fissioning) a heavy atom like
235
92 U the nucleons become more tightly bound. The differences in the binding
energies is emitted exoergically.
4-4
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
4-5
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
non-zero 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.
4-6
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 Q-values of the reactions shown in Section 4.4. With
these formulas, evaluate the Q-values.
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 Q-value (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 −→
 84 Be + 21 H




7
3


4 Be + 1 H


 6
4
3 Li + 2 He
Solution:
Consider the 9 Be(p,γ)10 B reaction. The Q-value 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
4-7
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
Q-value (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
 He + 2 H

2
1


 3
4
1 H + 2 He
What is the Q-value (in MeV) for each reaction?
Solution: Results are summarized in the table below.
Reaction
6
7
Li(n,γ) 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
Q-value (MeV)
type
7.250
0.0
−2.725
−2.361
4.783
exoergic
endoergic
endoergic
exoergic
14. Calculate the Q-values 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
22
22
0
0
11 Na −→ 10 Ne + −1 e + +1 e + ν. The Q-value 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
4-8
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 Q-value 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 Q-value
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 Q-value 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 + ν̄,
where the superscript + indicates that the Ni atom is produced as a positive
ion. The energy of excitation E ∗ = 2.205 MeV. Calculate the Q-value 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 +
ν̄,
or
60
27 Co
−→ 60
28 Ni* + ν̄.
Then the Q-value 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
27
(?) −→ 13 Al + +10 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
β − decay
38
16 S
β
+
decay
4
−→ 206
82 Pb + 2 He
−→ 38
17 Cl +
27
14 Si
−→
27
13 Al
+
0
−1 e
+ νe
0
+1 e
+ νe
electron capture
145
62 Sm
−→ 145
61 Pm + νe
neutron decay
∗
137
54 Xe
proton decay
108
52 Te
∗
1
−→ 107
51 Sb + 1 p
internal conversion
60 ∗
28 Ni
−→ 60
28 Ni +
1
−→ 136
54 Xe + 0 n
0
−1 e
(b) The total kinetic energy released is the Q-value 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.
5-1
5-2
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
!"
2E ∗
1+
−1
mn c2
#
$
" E∗ 1 −
%
E∗
.
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 ground-state
daughter nucleus must travel in opposite directions each with the same
amount of linear momentum. The Q-value of the decay reaction is E ∗
and must be equal to the sums of the kinetic energies of the photon and
ground-state nucleus. Hence it follows that Eγ < E ∗ , the difference equal
to the kinetic energy of the recoil ground-state 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 −
&
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
5-3
Chap. 5
Solving this quadratic equation for Eγ produces
!
#
"
2E ∗
2
Eγ = mn c −1 ± 1 +
.
mn c2
Only the + sign yields a positive real value for Eγ so that
Eγ = mn c2
$"
%
∗
1 + 2E 2 − 1
mn c
Since E ∗ is typically much less than mn√
c2 " thousands of MeV, then
defining ! ≡ (2E ∗ )/(mn c2 ) << 1, we have 1 + ! = 1 + (1/2)! − (1/8)!2 +
· · ·, and neglecting terms of order !2 and higher the above result can be
approximated by
&
'
∗
Eγ " E ∗ 1 − E 2 .
(P5.3)
2mn c
(c) Consider the decay of the first excited state of 60 Co (see Fig. 5.12) which
is E ∗ = 1.17 MeV above the ground state. The energy of the emitted
gamma ray upon deexcitation is, from Eq. (P5.3),
&
'
E∗
∗
Eγ " E 1 −
2M (60 Co)c2
&
= 1.17 MeV 1 −
1.17 MeV
2 × 60 × 936 MeV
'
= 1.17 [1 − 0.000010] MeV " 1.17 MeV.
The recoil kinetic energy of the ground-state 6 Co nucleus is
E(60 Co) = E ∗ − Eγ = 1.17 [0.000010] = 0.0117 keV.
3. What are the kinetic energies of each of the decay products in the decay
238
U −→ 234 Th + 4 He?
Solution:
The Q-value is found from Eq. (5.7) to be Qα = 4.266 MeV. Then from
Eq. (5.11)
E α = Qα
AD
234
= 4.266
= 4.194 MeV.
AD + Aα
234 + 4
Then ED = Qα − Eα = 0.072 MeV.
5-4
Radioactivity
Chap. 5
4. The radioisotope 224Ra decays by α emission primarily to the ground state of
220
Rn (94% probability) and to the first excited state 0.241 MeV above the
ground state (5.5% probability). What are the energies of the two associated
α particles?
Solution:
For the daughter left in an excited state, the Qvalue of Eq. (5.7) is modified to increase the rest
∗ 2
mass of M (220
86 Rn) by an amount E /c , namely
!
4
Qα1 = M (224
88 Ra) − [M (2 He)
" 2
∗ 2
+M (220
86 Rn) + E /c ] c .
With the masses in Ap. B we find
Qα1 = {224.0202020 − [4.00260325 + 220.0113841]} (u) × 931.5 (MeV/u)
−0.241 MeV = 5.548 MeV.
The kinetic energy of α1 is given by Eq. (5.11), namely
#
$
#
$
M (220
220
86 Rn)
E α 1 = Qα 1
#
5.548
= 5.449 MeV.
4
224
M (220
82 Rn) + M (2 He)
In a similar manner for alpha decay to the ground state (E ∗ = 0), we have
!
" 2
4
220
Qα2 = M (224
88 Ra) − [M (2 He) + M ( 86 Rn)] c
= {224.0202020 − [4.00260325 + 220.0113841]}931.5 (MeV/u)
= 5.789 MeV.
The kinetic energy of this alpha particle is
#
$
#
$
M (220
220
86 Rn)
E α 2 = Qα 2
#
5.789
= 5.686 MeV.
4
224
M (220
82 Rn) + M (2 He)
5. A particular radionuclide decays by α decay and emits an α particle with energy
4.7844 MeV and leaves the daughter nucleus in the ground state. The Q-value
for this decay is Qα = 4.8760 MeV. What is the atomic mass number of the
parent radionuclide?
Solution: First find the mass number of the daughter. Solve Eq. (5.11) for
AD to obtain
4Eα
AD =
= 222.
Qα − E α
Then the atomic mass number of the parent is AP = AD + 4 = 226.
Radioactivity
5-5
Chap. 5
6. The radionuclide 41 Ar decays by β − emission to an excited level of 41 K that
is 1.293 MeV above the ground state. What is the maximum kinetic energy of
the emitted β − particle?
Solution:
First find the Qβ− for this decay from Eq. (5.15)
with the atomic masses in Ap. B. The result is
!
" 2
∗ 2
41
Qβ− = M (41
18 Ar) − [M (19 K) + E /c ] c
= [40.9645008 − 40.96182597] × 931.5
−1.293 MeV
= 1.199 MeV.
The maximum kinetic energy the beta particle can have equals this Q-value,
i.e., (Eβ )max = 1.199 MeV.
7. As shown in Fig. 5.6, 64 Cu decays by several mechanisms. (a) List the energy
and frequency (number per decay) of all photons emitted from a sample of
material containing 64 Cu. (b) List the maximum energy and frequency of all
electrons and positrons emitted by this sample.
Solution:
64
29Cu
(12.7 h)
β- (39.0%)
Qβ- = 578.7
EC (0.5%)
1345.8 keV
EC (43.1%)
β+ (17.4%)
Qβ+ = 653.1
γ (0.5%)
64
28Ni
64 Zn
30
(stable)
(stable)
(a) From the above decay scheme, we see that a 1.3458-MeV gamma ray is
emitted with a frequency of 0.005. Although not emitted in the decay
process, there are also annihilation photons produced in the source material arising from the annihilation of the emitted positrons. Each positron
annihilates an ambient electron producing two gamma photons each of
energy 0.511 MeV. These annihilation photons appear with a frequency
of 2 × 0.174 = 0.348 per decay.
(b) Beta particles are emitted with a maximum energy equal to the Q-value of
the decay. Thus, we have β − particles with maximum energy 0.5787 MeV
and frequency 0.390, and β + particles with maximum energy 0.6531 MeV
and frequency 0.174.
Eγ MeV
Freq. (%)
(Eβ )max MeV
Freq. (%)
1.3458
0.511
0.005
0.348
0.5787
0.6531
0.390
0.174
5-6
Radioactivity
Chap. 5
8. What is the value of the decay constant and the mean lifetime of 40 K (half-life
1.29 Gy)?
Solution:
The decay constant is
λ=
ln 2
ln 2
=
= 5.37 × 10−10 y−1 .
T1/2
1.29 × 109 y
The average lifetime is t = 1/λ = 1.86 × 109 y.
9. From the energy level diagram of Fig. 5.12, what are the decay constants for
electron capture and positron decay of 22 Na? What is the total decay constant?
Solution:
The total decay constant is
λ=
ln 2
ln 2
=
= 0.2664 y−1 = λβ+ + λEC .
T1/2
2.602 y
From Fig. 5.12, the probability of β + decay is 0.8984 +0.00006 = 0.8990. Thus,
the decay constant for positron decay is λβ+ = 0.8990 × λ = 0.2395 y−1 .
From Fig. 5.12, the probability of electron capture is 0.101. Thus the decay
constant for positron decay is λβ+ = 0.101 × λ = 0.0269 y−1 . The total decay
constant is λ = λβ+ + λEC = 0.2664 y−1 .
10. The activity of a radioisotope is found to decrease by 30% in one week. What
are the values of its (a) decay constant, (b) half-life, and (c) mean life?
Solution:
(a) From the radioactive decay law A(t)/A(0) = exp(−λt), upon solving for t,
the time required to reach a specified value of A(t)/A(0) is
t=−
1
ln [A(t)/A(0)] .
λ
Here we are given A(t = 1 wk)/A(0) = 0.7. From the above result we find
λ=−
1
ln(0.7) = 0.357 wk−1 = 0.0510 d−1
1 wk
= 2.12 × 10−3 h−1 = 5.90 × 10−7 s−1 .
(b) T1/2 = ln 2/λ = 1.18 × 106 s = 326.5 h = 13.6 d = 1.94 wk.
(c) t = 1/λ = 1.69 × 106 s = 471 h = 19.6 d = 2.80 wk.
Radioactivity
5-7
Chap. 5
11. The isotope 132 I decays by β − emission to 132Xe with a half-life of 2.3 h. (a)
How long will it take for 7/8 of the original number of 132 I nuclides to decay?
(b) How long will it take for a sample of 132I to lose 95% of its activity?
Solution:
(a) After 3 half-lives, the activity is 1/23 = 1/8 of the original activity. Thus
the time to obtain 1/8 of the original activity is t = 3 × T1/2 = 6.9 h.
(b) From the radioactive decay law A(t)/A(0) = exp(−λt), upon solving for t,
the time required to reach a specified value of A(t)/A(0) is
!
"
1
A(t)
t = − ln
.
λ
A(0)
In this problem λ = ln 2/T1/2 = 0.3014 h−1 and A(t)/A(0) = 0.05 Thus
from the above equation we have
t=−
12. How many grams of
32
1
ln(0.05) = 9.94 h.
0.3014 h−1
P are there in a 5 mCi source?
Solution:
Since the activity A is related to the mass m of a radionuclide source by
mNa
,
A
and, hence, the mass of the radionuclide is
A ≡ λN = λ
m (g) =
AA
.
λNa
(P5.4)
From Ap. A the half-life of 32 P is 14.28 d so that λ = ln 2/T1/2 = 5.618 ×
10−7 s−1 . Then from Eq. (P5.4) we find the 32 P mass is
m(32 P) =
(5 × 10−3 Ci × 3.7 × 1010 Bq/Ci)(32 g/mol)
(5.618 × 10−7 s−1 )(6.022 × 1023atoms/mol)
= 1.75 × 10−8 g = 17.5 ng.
13. How many atoms are there in a 1.20 MBq source of (a)
24
Na and (b)
238
U?
Solution:
Because A ≡ λN we have N (atoms) = A(Bq)/λ(s−1 ).
(a) For 24 Na we find from Table A.4 that T1/2 = 14.96 h = 5.385 × 104 s.
Then λ = ln 2/T1/2 = 1.287 × 10−5 s−1 Thus the number of atoms of 24 Na
for a 1.20 MBq source is
N=
A
1.200 × 106
=
= 9.32 × 1010 atoms.
λ
1.287 × 10−5
5-8
Radioactivity
Chap. 5
(b) For 238 U we find from Table A.4 that T1/2 = 4.468×109 y = 1.410×1017 s.
Then λ = ln 2/T1/2 = 4.916 ×10−18 s−1 Thus the number of atoms of 238 U
for a 1.20 MBq source is
N =
A
1.200 × 106
=
= 2.44 × 1023 atoms.
λ
4.916 × 10−18
14. A very old specimen of wood contained 1012 atoms of 14 C in 1986. (a) How
many 14 C atoms did it contain in 9474 B.C.? (b) How many 14 C atoms did it
contain in 1986 B.C.?
Solution:
For 14 C the half-life is 5730 y so that its decay constant is λ = ln 2/T1/2 =
1.210 × 10−4 y−1 . From the radioactive decay law N (t) = N (0) exp[−λt]. In
this problem we identify the year 1986 as t = 0 with earlier times being negative.
(a) For t = −1986 − 9474 = −11, 460 y, the number of
specimen at 9474 BC is
14
C atoms in the
N (−11, 460 y) = 1012 exp[+1.210×10−4 ×11, 460] = 4.00 × 1012 atoms.
(b) For t = −1986−1985 = −3972 y, the number of 14 C atoms in the specimen
at 1986 BC is
N (−3972 y) = 1012 exp[+1.210 × 10−4 × 3972] = 1.62 × 1012 atoms.
15. The following count rate data from a radioactive isotope were measured.
t (min)
3.0
6.0
9.0
12.0
18.0
cnts/min
9302
7932
6901
5853
4397
t (min)
24.0
30.0
36.0
42.0
48.0
cnts/min
3200
2268
1679
1277
962
t (min)
54.0
60.0
66.0
72.0
78.0
cnts/min
693
477
366
277
165
From a semilog plot of these data estimate (a) the decay constant for the
radioisotope, (b) the half-life, and (c) the count rate at t = 0. Explain why all
the data points do not lie exactly on the straight as predicted by Eq. (5.45).
Solution:
The resulting plot is shown below.
Radioactivity
5-9
Chap. 5
Method 1: The mean count rate at time t is
CR(t) = CR(0)e−λt
from which
λ=
ln CR(t1 ) − ln CR(t2 )
.
t2 − t1
Pick t1 = 72 min and t2 = 3 min (or any other pair of times), and find the
count rates at these times on the line through the data. The decay constant is
evaluated as
ln 9350 − ln 270
λ"
= 0.0514 min−1 .
72 − 3
The half-life is then T1/2 = ln 2/λ = 13.5 min. The initial count rate is
found from extrapolation of the straight line through the data to be CR(0) "
11, 000 counts/min.
Method 2: A better method is to fit a line through the data by the least
squares method. Then taking the logarithm, in base 10, of the exponential
decay formula, one finds
log10 CR(t) = log10 CR(0) − [λ log10 e]t.
A least squares fit to the data gives log10 CR(t) = 4.045 − 0.02264t, from
which the initial count rate is CR(0) = 104.045 = 11, 092 counts/min and
the decay constant is λ = 0.02264/ log10 e = 0.5213 min−1 . The half-life is
then T1/2 = ln 2/λ = 13.30 min.
The data do not lie exactly on the straight line because radioactive decay is a
stochastic or random process. Equation (5.45) and the exponential decay law
describe only the average or expected number of atoms or activity at time t.
5-10
Radioactivity
Chap. 5
16. The following count rates from a radioactive sample that contains two radionuclides were measured.
t (min)
cnts/min
t (min)
cnts/min
t (min)
cnts/min
0.0
3.0
6.0
9.0
12.0
18.0
29770
19097
12548
8600
6189
3346
24.0
30.0
36.0
42.0
48.0
54.0
2155
1438
1048
802
592
463
60.0
66.0
72.0
78.0
334
256
183
167
From a semilog plot of these data estimate (a) the decay constants for the
radioisotopes, (b) their half-lives, and (c) their initial count rates at t = 0.
Solution:
The count rate data are plotted as cicles as shown below. The analysis is done
in two steps.
Step 1: First fit a least-squares line to the data that has negligible contributions from the faster decaying radionuclide, here the last six data points are
used. Then use “method 2” of the previous problem, to analyse the resulting
fit to the slower decaying radionuclide (here identified by the subscript 2). The
fit (shown by the solid line through the last six data points) is
log10 CR2 (t) = 3.698 − 0.01939t.
Radioactivity
5-11
Chap. 5
Thus, CR2 (0) = 103.698 = 4989 counts/min. Then, λ2 = 0.01939/ log10 e =
0.04465 min−1 and (T1/2 )2 = ln 2/λ2 = 15.52 min.
Step 2: Subtract the contribution of radionuclide 2 from the given data, i.e.,
CR1 (t) = CRmeas (t) − CR2 (0) exp(−λ2 t).
These values are shown as squares on the plot. Then, fit a least-squares
line to these data that yields CR1 (t) = 4.397 − 0.07477t. Then CR1 (0) =
104.397 = 24, 946 counts/min, and λ1 = 0.07477/ log10 e = 0.1722 min−1 .
The halflife of the faster decaying component is then (T1/2 )1 = ln 2/λ1 =
ln 2/0.1722 = 4.03 min.
17. For radioisotopes with half-lives of many years, it is impractical to measure
the exponential decay curve. For such radionuclides one must use the absolute
method for finding the half-live. To demonstrate this method, consider a 1-g
sample of pure 235 U. In a one-hour count, it is inferred that sample experienced
∆N = 2.88 × 108 decays. From this information, estimate the half-life of 235U.
Solution:
In the one-gram 235 U sample there are about N = Na /A = 6.022 × 1023/235 =
2.56 × 1021 atoms. From the fundamental definition of the decay constant, one
has, from Eq. (5.32),
λ#
∆N/N
2.88 × 108 /3600
=
= 3.125 × 10−17 s−1 .
N
2.56 × 1021
Thus, T1/2 = ln 2/λ = ln 2/3.125 × 10−17 = 2.22 × 1016 s = 7.03 × 108 y,
which is suspeciously close to the accepted value.
18. A 6.2 mg sample of 90 Sr (half-life 29.12 y) is in secular equilibrium with its
daughter 90 Y (half-life 64.0 h). (a) How many Bq of 90 Sr are present? (b) How
many Bq of 90 Y are present? (c) What is the mass of 90 Y present? (d) What
will the activity of 90 Y be after 100 y?
Solution:
The decay chain of concern is
90
Sr −→
29.12 y
90
Y −→
64.0 h
90
Zr (stable).
First some preliminary calculations. The relevant decay constants are
λSr =
ln 2
ln 2
=
= 7.543 × 10−10 s−1 .
T1/2
(29.12 y)(3.1557 × 107 s/y)
and
λY =
ln 2
ln 2
=
= 3.009 × 10−6 s−1 .
T1/2
(64.0 h)(3600 s/h)
5-12
Radioactivity
The number of atoms of
NSr =
90
Chap. 5
Sr initially present is
mSr Na
(0.0062 g)(6.022 × 1023 atoms/mol)
=
= 4.149 × 1019 atoms.
ASr
90 g/mol
(a) The activity of
90
Sr is ASr = λSr NSr = 3.129 × 1010 Bq.
(b) In secular equilibrium ASr = AY = λSr NSr = λY NY . Thus
AY = ASr = 3.129 × 1010 Bq.
(c) Because NY = mY Na /AY = AY /λY we find
mY =
AY AY
(3.129 × 1010)(90)
=
= 1.55 µg.
λY Na
(3.009 × 10−6 )(6.022 × 1023 )
(d) After 100 y
A(100 y) = ASr (100 y) = ASr (0) exp[−λSr 100]
= 3.129 × 1010 (Bq) exp[−7.543 × 10−10 s−1 × 3.156 × 109 s]
= 2.89 × 109 Bq.
19. A sample contains 1.0 GBq of 90 Sr and 0.62 GBq of 90 Y. What will be the
activity of each nuclide (a) 10 days later and (b) 29.12 years later?
Solution:
From data in Ap. D, we find
Thus
90
Sr has a half-life of 29.12 y, and
90
Y 64.0 h.
λSr ≡ λ1 = ln 2/T1/2 = 6.517 × 10−5 d−1 = 7.543 × 10−10 s−1
λY ≡ λ2 = ln 2/T1/2 = 0.260 d−1 = 3.009 × 10−6 s−1 .
(a) The 90 Sr activity decays exponentially, and the
Eq. (5.58) (after multiplying by λ), i.e.,
90
Y after is given by
A1 (t) = A1 (0)e−λ1 t
A2 (t) = A2 (0)e−λ2 t +
λ2 A1 (0) −λ1 t
[e
− e−λ2 t ].
λ2 − λ1
With A1 (0) = 1 GBq and A2 (0) = 0.62 GBq, these activities are evaluated
at t = 10 d as
A1 (10 d) = 1 × e−λ1 10 = 0.99935 GBq $ 1 GBq
λ2 A1 (0) −λ1 10
A2 (10 d) = 0.62e−λ210 +
[e
− e−λ2 10 ].
λ2 − λ1
$ 0.62e−λ210 + 1 × [1 − e−λ2 10 ] = 0.0461 + 0.926 = 0.972 GBq.
Thus the total activity = A1 (10 d) + A2 (10 d) = 1.972 GBq.
Radioactivity
5-13
Chap. 5
(b) After several weeks, 90 Y comes into secular equilibrium with
ing a sealed source). Thus
90
Sr (assum-
A1 = A2 = A1 (0)e−λ1 t
so that the total activity of the source at t = 29.19 y (= one half-life) is
A(29.12 y) = A1 + A2 = 2A1 = 2A1 (0)e−λt = 2A1 (0)(1/2) = 1 GBq.
20. Consider the three component decay chain discussed in Section 5.6.2. Because
the number of atoms must always be conserved, show that N1 (t) + N2 (t) +
N3 (t) = N1 (0) + N2 (0) + N3 (0) for all t.
Solution:
From Eqs. (5.57) to (5.9) one has
N
!
Ni (t) = N1 (0)e−λ1 t + N2 (0)e−λ2 t +
i=1
N3 (0) + N2 (0)[1−e−λ2 t ] +
λ1 N1 (0) −λ1 t
[e
− e−λ2 t ] +
λ2 − λ1
N1 (0)
[λ2 (1−e−λ1 t )−λ1 (1−e−λ2 t )].
λ2 − λ1
Collect terms proportional to Ni (0), i = 1, 2, 3 to obtain
N
!
i=1
"
#
−λ1 t
Ni (t) = N1 (0) e
1+
#
e−λ2 t −
λ1
λ2
−
λ2 − λ1
λ2 − λ1
λ1
λ1
++
λ2 − λ1
λ2 − λ1
$%
= N1 (0) + N2 (0) + N3 (0).
$
+
&
'
+ N2 (0) e−λ2 t + 1 − e−λ2 t + N3 (0)
21. Consider the three component decay chain of Section 5.6.2 in which N2 (0) = 0.
Derive an expression for the time tmax at which N (t) reaches a maximum.
Solution:
From Eq. (5.58) with N2 (0) = 0 one has
N2 (t) =
At the maximum
)
λ1 N1 (0) ( −λ1 t
e
− e−λ2 t .
λ2 − λ1
)
dN (t)
λ1 N1 (0) (
=0=
−λ1 e−λ1 t + λ2 e−λ2 t
dt
λ2 − λ1
from which one obtains
λ1 e−λ1 t = λ2 e−λ2 t
or
eλ2 λ1 )t = λ2 /λ1 .
5-14
Radioactivity
Chap. 5
Take the logarithm of both sides to find
tmax =
1
ln
λ2 − λ1
!
λ2
λ1
"
.
22. Consider the following β − decay chain with the half-lives indicated,
210
Pb −→ 210Bi −→
22 y
210
5.0 d
Po.
A sample contains 30 MBq of 210 Pb and 15 MBq of 210Bi at time zero. (a)
Calculate the activity of 210 Bi at time t = 10 d. (b) If the sample were originally
pure 210Pb, how old would it have been at time t = 0?
Solution:
From the half-lives we find the decay constants are
λP b ≡ λ1 = ln 2/T1/2 = 8.63 × 10−5 d−1 = 9.99 × 10−10 s−1
λBi ≡ λ2 = ln 2/T1/2 = 0.1386 d−1 = 1.605 × 10−6 s−1 .
From Eqs. (5.58) and (5.59) we have
A1 (t) = A1 (0)e−λ1 t
A2 (t) = A2 (0)e−λ2 t +
(P5.5)
λ2 A1 (0) −λ1t
[e
− e−λ2 t ].
λ2 − λ1
(a) As shown in the figure to the right,
we are given the activities at t = 0
and asked to find the activity of the
first daughter 10 days later. Specifically, we are given A1 (0) = 30 MBq
and A2 (0) = 15 MBq we find from
Eq. (P5.6)
A2 (10 d) = 15e−λ2 10 +
(P5.6)
Ai (t)
given
"$
#
!#
!
#
$
""#
seek
i=2
i=1
0
æ t = 10 d
30λ2
[e−λ1 10 − e−λ2 10 ].
λ2 − λ1
= 3.750 + 30.019 [0.99914 − 0.25007] = 26.2 MBq.
!t
Radioactivity
5-15
Chap. 5
(b) Here we change our time scale. We are given the
parent and daughter activities at some time to
(which the problem statement calls t = 0 but we
call to ) as shown in the figure to the right. Specifically, A1 (to ) = 30 MBq and A2 (to ) = 15 Mbq.
We then seek, in this new time scale, A1 (0).
Also we know A2 (0) = 0. From Eq. (P5.5),
A1 (to ) = A1 (0)e−λ1 to from which we obtain
A1 (0) = A1 (to )e+λ1 to . Substitute this result into
Eq. (P5.6) with A2 (0) = 0 to obtain
A2 (to ) =
=
Ai (t)
given
"" $
#
!#
i=2
i=1
!
"
0 to
æ
!t
λ2 A1 (to )eλ1 to −λ1to
[e
− e−λ2 to ]
λ2 − λ1
λ2 A1 (to )
[1 − e−(λ2 −λ1 )to ].
λ2 − λ1
(P5.7)
Finally, solving Eq. (P5.7) for to and substitution of data gives
!
"
1
A2 (to ) λ2 − λ1
to = −
ln 1 −
= 5.00 d.
λ2 − λ1
A1 (to )
λ2
23. A 40-mg sample of pure 226Ra is encapsulated. (a) How long will it take for
the activity of 222 Rn to build up to 10 mCi? (b) What will be the activity of
222
Rn after 2 years? (c) What will be the activity of 222 Rn after 1000 y?
Solution: The decay chain is
226
Ra −→
222
1600 y
Rn −→
3.82 d
218
Po −→ · · ·
Since λ = ln 2/T1/2 , we find λ1 = 4.33 × 10−4 y−1 = 1.37 × 10−11 s−1 and
λ2 = 66.28 y−1 .
(a) First find the initial activity of the
226
Ra parent.
m226 Na
(0.04)(6.022 × 1023)
= 1.37 × 10−11
A226
226
= 1.46 × 109 Bq = 39.55 mCi.
A1 (0) = λ1 N1 (0) = λ1
The activity of the
Initially, there is no
222
222
Rn is obtained my multiplying Eq. (5.59) by λ2 .
Rn so we obtain
A2 (t) =
λ2 A1 (0) −λ1t
[e
− e−λ2 t ].
λ2 − λ1
The time t for A2 (t) = 10 mCi is of the order of a few days since in
about two weeks, the 222 Rn will come into secular equilibrium with its
5-16
Radioactivity
Chap. 5
parent (about 40 mCi). For such a small t, exp(−λ1 t) " 1, and, because
λ2 >> λ1 , the above result can be approximated as
A2 (t)(= 10 mCi) " A1 (0)[1 − exp(−λ2 t)].
Taking the logarithm of both sides and solving for t, we obtain
!
"
1
A2 (t)
t = − ln 1 −
= 0.0044 y = 1.61 d = 38.5 h.
λ2
A1 (0)
(b) The first daughter 222 Rn starts with zero initial activity but, after a couple
of weeks, comes into secular equilibrium with its parent. Thus after 2 y,
A2 (2 y) = A1 (2 y) " A1 (0) = 39.55 mCi. = 1.46 GBq.
(c) After 1000 y
A2 (1000 y) = A1 (1000 y) = A1 (0)e−λ1 1000y = 25.65 mCi = 0.949 GBq
24. Tritium (symbol T) is produced in the upper atmosphere by neutron cosmic
rays primarily through the 14 N(n,T)12 C and 16 O(n,T)14 N reactions. Tritium
has a half-life of 12.3 years and decays by emitting a β − particle with a maximum energy of 18.6 keV. Tritium exists in nature almost exclusively as HTO,
and in the continental surface waters and the human body it has an atomic
T:H ratio of 3.3 × 10−18 . Since the human body is about 10% hydrogen by
weight, estimate the tritium activity (Bq) in a 100-kg person.
Solution:
First find the number of atoms of hydrogen and tritium. The mass of hydrogen
mH in a 100-kg person is 0.1 × 105 g = 104 g. The number of hydrogen atoms
NH in the person is thus
NH =
mH Na
(105 )(6.022 × 1023)
=
= 5.975 × 1027 atoms.
AH
1.00794
Then from the given atomic abundance of 1 H:3 T, the number of tritium atoms
is
NT = 3.3 × 10−18 NH = 1.972 × 1010 atoms.
This number of tritium atoms has an activity AT = λT NT . The decay constant
for tritium is
ln 2
λT =
= 1.786 × 10−9 s−1 .
T1/2
Thus the tritium activity in the 100-kg person is
AT = λT NT = (1.786 × 10−9 s−1 )(1.972 × 1010 atoms) = 35.2 Bq.
NOTE: it would be incorrect to say there are (105 g)(3.3×10−18) = 3.3×10−14 g
of tritium. Since T is three times more massive than H, the T:H mass ratio is
3 × (3.3 × 10−18 ) = 9.9 × 1018, thereby giving a tritium mass in the body of
9.9 × 10−14 g.
Radioactivity
5-17
Chap. 5
25. The global inventory of 14 C is about 8.5 EBq. If all that inventory is a result
of cosmic ray interactions in the atmosphere, how many kilograms of 14 C are
produced each year in the atmosphere?
Solution:
In equilibrium, the production rate of
Thus the production rate R is
14
C on earth must equal its decay rate.
R = decay rate = λN = 8.5 × 1018 Bq = 2.68 × 1026 atoms/y.
The mass of 14 C produced per year, Ṁ , is related to the global
λN = 2.68 × 1026 decays per year by
Ṁ =
14
C activity
(λN )A14
= 6.24 kg/y.
Na
26. The average mass of potassium in the human body is about 140 g. From the
abundance and half-life of 40 K (see Table 5.2), estimate the average activity
(Bq) of 40 K in the body.
Solution:
From Table 5.2, the atomic abundance of
40
K is f40 = 0.000117.
The number of potassium atoms in the body NK is
NK =
The number of
mK Na
(140)(6.022 × 1023 )
=
= 2.156 × 1024 atoms.
AK
39.0983
40
K atoms is
N40 = 0.000117NK = 2.523 × 1020 atoms.
This number of
for 40 K is
40
K atoms has an activity A40 = λ40 N40 . The decay constant
λ40 =
Thus the
40
ln 2
= 1.73 × 10−17 s−1 .
T1/2
K activity in the body is
A40 = λ40 N40 = (1.73 × 10−17)(2.523 × 1020 ) = 4, 360 Bq.
27. The naturally occurring radionuclides 222 Rn and 220Rn decay by emitting α
particles which have a range of only a few centimeters in air. Yet there is considerable concern about the presence of these radioisotopes in interior spaces.
Explain how these radioisotopes enter homes and workspaces and why they are
considered hazardous to the residents.
Solution:
Radon is a noble gas and, consequently, becomes airborne after escaping from
the soil and seeping into houses through cracks in the basement floors. The
5-18
Radioactivity
Chap. 5
buildup of radon in homes can become several times higher than that in the
ambient outdoor air. Breathing radon itself is or little concern since it is
almost immediately exhaled before it can decay. However, as it decays in the
air, its radioactive daughters adhere to dust and aerosol particles, which when
inhaled, become trapped in the surface lung tissues. The subsequent alpha
decay of these daughters is thought to cause significant damage to the cells
in these tissues, endowing the damaged cells with a predisposition for later
turning cancerous.
28. An ancient charcoal sample was found that had a specific acitivity of 14 C of 1.8
pCi/g. Modern samples of charcoal have a specific activity of about 7.5 pCi/g.
How old is the ancient charcoal?
Solution:
Assume that the ancient charcoal initially had the same equilibrium concentration of 14 C as present day wood and charcoal. Then the activity As (t) of
the ancient sample would decay as
As (t) = As (0)e−λt
where λ is the decay constant for 14 C and t is the age of the sample. Then
!
"
!
"
1
As (0)
5730
7.5
t = ln
=
ln
! 11, 800 y.
λ
As (t)
ln 2
1.8
29. Charcoal found in a deep layer of sediment in a cave is found to have an atomic
14
C/12 C ratio only 30% that of a charcoal sample from a higher level with a
known age of 1850 y. What is the age of the deeper layer?
Solution:
Let R denote the atomic ratio of 14 C to 12 C. We assume the same initial ratio
was incorporated into both layers when they were formed, i.e., Rd (0) = Rh (0).
Then the present ratio in the deeper layer to that in the higher layer is
Rd (td )
exp(−λtd )
=
= e−λ(td −th ) ,
Rh (th )
exp(−λth )
where th and td are the ages of the higher and deeper layers, respectively.
Solving this equation for td gives
#
$
1
Rd (td )
td = th − ln
λ
Rh (th )
For 14 C, λ = ln 2/T1/2 = 1.21 × 10−4 y−1 , and for the given information that
R(td )/R(th ) = 0.30 and th = 1850 y, Eq. (P5.8) gives the desired age as
td = 1850 −
1
ln(0.30) = 11, 800 y.
1.21 × 10−4
Radioactivity
30.
5-19
Chap. 5
238
Pu, an alpha-particle emitter, has been used as a thermal power source, as
described in Table 12.2. What is the energy recoverable as heat per decay?
Solution:
238
Pu decays by alpha-particle emission to
using Eq. 5.7 and Table B.1, namely
234
U. The Q-value is calculated
234
4
Qα /c2 ! M (238
92 Pu) − [M ( 92 U) + M (2 He)]
Qα = 931.5[238.0495534 − 234.0409456 − 4.0026032497] = 5.593 MeV.
Depending on the configuration of the power source, all of this energy which
appears as kinetic energy of the products may be dissipated as heat. If only
the kinetic energy of the alpha particle is useful, the energy dissipated as heat
is given by Eq. 5.11, namely,
!
"
AU
E α ! Qα
= 5.50 MeV,
AU + Aα
the value given in Table 12.2.
5-20
31.
Radioactivity
Chap. 5
90
Sr, in secular equilibrium with its daughter 90 Y, both beta-particle emitters,
has been used as a thermal power source, as described in Table 12.2. What is
the energy recoverable as heat per decay of 90 Sr?
Solution:
This problem may be addressed according to the Q-value methodology of Sec.
5.3.3 and Eq. 5.14. For decay of 90 Sr,
!
"
90
Qβ− = c2 M (90
38 Sr) − M (40 Y) = 0.546 MeV.
For decay of
90
Y,
!
"
90
Qβ− = c2 M (90
40 Y) − M (40 Zr) = 2.28 MeV.
However, these Q-values reveal only the maximum possible kinetic energies of
beta particles, with negligible kinetic energy of the recoil nucleus. The values
agree with maximum energies tabulated in Appendix B. There is no simple
way to determine the mean energy of the beta particle or anti-neutrino in beta
decay. Here we must rely on data in Appendix B. The mean energies are 0.196
MeV for 90 Sr decay and 0.935 MeV for 90 Y, for a total of 1.13 MeV, which is
the value cited in Table 12.2 for available energy.
32. The text discusses 14 C dating of biogenic materials, useful for ages up to about
50,000 years. Another dating method, useful for dating rock of volcanic origin
of ages 100,000 to 4 billion years, is the 40 K-40 Ar method. The former nuclide
decays to the latter in 10.7% of its decays. Argon is expelled from molten rock
but, after solidification remains in the rock as a stable nuclide along with the
slowly decaying 40 K. In a rock specimen under analysis, the rock is melted and
the 40 Ar is collected and measured using mass spectrometry. The 40 K in the
rock is measured using flame photometry or atomic absorption spectrometry.
What is the atomic ratio of 40 Ar:40 K at rock ages of 105 , 107 , and 109 years.
Solution:
From Eq. 5.75, accounting for the branching ratio of 0.101 in
atomic ratio is given by.
40
K decay, the
!
"
NA
1 − e−λK t
= 0.101 −λ t = 0.101 eλK t − 1 .
K
NK
e
40
K has a half-life of 1.28 billion years, corresponding to a decay constant
λK = 5.415 × 10−10 = y−1 . It follows that, for times 105 , 107 , and 109 years,
the atomic ratios are, respectively, 5.46 × 10−6 , 5.48 × 10−4 , and 7.26 × 10−2 .
Chapter 6
Binary Nuclear
Reactions
PROBLEMS
1. For each of the following possible reactions, all of which create the compound
nucleus 7 Li,
7
Li + γ




6 Li + n
1
n + 6 Li −→ 7 Li∗ −→ 6 He + p

5

He + d


3
H+α
calculate (a) the Q-value, (b) the kinematic threshold energy, and (c) the minimum kinetic energy of the products. Summarize your calculations in a table.
Solution:
For a binary, two-product reaction x +X −→ y +Y we use the following results
from Ch. 6.
(a) The Q-value 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
Exth # − 1 +
Q
mX
and
ExC # 1.2
Zx ZX
1/3
Ax
1/3
+ AX
.
(c) The minimum KE of the products is (Ey + EY )min = Q + (Exth )min .
Results are tabulated below.
6-1
6-2
Binary Nuclear Reactions
Reaction
6
Li(n,γ)7 Li
6
Li(n,n)6 Li
6
Li(n,p)6 He
6
Li(n,d)5 He
6
Li(n,t)4 He
ExC
Q (MeV)
7.250
0
−2.725
−2.361
4.783
0
0
0
0
0
Chap. 6
Exth
condition
(Ey + EY )min
0
0
3.179
2.755
0
none
none
Ex > 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 Q-value, (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, two-product reaction x +X −→ y +Y we use the following results
from Ch. 6.
(a) The Q-value 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 right-hand side of the Q-value
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
Be(p,γ)10 B
9
Be(p,α)6 Li
9
Be(p,d)8 Be
9
Be(p,n)9 B
9
Be(p,αn)5 Li
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
6-3
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
16
19 ∗
H + O −→ F −→ 18
O+p


16
N + 3 He
For each of these reactions calculate (a) the Q-value, (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, two-product reaction x +X −→ y +Y we use the following results
from Ch. 6.
(a) The Q-value 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
He + 11 B −→ 15 N∗ −→ 14
C + 1H
3
H + 12 C −→ 15 N∗ −→ 14 C + 1 H
' 11
B + 4 He
2
H + 13 C −→ 15 N∗ −→ 12
C + 3H
2
H + 13 C −→ 15 N∗ −→ 14 N + 1 n
1
n + 14 N −→ 15 N∗ −→ 11 B + 4 He
6-4
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
Q-value
(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
4π#o 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,
4π#o 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.
4π#o 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
+ 131/3
21/3
= 0.
min(Ey + EY ) = Q + (Exth )min = 1.311 + 1.992 = 3.305 MeV.
Binary Nuclear Reactions
6-5
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
+ 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
11/3
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
#
Ey =
±
$
mx my Ex
cos θy
(my + mY )2
$
%
&
mx my Ex
mY − mx
mY Q
2
cos θy +
Ex +
. (P6.1)
(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,
6-6
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
"
"
!
M me EM
M me EM
Ee =
cos θe +
cos2 θe + 0
(me + M )2
(me + M )2
=
!
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 2-MeV 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)
#√
$2
!
1
E cos θs + E(A2 − 1 + cos2 θs )
2
(A + 1)
E! =
$2
!
1 #√
2 cos(45) + 2(144 − 1 + cos2 (45)) = 1.90 MeV.
2
13
=
9. The first nuclear excited state of 126 C is 4.439 MeV above the ground state.
(a) What is the Q-value 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 8-MeV neutron scattered inelastically from
this level at 45 degrees?
Solution:
(a) The Q-value 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)
%
&
%
&
mn
1
Enth = − 1 +
Q =− 1+
(−4.439) = 4.81 MeV.
MC
12
(c) From Eq. (6.25)
E! =
#√
$2
!
1
E cos θs + E(A2 − 1 + cos2 θs ) + A(A + 1)Q
2
(A + 1)
Binary Nuclear Reactions
=
6-7
Chap. 6
#2
"
1 !√
8 cos(45) + 8(144 − 1 + cos2 (45)) + 12(13)(−4.439)
2
13
= 3.22 MeV.
10. An important radionuclide produced in water-cooled nuclear reactors is 16 N
which has a half-life 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)
$
%
$
%
mx
1
th
Ex # − 1 +
Q=− 1+
(−9.638) = 10.24 MeV.
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 Q-value 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 Q-values 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
6-8
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 Q-value, (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 Q-value 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 neutron-producing reactions caused by incident 5.5MeV α particles.
α + 7 Li −→ 10 B + n
α + 9 Be −→ 12 C + n
(a) What is the Q-value 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
6-9
Chap. 6
Solution:
(a) Using the masses in Appendix B, the Q-values 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
!
!
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
!
!
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
"
"
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
%
E"
1
A(A + 1)Q
2
2
=
cos θs ± (A − 1 + cos θs ) +
,
E
(A + 1)2
E
6-10
Binary Nuclear Reactions
Chap. 6
The maximum scattered neutron energy occurs for θs = 0. The above equation
becomes
!
#2 $
"
√
%2
!
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
!
Emax
$√
%2
1+∆+1
=E
.
A+1
The minimum energy of the scattered neutron occurs for θs = π and Eq. (6.25)
becomes
!
#2
"
!
Emin
1
(A + 1)Q
=
−1
±
A
1
+
.
E
(A + 1)2
AE
Again the + sign gives the only meaningful result, so that
!
Emin
=E
$√
1+∆−1
A+1
%2
.
15. How many elastic scatters, on the average, are required to slow a 2-MeV 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,
E!
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
6-11
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 Ŝn = 2.3 × 1012 n/(s g). Thus, 1 mg
of 252 Cf has a source strength Sn = 10−3 Ŝn = 2.3 × 109 n/s.
17. In a particular neutron-induced 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 Q-value 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.
6-12
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 Q-value 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 half-lives 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
6-13
(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 half-lives 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.
6-14
Binary Nuclear Reactions
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 D-D 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 D-D fusion energy potential in an 8-oz 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 8-oz 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
6-15
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,
coal-fired 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
6-16
Binary Nuclear Reactions
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 6-cm
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 half-thickness for 1-MeV photons
in (a) water, (b) iron, and (c) lead.
Solution:
The half-thickness is given by Eq. (7.9). Thus, from the data of Ap. C we
obtain
7-1
7-2
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
7-3
Chap. 7
4. A material is found to have a tenth-thickness of 2.8 cm for 1.25-MeV gamma
rays. (a) What is the linear attenuation coefficient for this material? (b) What
is the half-thickness? (c) What is the mean-free-path length for 1.25-MeV
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}
7-4
Radiation Interactions
=
i
!
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
!
exp(−µj tj ).
j=1
7. Show the details to obtain Eq. (7.8)
Solution:
Integrate by parts to obtain
" ∞
#∞ " ∞
#
x = µt
xe−µt x dx = −xe−µt x # +
e−µt x
0
0
0
" ∞
#∞
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
$
%
&
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
$
%
&
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
7-5
Chap. 7
9. Calculate the linear interaction coefficients in pure air at 20◦ C and 1 atm
pressure for a 1-MeV 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 gamma-ray data in Ap. C to obtain
!
! " µ # ! " µ # " ρi #
! "µ#
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
!
! fi σ i
t
µair
Ni σti = ρair Na
n =
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 mean-free-path 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 .
7-6
Radiation Interactions
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 mean-free-path 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 3-MeV photons with intensity 108 cm−2 s−1 irradiates a small sample
of water. (a) How many photon-water 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
!
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
7-7
Chap. 7
(b) The positron production rate equals the rate of pair production interactions in the water sample. Thus
" #
!β+ = 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 1-h 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
!t ∆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 2-mCi source of 60 Co is placed in the center of a cylindrical water-filled 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)
7-8
Radiation Interactions
Chap. 7
From Ap. C we find for 1.25-MeV 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 100-keV photon, (b) a 1-MeV photon, and (c) a 10-MeV
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 " 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 " . From Eq. (7.33) we then have
!
"
1
T = E − E" = E 1 −
.
1 + (me c2 /E)(1 − cos θs )
By inspection, T is greatest when θs = π, so that
!
"
!
"
1
me c2
Tmax = E 1 −
=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 27-keV neutrons that pass through
a 10-cm thick slab without interaction. (b) What is this fraction for 28-keV
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
7-9
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) 27-keV 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 27-keV neutrons transmitted through the slab
is
Fraction transmitted = exp[−Σt x] = 0.712.
(b) 28-keV 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 28-keV 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
(−dE/ds)rad
EZ P b ! me "2
(5)(82) 1
#
=
= 0.59.
(−dE/ds)coll
700
M
700
1
18. About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV
electrons, (b) 2.5-MeV protons, and (c) 10-MeV 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 2.5-MeV 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.
7-10
Radiation Interactions
Chap. 7
(b) The range in aluminum of a 2.5-MeV 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 10-MeV
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 10-MeV 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.
Rp (2.5 MeV)
mp zα2
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 10-MeV tritium nucleus in air.
Solution:
First find the kinetic energy of a proton with the same speed as a 10-MeV
triton. Classical mechanics, appropriate for these energies, gives
(1/2)mp vp2
Ep
mp
1
=
=
= .
Et
(1/2)mt vt2
mt
3
The proton energy with the same speed as a 10-MeV triton is, thus, Ep =
Et /3 = 3.333 MeV.
Now find the range in air of a 3.333-MeV 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
7-11
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
7-12
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 tube-gas 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 tube-gas ion thereby neutralizing
it. The resulting positively charged quench-gas 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
8-1
8-2
Radiation Detection
Chap. 8
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
low-energy 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 1-MeV 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
"
!
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 )
Radiation Detection
8-3
Chap. 8
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 ) =
!
to
λN (t) dt = λN (0)
0
!
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
to = − ln 1 −
=−
ln(1 − 0.9) = 53.0µs.
λ
N (0)
4.35 × 106 s−1
6. In anthracene, scintillation photons have a wavelength of 447 nm. If 1-MeV 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
8-4
Radiation Detection
Chap. 8
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
full-energy 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 one-half the maximum height. From Ap. D, 137 Cs
emits a 0.662 MeV gamma ray. This the FWHM of the 0.662-MeV peak in a
NaI spectrum is
FWHM = Emax
Res(%)
8
= 0.662
= 0.00662 MeV.
100
100
9. Which detector has the greatest full-energy peak efficiency: NaI(Tl), HPGe,
Si(Li), or CZT for (a) 100-keV gamma rays, and (b) for 1-MeV gamma rays?
Justify your answer.
Solution:
The full-energy 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
Chap. 8
8-5
Photoelectric interactions almost always produce a full-energy 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 pair-production 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) 100-keV Photons:
For 100-keV 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 full-energy
peak efficiency. Here is it is the CZT detector.
(b) 1-MeV 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
full-energy deposition, giving them the largest full-energy efficiency.
8-6
Radiation Detection
Chap. 8
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 (half-life
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 air-kerma 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
K̇(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.
9-1
9-2
Radiation Doses and Hazards
Chap. 9
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
Ẋ = 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
K̇ = 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 K̇ equals the absorbed dose rate Ḋ. The dose equivalent equals the quality factor for
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
Ḣ = QF × Ḋ = 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
Chap. 9
9-3
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
Ẋ = 1.835 × 10−8 E
φo
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
K̇ = 1.602 × 10−10 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 K̇ equals the absorbed dose rate Ḋ. The dose equivalent equals the quality factor for
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
Ḣ = QF × Ḋ = 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.
9-4
Radiation Doses and Hazards
Chap. 9
4. What is the gamma-ray absorbed dose rate (Gy/h) in an infinite air medium at
a distance of 10 cm from a 1-mCi 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) !
µen
Ḋ(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
Ḋ =
(1.602 × 10−10 )(3.7 × 107 )
× 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
Ḋ =
(1.602 × 10−10 )(3.7 × 107 )
× 3600 × 0.02789 = 0.4736 mGy/h.
4π102
Radiation Doses and Hazards
9-5
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
(1.602 × 10−10 )S(Bq) !
µen
e
Ḋ(Gy/h) =
f
E
(MeV)
(cm2 /g) exp[−µF
i i
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
Ḋ =
(1.602 × 10−10)(3.7 × 107 )
× 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
Ḋ =
(1.602 × 10−10 )(3.7 × 107 )
× 3600 × 9.7389 × 10−5 = 1.654 µGy/h.
4π102
9-6
Radiation Doses and Hazards
Chap. 9
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:
Ẋ =
6CEN
,
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 Ẋ 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
[r(ft)]2
!
"#!
"2
C(Bq)
r(m)
=6
3.7 × 1010(Bq/Ci)
0.3048(m/ft)
Ẋapprox (R/h) =
=
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
Ẋexact (R/h) = 1.835 × 10−8 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
Ẋexact (R/h) = 209.36(µen/ρ)air (cm2 /g)
.
[r(ft)]2
The ratio of the approximate to exact exposure rates is thus
$
%−1
Ẋapprox /Ẋexact = 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 Ẋapprox is within ±25% of
Ẋexact for photon energies between 0.05 and 2 MeV.
Radiation Doses and Hazards
Chap. 9
Problem 9-6: 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
---------------------------------
9-7
9-8
Radiation Doses and Hazards
Chap. 9
7. A medium sized banana (1̃20 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 half-life 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 .
T1/2
(1.27 × 109 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
! E = 5.1 × 10−9 Sv/Bq.
9.4, the specific committed effective dose equivalent is H
Thus his committed dose is
! 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 alpha-particle 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
Ḋ = (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
Chap. 9
9-9
9. A male worker at a nuclear facility receives an accidental whole-body 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 person-Gy collective exposure of low
LET radiation. Here the collective population dose is 500, 000×0.005 Gy =
2.5 × 103 person-Gy. The expected number of radiogenic cancer deaths
resulting from the accidental exposure is, therefore,
1
2.5 × 103 person-Gy
(480 + 660)
= 143 deaths.
2
104 person-Gy
9-10
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 (whole-body) 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 radiation-induced cases of hereditary illness that
occurs in one generation as a result of this excess exposure to the parental and
grand-parental generations. (c) Compare this number to the natural incidence
of genetic illness in this population.
Solution:
(a) We assume a steady-state (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 person-Gy.
(b) From the estimates in Table 9.11, there are 3930 to 6700 cases in one
generation resulting from exposure of parental and grand-parental generations to one million person-Gy. 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 person-Gy = 133 to 226 cases.
106 person-Gy
(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
9-11
12. On September 13, 1987, a 137Cs source for radiotherapy was stolen from an
abandoned treatment facility in Goiâia, 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 Goiâia, 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 gamma-ray 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 well-approximated 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
Ḋ = 1.602 × 10−10E
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 long-term effects.
9-12
Radiation Doses and Hazards
Chap. 9
(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/person-Gy.
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/person-rad) × (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/person-Gy.
105 people × 0.1 Gy
Thus, the expected number of cancer deaths is (0.0570 cases/person-rad)×
(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 40-year old female worker receives an x-ray exposure of 5.2 rad (whole-body)
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 30-year old female is
(455 + 52)/105 per 10 rad of gamma-ray 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 whole-body 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 low-level exposure). If
we assume equal numbers of males and females, the average radiogenic
Radiation Doses and Hazards
Chap. 9
9-13
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
Ṅrad =
3.76 × 106 deaths
= 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 =
1
330 × 106 people
(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
Ṅnat =
67.4 × 106 deaths
= 922, 000 per year.
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 gamma-ray sources. For gamma-ray
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?
9-14
Radiation Doses and Hazards
Chap. 9
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:
!
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
9-15
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 non-smoking 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 radon-induced lung cancer if
she is a nonsmoker? (b) What is the probability if she smokes?
Solution:
The annual radon exposure is calculated as
!
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 non-smoking female has a radon-induced cancer mortality risk of 0.0088 per MBq h m−3 annual exposure. Thus her
probability of death from radon-induced 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 radon-induced cancer
mortality risk of 0.081 per MBq h m−3 annual exposure. Thus her probability of death from radon-induced 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.
9-16
Radiation Doses and Hazards
Chap. 9
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
radon-induced 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 radon-induced 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 radon-induced 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 radon-induced 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 radon-caused 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
−3
= 0.154 MBq h m .
Radiation Doses and Hazards
Chap. 9
9-17
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 radon-induced 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 whole-body 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 radiation-induced 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 non-radiation
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 non-radiation workers, namely
Annual dose limit =
=
deaths/y for non-radiation workers
risk of death from radiation/rem
10−4 y−1
= 4.3 rem/y " 5 rem/y.
0.23 × 10−4 rem−1
9-18
Radiation Doses and Hazards
Chap. 9
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 fission-neutron thermalization. How
many scatters with sodium, on the average, would it take for 2-MeV 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
10-1
10-2
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 hydrogen-oxygen gas mixtures
Graphite:
• reactor-grade (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?
Principles of Nuclear Reactors
10-3
Chap. 10
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 non-1/v factor
ga (To ) for 235 U at room temperature?
Solution:
Rearrangement of Eq. (10.5) yields
235
ga (To ) =
Σa
Σ235
a (Eo )
!
T
To
"1/2
2
σ 235
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
10-4
Principles of Nuclear Reactors
Chap. 10
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
ν 235σ 235
/N U
f N
= 234 234 U
235 /N U + σ 238 N 238/N U
σ a N /N + σ 235
a N
a
235 235 235
ν σf f
= 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
ρU Na
235 235
= (f 234 σ 234
σ a + f 239 σ 239
a +f
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
=
σ 235
a
ν 235σ 235
f
238 /N 235)
+ σ 238
a (N
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
10-5
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
1
2.4367 × 505.9
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
10-6
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
a
σ 235
a
H2 O
+ σ a (N H2 O /N 235)
=
592.6
= 0.5723
592.6 + 0.6644(1/1.5 × 10−3 )
Hence we have k∞ = !pηf " ηf = (2.080)(0.5723) = 1.190.
Principles of Nuclear Reactors
10-7
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
1
=
=
.
1 + L2H2 O (1 − f)Bc2
1 + 8.12(1 − 0.5723)Bc2
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, source-free, critical, uranium-fueled
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 surface-to-volume 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 non-leakage probabilities to increase. Hence, the reactor becomes
supercritical and the power increases.
10-8
Principles of Nuclear Reactors
Chap. 10
(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.
Principles of Nuclear Reactors
10-9
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 fuel-to-moderator 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 fuel-to-moderator
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 fuel-to-moderator mixture, i.e., r becomes very large. This value is
(k∞ )max = 2.080. The plot below shows how keff varies with r.
10-10
Principles of Nuclear Reactors
Chap. 10
NOTE: This is poorly conceived problem. The assumption !p ! 1 is valid
only for dilute mixtures, r << 1, in which neutrons are rapidly thermalized.
As the material approaches pure 235 U metal, there is little moderation and
fast-neutron cross section should be used. The correct MCNP Monte Carlo
results are shown in the above figure and are seen to increasingly differ from
Eq. (P10.1) for N 235 /N C greater than 10−4 .
17. Consider a homogeneous mixture of graphite and uranium enriched to 1% with
N M /N U = 500. Calculate k∞ for this mixture at room temperature. Ignore
234
U and assume ! = 1.
Solution:
In this mixture the isotope abundances of the two uranium isotopes are f235 =
0.01 and f238 = 1 − f235 = 0.99. The thermal utilization factor is thus
f=
f235 σ a + f238 σ a
.
M
U
f235 σ a + f238 σ a + σ M
a (N /N )
√
√
M
Here σ M
a = ( π/2)σa (Eo ) = ( π/2)(0.003861) = 0.0034217 b. Substitute
values into the above equation to obtain
f=
(0.01)(592.6) + (0.99)(2.382)
= 0.82883.
(0.01)(592.6) + (0.99)(2.382) + (0.0034217)(500)
The thermal fission factor is
η=
ν 235f235 σ 235
(2.437)(0.01)(505.9)
f
= 1.4882.
238 =
(0.01)(592.6)
+ (0.99)(2.382)
f235 σ 235
+
f
σ
238
a
a
Principles of Nuclear Reactors
10-11
Chap. 10
The resonance escape probability is given by Eq. (10.9). Substitution of values
into this equation gives
!
"
#(1−0.486)$
2.73 (0.99)/(500)
p = exp −
= 0.73004.
0.158
4.8
Thus k∞ = !pfη = (1)(0.730040(0.82883)(1.4882) = 0.90050.
18. Consider a homogeneous mixture composed of of graphite and uranium with
1.5% enrichment What is the optimum moderator-to-fuel ratio that yields the
largest value of k∞ at room temperature? Ignore 234 U and assume ! = 1.
Solution:
Use the technique of the previous problem and iterate using different values of
N M /N U to generate a table such as that below.
N M /N U
!
η
f
p
k∞
550
560
570
572
574
577
578
583
590
595
600
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.64599
1.64599
1.64599
1.64599
1.64599
1.64599
1.64599
1.64599
1.64599
1.64599
1.64599
0.85653
0.85430
0.85208
0.85164
0.85120
0.85054
0.85032
0.84922
0.84768
0.84659
0.84550
0.74168
0.74373
0.74573
0.74612
0.74651
0.74710
0.74729
0.74825
0.74958
0.75052
0.75144
1.04566
1.04581
1.04590
1.04591
1.04591
1.04592
1.04592
1.04591
1.04588
1.04583
1.04577
From these results the optimum value of (N M /NU )opt = 577.5 that gives
k∞ = 1.04591.
19. A homogeneous mixture of graphite and 3% eniched uranium at room temperature has an optimum N M /N U ratio of 828 that produces k∞ = !pηT f =
(1.0)(0.78642)(1.84117)(0.87640) = 1.26897. What is the dimension of a cube
of this material that produces a critical assembly?
Solution:
For criticality we require
keff =
!pηf
exp(−B 2 τT ) = 1.
1 + L2T B 2
2
From Eq. (10.16) L2T = lM
(1 − f) = 3070(1 − 0.87640) = 379.45 cm2 . For such
a dilute material, the Fermi age to thermal energies is that of the moderator,
10-12
Principles of Nuclear Reactors
Chap. 10
i.e., τT = 268 cm2 . Thus the criticality condition is
keff = 1 =
1.6939
exp(−368Bc2 ).
1 + 379.45Bc2
Solution for Bc2 by “trial and error” yields Bc2 = 3.2826 × 10−4 cm−2 . From
Table 10.10 the critical buckling for√a cube with side a is Bc2 = 3(π/a)2 from
which the critical dimension is a = 3π/Bc = 300.3 cm.
20. What is the optimum fuel-to-moderator ratio of a homogeneous mixture of 235 U
and (a) light water, (b) heavy water, (c) beryllium, and (d) graphite to produce
a mixture with the maximum k∞? Data: the thermal (0.0253 eV) absorption
cross sections for water, heavy water, beryllium and carbon are 0.664, 0.00133,
0.0092 and 0.0034 b per molecule or atom, respectively.
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
σ 235
a
+ σM
a (1/r)
=
2.4367 × 505.9
.
592.6 + σ M
a (1/r)
where r is the fuel-to-moderator ratio, i.e., r ≡ N 235/N M , and N M is the
atom/molecular density of the moderator. By inspection of the above expression, the maximum value of k∞ is obtained for a fully rich fuel-to-moderator
mixture, i.e., as r becomes very large. Note this value is (k∞ )max = 2.080 for
all four moderators. The plot below shows how keff varies with r for the four
moderators.
Principles of Nuclear Reactors
10-13
Chap. 10
NOTE: This is poorly conceived problem. The assumption !p ! 1 is valid
only for dilute mixtures, r << 1, in which neutrons are rapidly thermalized.
As the material approaches pure 235 U metal, there is little moderation and fastneutron cross section should be used. The correct MCNP Monte Carlo results
for carbon are shown in the above figure and are seen to differ significantly
from Eq. (P10.2) for N 235 /N C greater than 10−4 .
21. A spherical tank with a radius of 40 cm is filled with a homogeneous mixture of
235
U and light water. The mixture has a moderator-to-fuel ratio N H2 O /N 235
of 800. (a) What is k∞ of the mixture? (b) What is keff for this bare core?
Solution:
For fully enriched uranium fuel !p ! 1. Hence we need to calculate only f, η,
f
th
PNL
, and PNL
. Needed nuclear data are taken from Tables 10.1, 10.2 and 10.4,
(a) From Eq. (10.12)
η = ν 235
σ 235
f
σ 235
a
= 2.4367
505.9
= 2.0802.
592.6
From Eq. (10.11)
f=
σ 235
a
+
σ 235
a
H2 O
σ a (N H2 O /N 235)
=
592.6
= 0.52717.
592.6 + 0.6644(800)
Hence we have k∞ = ηf = (2.0802)(0.52717) = 1.0966.
(b) To calculate the non-leakage probabilities, we need the core buckling,
which for a sphere is Bc2 = (π/R)2 = (π/40)2 = 0.006169 cm−2 . Then
f
PNL
is obtained from Eq. (10.15)
2
f
PNL
= e−Bc τ = e−(0.006169)(27) = 0.8466.
th
The PNL
is obtained from combining Eqs. (10.13) and (10.14), namely
th
PNL
=
1
1+
L2H2 O (1
− f)Bc2
=
1
= 0.9769.
1 + 8.12(1 − 0.52717)0.006169
f th
Thus keff = k∞ PNL
PNL = (1.0966)(0.8466)(0.9769) = 0.9069.
22. What should the radius of the tank in the previous problem be to produce a
critical configuration? What is the critical mass of 235U needed?
Solution:
(a) For criticality, it is necessary that
2
f th
keff = k∞ PNL
PNL = k∞
e−BC τH2 O
= 1.
1 + L2H2 O (1 − f)Bc2
10-14
Principles of Nuclear Reactors
Chap. 10
From the previous problem, k∞ = 1.0966 and f = 0.52717. From Table
10.4, we find τ H2 O = 27 cm2 and L2H2 O = 8.12 cm2 . The above equation
thus reduces to
2
e−27Bc
keff = 1.0966
=1
1 + 3.839Bc2
Solving this equation for Bc2 , we find Bc2 = 0.0029924 cm−2 . For a spherical
tank of radius R, Bc2 = (π/R)2 . Thus the critical radius is R = π/Bc =
57.43 cm.
(b) For a dilute fuel-to-water mixture of 1 : 800, the mass of water is M H2 O !
(4/3)πR3 ρH2 O . The ratio of fuel mass to water mass is
M 235
N 235A235
=
,
M H2 O
N H2 O AH2 O
where N 235 /N H2O is the fuel atom density to water molecular density.
From these two results we obtain
! 235 " 235
!
"
!
" 235
N
A
4 3 H2 O N 235
A
M 235 = M H2 O
=
πR
ρ
N H2 O AH2 O
3
N H2 O AH2 O
=
!
"
!
"
4
1
235
3
π57.43 (1.00)
= 12, 900 g = 12.9 kg.
3
800 18
23. Consider the lattice shown in Fig. 10.5 with the natural uranium fuel rods
replaced by ones that are 1% enriched. Prepare a table similar to that of Table
10.8 for this case. Ignore 234 U and assume $ remains unchanged.
Solution:
Using the methods described in Section 10.5, computer program was written
and produced the results shown below.
Pitch
a (cm)
!
η
f
p
k∞
12
14
16
18
19
20
21
22
24
26
28
1.027
1.027
1.027
1.027
1.027
1.027
1.027
1.027
1.027
1.027
1.027
1.488
1.488
1.488
1.488
1.488
1.488
1.488
1.488
1.488
1.488
1.488
0.972
0.961
0.948
0.934
0.926
0.917
0.909
0.900
0.881
0.861
0.840
0.740
0.805
0.848
0.878
0.890
0.900
0.909
0.917
0.930
0.940
0.948
1.102
1.182
1.228
1.253
1.259
1.262
1.263
1.261
1.252
1.234
1.217
Principles of Nuclear Reactors
10-15
Chap. 10
24. A control rod is dropped into a critical, source-free, uranium-fueled reactor
and the asymptotic period of the resulting exponentially decreasing power is
observed to be -200 s. (a) What is the value of keff of the reactor after the
control rod drop? (b) What was the reactivity insertion in dollars?
Solution:
NOTE: In the text the period is incorrectly given as −2 s. This is physically
impossible. The correct period should be −200 s.
(a) First find the change in keff needed to produce the given period. From
Eq. (10.33) we find δk = βτ /T , where for 235 U βτ = 0.083 s. Thus keff
after the rod insertion is
keff = 1 + δk = 1 +
βτ
0.083
=1+
= 0.999585.
T
−200
(b) The reactivity insertion, from Eq. (10.33), is
ρ($) =
ρ
1 keff − 1
1 0.999585 − 1
=
=
= −0.0639($).
β
β keff
0.0065 0.999585
25. What is the asymptotic period resulting from a reactivity insertion of (a) 0.08$
and (b) -0.08$ ?
Solution:
By definition ρ($) ≡ (keff − 1)/(βkeff ). Solving this for keff and substituting
ρ($) = ±0.08$ we find
keff =
1
1
=
= 1 ± 0.000520 ≡ 1 + δk.
1 − βρ($)
1 − 0.0065(± 0.08)
Thus for ρ($) = ±0.08($), we have δk = ±0.000520.
Finally, from Eq. (10.33) and from βτ = 0.083 s for
asymptotic periods are
T =
235
U, we find the two
βτ
0.083
=
= ±160 s.
δk
±0.000520
26. Following a reactor scram, in which all the control rods are inserted into a
power reactor, how long is it before the reactor power decreases to 0.0001 of
the steady-state power prior to shutdown?
Solution:
On page 345 it is discussed that, for large negative reactivity insertions into
a core, power cannot decrease any faster than the decay rate of the longest
lived delayed neutron precursors. For 235 U the mean lifetime of the longest
lived precursors is about 80 s. Thus, the smallest asymptotic period for a very
10-16
Principles of Nuclear Reactors
Chap. 10
subcritical reactor is T = −80 s. Asymptotically, the power decays exponentially, i.e., P (t) = P (0) exp[t/T ]. Solving this equation for t, and ignoring any
prompt drop in power, the time to decrease the power to 0.0001P (0) is
!
"
!
"
P (t)
0.0001P (0)
t = T ln
= (−80 s) ln
= 737 s = 12.3 min.
P (0)
P (0)
27. At time t = 0 a reactivity of 0.15$ is inserted into a critical reactor operating
at 100 W. How long is it before the reactor power reaches 1 MW?
Solution:
First find the resulting reactor period. For δk << 1, we have from Eq. (10.33)
and the fact that for 235U τ = 12.8 s
T =
βτ
βτ
βτ
τ
τ
12.8 s
=
=
=
"
=
= 85.3 s.
δk
keff − 1
keff ρ
keff ρ($)
ρ($)
0.15
Then from Eq. (10.32), which can be written as P (t) = P (0) exp[t/T ],
!
"
! 6
"
P (t)
16 W
t = T ln
= (85.3 s) ln
= 785 s = 13.1 min.
P (0)
100 W
28. A reactivity insertion into an initially critical reactor operating at steady state
causes the power to increase from 100 W to 10 kW in 6 minutes. What was
the value of the reactivity insertion in $?
Solution:
From Eq. (10.32), which can be written as P (t) = P (0) exp[t/T ], the period is
# ! P (t) "
# ! 104 W "
T = t ln
= (6 × 60 s) ln
= 78.2 s.
P (0)
100 W
Since keff " 1, we obtain from Eq. (10.33)
T =
βτ
βτ
βτ
τ
τ
=
=
=
"
,
δk
keff − 1
keff ρ
keff ρ($)
ρ($)
from which we obtain for a
235
ρ($) =
U-fueled reactor in which τ = 12.8 s
τ
12.8 s
=
= 0.164 $.
T
78.2 s
29. Explain how a decrease in the boiling rate inside a boiling water reactor affects
the reactivity of the reactor.
Solution:
Most cores are “undermoderated” so that if the amount of moderator is decreased, such as when water turns to vapor, keff decreases. By contrast, when
Principles of Nuclear Reactors
10-17
Chap. 10
the boiling rate decreases, there are fewer steam voids and hence more water
in the core, so keff increases. The explanation is as follows. With more water/moderator in the core, neutrons can slow to thermal energies more quickly
with less chance of being absorbed while slowing down, and thus the resonance
escape factor p increases. Also, more water means that more thermal neutrons
will be absorbed by the nonfuel and hence the thermal utilization factor f decreases. However, in undermoderated cores, the change in p is greater than
that in f, so that as the boiling rate decrease, fp and keff increase causing the
power to rise.
30. Explain why it is reasonable that
for neutron absorption.
135
Xe should have a very large cross section
Solution:
The nuclide 135
54 Xe has 81 neutrons and hence is one neutron shy of having
the “magic number” of 82 neutrons (see Section 3.2.6). The nucleus 136
54 Xe
is exceptionally stable since all the neutrons in the nucleus form completely
closed shells. For this reason, it is energetically very attractive for 135 Xe to
absorb a neutron to complete its neutron shell structure. In fact, 135 Xe has an
enormous resonance near thermal energies, giving it a microscopic cross section
for thermal neutrons larger than that for any other nuclide.
31. Following a reactor shutdown (φo → 0), show from Eq. (10.49) that the time
tm for 135 Xe to reach a maximum is given by
! "
#$
1
X(0)
tm = −
ln r 1 + (1 − r)
,
λI − λX
I(0)
where X(0) and I(0) are arbitrary 135 Xe and 135 I concentrations at shutdown
and r ≡ λX /λI . From this result show, that for an increase in 135 Xe following
shutdown, the initial values must satisfy
X(0)
λI
≤
= 1.38.
I(0)
λX
Solution:
First we solve for the 135 Xe and 135I transients following a reactor shutdown
at t = 0 to zero flux. The 135Xe and 135 I decay equations, Eqs. (10.46) and
(10.47) become for t > 0
dI(t)
= −λI I(t),
dt
dX(t)
= λI I(t) − λX X(t).
dt
(P10.2)
(P10.3)
The solution of the simple radioactive decay equation, Eq.(P10.3), gives
I(t) = I(0) exp[−λI t].
10-18
Principles of Nuclear Reactors
Chap. 10
Substitution of this result in Eq. (P10.4) and solving (see the general solution
of Eqs. (5.51) and (5.52))
X(t) = e−λX t
!"
0
t
#
!
λI I(t" )eλX t dt" + X(0)
= X(0)e−λX t +
%
λI I(0) $ −λI t
e
− e−λX t .
λX − λI
(P10.4)
These two results for the post shutdown transients X(t) and I(t), can also
be obtained directly from Eqs. (10.48) and (10.49) by setting φo = 0 in these
equations.
To find the time tmax following the shutdown at which X(t) reaches a maximum,
we differentiate Eq. (P10.5) and set the result to zero at t = tmax . The result
0=
!
#
λX λI I(0)
λ2 I(0)
−λI X(0) +
exp[−λX tmax ] − I
exp[−λI tmax ].
λX − λI
λX − λI
Multiply by eλI tmax and collect terms to obtain
exp[−(λI − λX )tmax ] =
(λI − λX )λX X(0) λX
X(0)
+
= (1 − r)r
+ r.
λ2I
I(0)
λI
I(0)
where r ≡ λX /λI # 0.724. Take the logarithm of both sides and solve for tmax
to get the desired result, namely
! &
'#
1
X(0)
tmax = −
ln r 1 + (1 − r)
.
(P10.5)
λI − λX
I(0)
Now consider under what conditions there is a rise in 135 Xe following the shutdown. Since λI > λX , we see that in Eq. (P10.6) the argument of the logarithm
must be < 1 to make tmax > 0, i.e.,
1 + (1 − r)
X(0)
1
<
I(0)
r
which, upon rearrangement gives the condition
(
)
X(0)
1
λI
<
=
# 1.38.
I(0)
r
λX
Thus to get a rise in the 135Xe following a shutdown, the initial X(0) and I(0)
concentrations must be such that
X(0)
< λI # 1.38.
λX
I(0)
Principles of Nuclear Reactors
Chap. 10
32. In a fast reactor, does
135
10-19
Xe produce feedback? Why?
Solution:
The absorption cross section of 135 Xe for fast neutrons is not large, and hence
the presence of 135 Xe produces negligible feedback.
33. Many phenomena produce reactivity feedback. Within orders of magnitude,
over what time interval would you expect the following reactivity feedback
causes to take effect? (a) temperature increase in the fuel, (b) temperature
increase in the moderator/coolant, (c) 135 Xe increase, (d) fuel burnup, (e)
boiling in the core, and (f) increased coolant flow through the core.
Solution:
(a) Fuel Temperature: Since the fission fragments transfer their initial kinetic energy to thermal energy of the fuel matrix within 10−12 s (see Section 6.6), the fuel temperature responds to changes in the fission rate or
power and produces feedback within a very small fraction of a second.
(b) Moderator/Coolant Temperature: The thermal energy deposited in
the fuel material takes time to be conducted to the fuel rod surface, the
time depending on the diameter of rods, properties of the fuel-clad interface, and the conductivities of the fuel and cladding. Typically, it take
several to tens of seconds for the moderator/coolant to change its temperature and produce feedback.
(c)
135
Xe Poisoning: The transient effects of 135 Xe are determined by the
half-lives of 135 Xe (6.7 h) and 135I (9.2 h). Thus, normal 135 Xe feedback
effects take several half-lives to be seen. For example, from Fig. 10.13,
135
Xe takes about 30 h to reach equilibrium. Following shutdown, however,
135
Xe can build up rapidly, and, within 10 to 30 minutes, can poison-out
the core.
(d) Fuel Burnup: In a power reactor, fuel typically stays in the core for
three to four years before two-thirds of the initial 235U is consumed. Thus
fuel-burnup feedback effects are of the order of weeks to months.
(e) Coolant Boiling: The time constant for this effect is about the same as
for coolant temperature changes, namely of the order of tens of seconds.
(f) Increased Coolant Flow: With increased coolant flow, there is less
boiling (if a BWR) or lower coolant temperature (if a PWR). Such changes
produce feedback effects with time constants of several to many seconds.
10-20
Principles of Nuclear Reactors
Chap. 10
Chapter 11
Nuclear Power
PROBLEMS
1. In a BWR or PWR, steam is generated with a temperature of about 290 ◦ C.
If river water used to receive waste heat has a temperature of 20 ◦ C, what
is the maximum possible (ideal) conversion efficiency of the reactor’s thermal
energy into electrical energy? Nuclear power plants typically have conversion
efficiencies of 34%. Why is this efficiency less than the ideal efficiency?
Solution:
The maximum possible thermodynamic efficiency is the Carnot efficiency given
by Eq. (11.1), which, for the given temperatures, yields
η=
Tin − Tout
(290 + 273) − (20 + 273)
=
= 48%.
Tin
(290 + 273)
In practice the actual conversion efficiency is quite a bit less, typically around
34%. One may be tempted to think a power plant using a steam/water cycle
must have huge energy losses. However, relatively little energy is lost in the
piping and turbines or is consumed by the various pumps. Moreover, the
generator converts rotational energy to electrical energy with high efficiency.
The principal reason a the efficiency of a BWR or PWR power plant is less
than the ideal Carnot efficiency is because of the irreversible nature of how
energy is transferred from the hot fuel to the water coolant. To obtain the
ideal Carnot efficiency in a closed cycle, all energy transfer must be between
elements at infinitesimally different temperatures. In a reactor, the average fuel
temperature is much higher than the water/steam coolant so that the energy
transfer from the fuel to the coolant is done in a very irreversible manner,
thereby reducing the overall thermal efficiency of the water/steam cycle.
11-1
11-2
Nuclear Power
Chap. 11
2. A 1000 MWe nuclear power plant has a thermal conversion efficiency of 33%.
(a) How much thermal power is rejected through the condenser to cooling
water? (b) What is the flow rate (kg/s) of the condenser cooling water if the
temperature rise of this water is 12 ◦ C? Note: specific heat of water is about
4180 J kg−1 C−1 .
Solution:
(a) By definition the thermal conversion efficiency η = Pe /Pt where Pt is the
thermal power used to create the electrical power Pe . The rate of heat
rejected to the cooling water is thus
!
"
!
"
1
1
Pr = Pt − Pe =
− 1 Pe =
− 1 1000 = 2030 MW.
η
0.33
(b) The thermal power Pr (W = J s−1 ) needed to increase the coolant by ∆T
(C) flowing at a rate Ṁ (kg/s) is Pr = cp Ṁ ∆T where cp (J kg−1 C−1 ) is
the specific heat of the cooling water. Thus. the coolant flow rate is
Ṁ (kg/s) =
Pr (J s−1 )
2030 × 106 (J s−1 )
=
−1
−1
cp (J kg C ) ∆T (C)
4180 (J kg−1 C−1 ) 12 (C)
= 40, 470 kg/s = 145 × 106 kg/h.
3. What are the advantages and disadvantages of using (a) light water, (b) heavy
water, and (c) graphite as a moderator in a power reactor.
Solution:
(a) Light Water:
advantages:
• It is very inexpensive.
• It has a large ξ = 0.92, so only about 20 scatters are needed to
thermalize fission neutrons. Neutrons slow near their place of birth
(small Fermi age τ = 27 cm2 ) so small compact cores can be built.
• It can be used as a coolant.
• It can be used in a BWR to produce the steam for the turbine.
• Its chemical inertness minimizes corrosion problems.
disadvantages:
• It boils at relatively low temperatures, and hence an expensive pressure vessel surrounding the core is needed.
• Its thermal absorption cross section is relatively large (σa = 0.664 b)
which tends to reduce the thermal utilization factor f.
• Enriched uranium must be used to obtain keff > 1.
• It cannot be used as a structural material as can solid moderators.
Nuclear Power
Chap. 11
11-3
• It produces the short-lived 16 N as an activation product.
• Radiation can dissociate water molecules to produce an explosive mixture of oxygen and hydrogen.
(b) Heavy Water:
advantages:
• It has a moderate ξ = 0.51, so about 36 scatters are needed to thermalize fission neutrons (see Table 6.1). Its very large thermal diffusion
length (L = 170 cm) and moderate Fermi age τ = 131 cm2 ) requires
a large core to achieve criticality.
• Its thermal absorption cross section is extremely small (σa = 1.2 mb)
which produces a large thermal utilization factor f. Can use natural
uranium to get keff > 1.
• It can be used as a coolant; more often, however, light water in pressure tubes containing the fuel is used as coolant.
• Its chemical inertness minimizes corrosion problems.
disadvantages:
• It is moderately expensive, and thus a “tight” system is required to
minimize losses.
• It boils at relatively low temperatures; hence, an expensive pressure
vessel (or thousands of pressure tubes containing fuel and coolant)
are required.
• It cannot be used as a structural material as can solid moderators.
• It produces 3 H as an activation product.
• Radiation can dissociate water molecules to produce explosive mixture of hydrogen and oxygen.
(c) Graphite:
advantages:
• It has a relatively large ξ = 0.716, so about 115 scatters are needed
to thermalize fission neutrons. Its large thermal diffusion length (L =
59 cm) and very large Fermi age τ = 368 cm2 ) requires a very large
core to achieve criticality.
• Its thermal absorption cross section is extremely small (σa = 3.4 mb)
which produces a large thermal utilization factor f. Can use natural
uranium to obtain keff > 1.
• It can be used as a structural element
• It remains solid to very high temperatures (>
∼ 2500 F) and thus no
pressure vessel is needed.
• It produces no radioactive activation products.
• It is very resistant to radiation damage.
11-4
Nuclear Power
Chap. 11
disadvantages:
• It requires large core to achieve keff > 1.
• Reactor-grade graphite (very pure) is moderately expensive.
• It is chemically reactive with oxygen; thus, one must use carbon dioxide or helium as gas coolants, or water in pressure tubes.
• Graphite stores energy (Wigner effect) and must periodically be carefully annealed to remove this stored energy to prevent spontaneous
catastrophic release.
4. Why are the blades of a low-pressure turbine larger than those of a highpressure turbine?
Solution:
For the same mass flow rate in both turbines, the volume flow rate is necessarily
larger in the low-pressure turbine than in the high-pressure turbine. To accommodate the larger volume flow rate, the volume and blades of the low-pressure
turbine must be larger.
5. Why can a heavy-water moderated reactor use a lower enrichment uranium
fuel than a light-water moderated reactor?
Solution:
Because the thermal absorption cross section is extremely small (σa = 1.2 mb)
for heavy water, it yields a large thermal utilization factor f, larger than that
for any other moderator at the optimum fuel-to-moderator ratio (see Table
10.3).
Also, heavy-water moderated reactors have a very large optimum moderatorto-fuel ratio compared to light-water reactors (see Table 10.3). Neutrons, thus,
have a much better chance of slowing to thermal energies without encountering
238
U and being absorbed while slowing. As a consequence, heavy water reactors
have much larger resonance escape probabilities p than do light-water reactors.
To compensate for lower values of f and p in light-water reactors, these reactors
use enriched uranium to increase the thermal fission factor η. As discussed in
Section 10.4, only heavy-water and graphite moderated reactors can be made
critical with natural uranium. Light-water moderated reactors must use enriched uranium.
Nuclear Power
Chap. 11
11-5
6. Explain whether the turbine room of a BWR is habitable during normal operation.
Solution:
In any water-moderated reactor, the radionuclide 16 N is produced via the
16
O(n,p)16 N reaction. This radioisotope has a half-life of 7.13 s and emits
very energetic photons with energies of 6.1 and 7.1 MeV (see Appendix D).
Since the 16 N is carried along with the primary water, any piping carrying
the primary water/steam leaving the reactor vessel must be heavily shielded or
placed in chambers not accessible by plant personnel during operation to avoid
excessive radiation doses being received by workers.
In a BWR the primary water includes the steam leaving the reactor and passing
through the turbines. To avoid costly shielding of the piping and turbines, these
plants simply close the turbine room to workers when the reactor is operating.
7. If the demand on the generator increases (i.e., a greater load is placed on the
turbine), explain what happens to the reactor power of (a) a PWR and (b) a
BWR if no operator-caused reactivity changes are made. Which reactor follows
the load?
Solution:
(a) PWR: To prevent slowing of the turbine when a greater load is applied to
the turbine shaft, the main steam valve to the turbine is opened more to allow more steam to flow into the turbine. This increased steam flow results
in a decrease of the main steam pressure in the steam generators thereby
allowing an increase in the boiling in the steam generators. This increased
boiling removes more heat from the primary water thereby decreasing the
temperature of the primary water leaving the steam generators and reentering the reactor. Because PWRs have a negative temperature coefficient,
the cooler entering primary water causes keff to increase and the reactor
power to rise. Thus, in a PWR, the reactor power follows the turbine load.
(b) BWRs: The decreased main steam pressure, following an increased flow
of steam through the turbine in response to an increased load demand,
causes more boiling to occur in the reactor (where the steam is produced).
This increased boiling increases the water void-to-liquid fraction in the
core, thereby reducing the amount of water moderator in the core. Since
BWR cores are slightly undermoderated, this decrease in the amount of
moderator in the core causes keff to decrease and, consequently, the reactor
power also decreases. Thus, a BWR does not follow the turbine load
directly. However, increased power demand is met by increased circulation
flow (Fig. 11.20), which improves moderation and thus leads to increased
reactor power.
11-6
Nuclear Power
Chap. 11
8. Although the steam cycle is simpler in a BWR, explain why the capital costs
of BWR and PWR plants are very competitive.
Solution:
The simplicity of a single steam/water cycle in a BWR compared to the double
primary/secondary cycles in a PWR is offset by the increased complexity of the
BWR components. Although a BWR does not require steam generators and a
pressurizer, the reactor pressure vessel is much more complex. A BWR vessel
must include steam dryers in the top and penetrations for control rods and
instrumentation in the bottom of the vessel. Two-phase flow through the core
requires a very detailed thermal-hydraulic design. For example, fuel assemblies
are shrouded (Fig. 11.21) to prevent cross flows. Furthermore, differing sized
orifices are required in the entries to coolant channels to adjust flow rates to
assure that steam quality is uniform across the core. More important, A BWR
core has a power density (W/liter) about half that of PWRs, and, for the
same power, a BWR must therefore have a large core with a greater loading
of expensive uranium. Soluble poisons such as boric acid cannot be used in
BWRs for routine reactivity control. Rather, to regulate reactivity and allow
a BWR to follow the load, complex jet pumps are needed to control the flow
and boiling in the core.
Also because the primary coolant flows throughout the entire plant, greater
efforts must be made to control the water chemistry to remove entrained radionuclides and activation products in the coolant. The 16 N activation product
produced in all water moderated reactors requires special shielding and radiation control procedures for BWRs that PWRs do not. For example, turbine
rooms must be closed to personnel during BWR plant operation.
Finally, the possibility of an accident from a piping break that leads to a
depressurization of the reactor is more problematic for a BWR since there
is far more piping containing the primary water and steam.
9. Explain the advantages and disadvantages of using helium instead of water as
a coolant for a power reactor.
Solution:
Solution:
advantages:
• Helium does not change phase and, consequently, it does not require
expensive pressure vessels as do water-cooled reactors.
• After passing through the reactor core where it is heated to temperatures far above those in water cooled reactors, it can be sent directly
to a gas turbine which is physically much smaller and less expensive
than steam turbines of the same capacity.
• The hot helium gas can also be used to operate compressor turbines
to maintain the helium flow through the reactor.
• Helium is chemically inert and thus presents no corrosion problems.
Nuclear Power
Chap. 11
11-7
disadvantages:
• As a gas, it has negligible affect on the neutrons in the core. Thus,
another material must be used as a moderator. In He cooled reactors,
graphite is usually used as the moderator.
• Helium is moderately expensive.
• For accidents involving the breach of primary the primary flow, there
must be special fuel designs to prevent release of fission products resulting from fuel failure. This usually means the fissile fuel must
be small microspheroids encased in sufficient graphite to absorb any
residual heat following an emergency shutdown or loss of coolant accident.
10. Small modular reactors employed several passive safety features. List five such
systems. Would any of them work on a space reactor? Explain.
Solution:
(a) In the integrate MSR, an electrically cooled fuse (drain) plug is located
at the bottom of the reactor vessel. With loof of power, this plug melts and
gravity drains the molten salt fuel into an emergency dump tank designed to
remove the decay heat.
(b) Also in the integrated MSR, the reactor vessel is surrounded by a thick
blanket of solid salt. If the reactor vessel becomes hot than normal, as in a
emergency, the solid salt melts, thereby, removing heat from the reactor vessel.
(c) Surround the integrated reactor vessel with a large volume of water. Natural
convection and eventually boiling remove the decay passively. Large abovevessel reservoirs of water can also be used as makeup water.
(d) The SC-HTGR has a massive graphite core that has a large heat capacity,
very high melting point and a low power density. In an emergency the core
inteself can safely absorb the decay heat. This heat is then conducted to the
vessel walls where cooling water surrounding the vessel removes additional heat
by natural convection.
(e) The SMR-160 uses natural water convection for both operating and emergency modes
For space reactors there is no gravity and passive systems that rely on gravity
flows such as water makeup and convection will not work. Nongravitation
methods such as heat conduction can work.
11. Many small modular plants can use wet or dry cooling towers to get rid of waste
heat. Why does using water increase the plant’s thermal conversion efficiency?
Solution:
Dry cooling towers reject heat at a higher temperature and thus the thermodynamic ∆T is smaller and so the ideal Carnot efficiency which limits the actual
conversion efficiency is smaller.
11-8
Nuclear Power
Chap. 11
12. Discuss the economics of small modular plants.
Solution:
This best left as a classroom discussion topic. But items of interest should
include (a) the losss of econoomy of scale so the cost per kWh of capacity of
a SMR is higher than a Gen III system, (b) the captital outley by a utility
per SMR is far less than a Gen III system, (c) multiple SMRs can be added
to a plant over time as the need for capacity grows, (d) the SMR have fewer
compents because of their simplified designs, (e) siting for a SMR is much more
flexible since dry cooling can ofter be used.
13. During the April 1986 accident in the 1000-MWe RMBK Chernobyl reactor,
the water in the cooling tubes of the graphite-moderated reactor was allowed,
through operator error, to boil into steam and cause a supercritical, run-away
chain reaction. The resulting energy excursion resulted in the destruction of
the reactor containment and a large amount of the fission products in the fuel
elements were released to the environment as the reactor containment ruptured.
Explain why the boiling in the cooling tubes led to supercriticality.
Solution:
In this type of water-cooled graphite-moderated reactor, the relatively large
thermal-neutron absorption cross section for water compared to that for graphite
(0.664 b versus 3.4 mb) causes water to act as neutron absorber or poison (decreasing the thermal utilization factor f). The water’s contribution to neutron
moderation is negligible. Thus, if the water boils, there is less liquid water in
the core, and f increases causing keff to become greater than unity thereby letting the power increase. If the boiling occurs suddenly in many pressure tubes
simultaneously, the reactor can become superprompt critical, as happened in
the Chernobyl accident.
14. How many years are required for the initial activity to decrease by a factor of
1010 for (a) 137 Cs, (b) 90 Sr, and (c) 239Pu?
Solution:
From the radioactive decay law, Eq. (5.35), we have A(t)/Ao = e−λt from which
the decay time is
!
"
!
"
!
"
T1/2
T1/2
1
A(t)
A(t)
1
t = − ln
=−
ln
=−
ln
= 33.219 T1/2 .
λ
Ao
ln 2
Ao
ln 2
1010
Using the half-life data from Appendix D and Table C.1 we obtain:
(a)
137
(b)
90
(c)
239
Cs: t = 33.219 × 30.0 y = 997 y.
Sr: t = 33.219 × 29.12 y = 967 y.
Pu: t = 33.219 × 24, 110 y = 800, 900 y.
Nuclear Power
11-9
Chap. 11
15. Over a period of one year what mass (in kg) of fission products is generated by
a 1000 MWe power reactor?
Solution:
Assume the power plant has an electrical conversion efficiency η ! 33% . Then
the thermal energy generated in ∆t = 365.35 d is Eth = (Pe /η)∆t = 1.107×106
MWd. From page 148, we find 1 MWd of energy is released from the thermal
fission of 1.05 g of 235 U.
The mass of fission products created almost equals the mass of 235 U fissioned
(the very small difference being the mass-equivalent of the fission energy released). Thus, the mass of fission products created in one year of continuous
operation is
M (235 U) = Eth (MWd) × 1.05 (g/MWd) = 1160 kg.
Note: Table 11.4 indicates a typical 1000 MWe plant generates about 873 kg
of fission products. However, a nuclear power plants typically has a capacity
factor of about 75%. Thus, for continuous operation, the annual fission product
generation would be about 873/0.75 = 1160 kg, a value in agreement with our
calculated value above.
16. Discuss possible environmental, technical and political problems associated
with each of the disposal options listed in Table 11.6.
Solution:
This is best left as a classroom discussion topic.
17. A nuclear drive in a submarine delivers 25,000 shaft horse power at a cruising
speed of 20 knots (1 knot = 1.15 miles/h). If the power plant has an efficiency
of 25%, how much (in kg) of the 235U fuel is consumed on a 40,000 mile trip
around the world?
Solution:
The voyage would take ∆t = 40000 (mi)/(20 × 1.15 (mi/h)) = 1, 739 h =
72.46 d. The thermal energy required for the voyage is
Eth =
=
Pshaft (HP)
× 745.7 (W/HP) × ∆t
η
25, 000 (HP)
× 745.7 (W/HP) × 72.46 (d) = 5, 403 MWd.
0.25
From page 149 we find that for 1 MWd of fission energy, 1.24 g of
be consumed. Thus the mass of 235 U consumed on the voyage is
M (235U) = Eth × 1.24 = 6.70 kg.
235
U must
11-10
Nuclear Power
Chap. 11
18. Reactors for naval vessels are designed to have very long lifetimes without
the need to refuel. Discuss possible techniques that can be used to maintain
criticality over the core lifetime as 235 U is consumed.
Solution:
For a long lifetime, a large initial mass of 235 U is needed, an amount in excess of
the critical mass equal to the anticipated fuel burnup over the design lifetime.
Naval reactors are of the PWR type and can employ the following methods
to maintain criticality despite the large variation in fissile mass over the core
lifetime:
• incorporate large amounts of burnable poison in the fuel assemblies so
that as the fuel is consumed (a negative reactivity effect) so is the poison
(a positive reactivity effect). Thus, the two effects tend to compensate for
each other.
• use very enriched uranium to obtain a large initial mass of fissile fuel in
the relatively small core.
• use initially high concentration of soluble poisons, such as boric acid, in
the water moderator/coolant.
• have a large number of control rods to adjust the reactivity over a wide
range.
19. A thermal nuclear rocket using hydrogen as the propulsive gas operates for one
hour at a thermal power of 4000 MW and a temperature of 2700 K. Estimate
(a) the amount (in grams) of fissile material consumed, (b) the specific impulse
of the engine.
Solution:
(a) The fission energy released is Eth = Pth ×∆t = 4000 (MW)×(1/24) (d) =
166.7 MWd. Since 1 MWd of fission energy requires the consumption of
1.24 g of 235U (see p 150), the mass of 235 U consumed by the rocket is
M (235U) = Eth × 1.24 g/MWd = 206.7 g
235
U consumed.
(b) At an exhaust temperature of 2700 K, the hydrogen is in the form of a
diatomic molecule H2 . The mass of an exhaust hydrogen molecule is
Me = 2 × M (1 H) = 2(1.0078 u)(1.661 × 10−27 kg/u) = 3.347 × 10−27 kg.
From page 342, the specific impulse is
"
!
8kT
8(1.381 × 10−23 J K−1 )(2700 K)
Isp = ve =
=
πMe
π(3.347 × 10−27 kg)
= 5326 m/s = 5.3 km/s.
Nuclear Power
11-11
Chap. 11
20. Estimate the specific impulse Isp for (a) a chemical rocket burning hydrogen
and oxygen at a temperature of 4000 K, and (b) a thermal nuclear rocket
emitting hydrogen at 3000 K.
Solution:
(a) The mass of an exhaust water molecule is
Me = 2M (1 H) + M (16 O) = 18 u
= (18 u)(1.661 × 10−27 kg/u) = 2.990 × 10−26 kg.
From page 342, the specific impulse is
"
!
8kT
8(1.381 × 10−23 J K−1 )(4000 K)
Isp = ve =
=
= 2.16 km/s.
πMe
π(2.990 × 10−26 kg)
(b) At an exhaust temperature of 3000 K, the hydrogen is in the form of a
diatomic molecule H2 . The mass of an exhaust hydrogen molecule is
Me = 2 × M (1 H) = 2(1.0078 u)(1.661 × 10−27 kg/u) = 3.347 × 10−27 kg.
From page 342, the specific impulse is
"
!
8kT
8(1.381 × 10−23 J K−1 )(3000 K)
Isp = ve =
=
= 5.61 km/s.
πMe
π(3.347 × 10−27 kg)
21. A nuclear rocket propulsion system uses an ion drive to accelerate mercury
atoms to energies of 50 keV. Estimate the specific impulse of this drive.
Solution:
For a 50-keV mercury ion, classical mechanics may be used. Thus the speed
and the specific impulse of a 50-keV mercury ion (A = 200.6 u) is
"
"
2E
2(50, 000 eV)(1.602 × 10−19 J/eV)
ve =
=
MHg
(200.6 u)(1.661 × 10−27 kg/u)
= 2.19 × 105 m/s = 219 km/s.
11-12
Nuclear Power
Chap. 11
Chapter 12
Fusion Reactors and Other
Conversion Devices
PROBLEMS
1. In Section 6.7 some fusion reactions that do not produce neutrons as a reaction
product are shown. Discuss the advantages and disadvantages of using such
reactions to convert nuclear energy to electricity.
Solution:
Advantages:
(1) There would be no fast neutron damage to the chamber walls and hence
the components will have longer lifetimes.
(2) There would be no neutron activation and production of readioactivity in
the components.
(3) Both fusion reaction products can be used to heat the plasma.
Disadvantages:
(1) Need higher plasma temperatures to overcome increased Coulombic repulsion between the reactants. Hence there will be higher bremsstrahlung losses.
(2) It would be harder to extract energy to make electricity.
(3) New fuel can not be created with the fusion products.
2. Starting with Eq. 12.17 verify that, for an ICF D-T plasma at kT = 50 keV,
2
the ρR criterion is ρR >
∼ 3 g/cm .
Solution:
First note the unit conversion that 1 J= 1 N m = 1 kg m2 s−2 = 107 g cm2
s−2 . Now express the plasma kinetic temperature of 50 keV in these units.
kT = (50 keV)(1.602×10−16 J/keV)(107 (g cm2 s−2 )/J) = 8.01×10−8 g cm2 s−2 .
The average reactant mass is
m=
1
(2 + 3u)(1.666 × 10−24 g/u) = 4.15 × 10−24 g.
2
12-1
12-2
Fusion Reactors and Other Conversion Devices
Chap. 12
From Fig. 12.1, the reaction rate parameter !vσ" is seen to be about 9 × 10−16
cm3 s−1 at a kinetic temperature of kT = 50 keV. Finally, substitution of these
values into Eq. (12.17) gives
!
√
4
mkT
4
(4.15 × 10−24 g)(8.01 × 10−8 g cm2 s−1 )
ρR >
= 2.56 g/cm2 .
∼ !vσ" =
9 × 10−16 cm3 s−1
3. Derive an expression for the ratio of the triple product nT τE for ICF to that
for magnetically confined fusion. For a plasma at kT = 50 keV, estimate the
value of this ratio.
Solution:
From Eq. (12.16) for ICR, nτE & 4/!vσ", so that the triple product nτE T &
4T /!vσ". Then, with the use of Eq. (12.11), the ratio of the two triple products
is
(ni τE T )ICF
4T /!vσ"
4Ec
=
=
.
(ni τE T )M C
12kT 2 /(Ec !vσ")
12kT
For a D-T plasma, Ec = 3500 keV, and for kT = 50 keV, this ratio is
(ni τE T )ICF
4Ec
4 × 3500 keV
=
.=
= 23.
(ni τE T )M C
12kT
12 × 50 keV
4. If all the neutrons produced in a D-T reactor were slowed and absorbed in a
natural lithium medium, what is the thermal energy that would be produced
per incident 14.1 neutron? How many tritium atoms would be produced per
incident thermal neutron?
Solution:
From Table C.1, it is seen that in natural lithium almost all neutrons are
absorbed by 6 Li in the 63 Li(n, α)31 H reaction. The Q-value for the reaction is
found to be
"
#
Q = M (6 Li) + mn − M (3 H) − M (4 He) c2 = 4.78 MeV.
Thus, in addition to the 14.1 MeV initial neutron kinetic energy being deposited
in the lithum, an additional 4.78 MeV of energy is also generated in the lithium
medium. Thus
Etot /neutron = 14.1 + 4.78 = 18.8 MeV.
Also, for each (n, α) reaction one new atom of 3 H fuel is generated.
Fusion Reactors and Other Conversion Devices
Chap. 12
12-3
5. For an ICF device, estimate the energy gain, i.e., fusion energy to incident
laser energy, that would be needed with present day lasers (1.5% electric-tolight conversion efficiency) to achieve break-even. Assume the conversion of
captured fusion energy to electrical energy has an efficiency of 32%.
Solution:
The thermal fusion thermal energy Efus needed to produce the laser energy is
Elaser /(0.015 × 0.32) " 208Elaser . Thus, the needed energy gain is
G=
Efus
" 201.
Elaser
6. A pellet for ICF contains 4 mg of a 50:50 mixture of deuterium and tritium. If
all of the atoms fused, how much energy would be released? How many gallons
of gasoline is this equivalent to? The heat content of gasoline is 1.42 × 108
J/gallon.
Solution:
The number of atoms of deuterium nD and tritium nT are
nD = nT =
1 mNa
1 (0.004)(6.02223)
=
= 4.82 × 1020 atom pairs.
2 A
2
2.5
For each D-T fusion reaction, 17.6 MeV of energy is released. Thus, the fusion
energy release by fusing all the atoms is
Efus = (17.6 MeV/fusion)(4.82 × 1020 fusions)(1.602 × 10−13 J/MeV)
= 1.35 × 108 J.
This corresponds to
Efus
" 1 gallon of gasoline.
1.42 × 108 J/gal
7. Discuss the advantages and disadvantages of a fusion power plant to a conventional coal-fired power plant.
Solution:
This is an open ended question and best left to classroom discussion.
8. The energy of decay alpha particles from 226 Ra is shown in Fig. 5.2. What
initial mass of 226 Ra is needed to provide a thermal power of 100 W(t) after
10 years?
Solution:
Assume that, in the 10 years, 226 Ra has come into secular equilibrium with
its radioactive decay products through 214 Po. Assume further that only alpha
and beta decay contribute to the thermal power. From data in Appendix D,
12-4
Fusion Reactors and Other Conversion Devices
Chap. 12
the following energy, degraded to thermal energy, is released per decay of the
parent 226 Ra.
Nuclide
226
Ra
Rn
218
Po
214
Pb
214
Bi
214
Po
Total
222
MeV
4.777
5.489
6.001
0.293
0.658
7.687
24.905
For a thermal power of 100 J/s, the decay rate or activity A of the members
of the chain in secular equilibrium must be
A=
100 J/s
= 2.506 × 1013 Bq.
1.602 × 10−13 J/MeV × 24.905 MeV/decay
The decay constant for 1600-y 226 Ra is 1.373 × 10−11 s−1 . Thus, the number
of atoms of 226 Ra is N = A/λ = 1.83 × 1024 atoms. The mass of 226Ra needed
for a thermal power of 100 W is
m=
1.83 × 1024 atoms × 226 g/mol
= 685 g.
6.023 × 1023 atoms/mol
9. Show that 90 Sr in secular equilibrium with its daughter 90 Y has (a) a specific
activity of 136 Ci/g, and (b) a specific thermal power of 0.916 W/g.
Solution:
(a) For 29.12-y 90 Sr, the decay constant is 7.543 ×10−10 s−1 . One gram of
90
! = m(g) Na (atoms/mol)/A(g/mol) = 6.022 × 1023 /90 =
Sr consists of N
21
6.69 × 10 atoms. The specific decay rate is thus
! = 5.05 × 1012 Bq/g = 136 Ci/g.
λN
(b) From Appendix D, the combined energy release from the decay chain is
1.130 MeV = 1.81 ×10−13 J. Thus, the specific thermal power is
P! = 1.81 × 10−13 J/decay × 5.047 × 1012 decays/g s = 0.92 W/g.
Fusion Reactors and Other Conversion Devices
Chap. 12
12-5
10. What initial mass of 210Po is needed in a space RTG with a 19% thermal to
electric conversion efficiency that is to have a power output of 65 W(e) one
year after the start of the mission?
Solution:
From Table 12.2, 210 Po has a half life of 138.4 d and a specific thermal power
of 144 W/g. To achieve an electrical power of Pe = 65 W after one year (t),
the initial thermal power Pt (0) would need to be
Pe (1 y) λt
e
η
!
"
65
ln 2 × 365.25
=
exp
= 2130 W.
0.19
138.4
Pt (0) = Pt (1 y)eλt =
The required initial mass of
210
m=
Po is thus
2130 W
= 14.8 g.
144 W/g
11. Many thermal energy conversion devices, such as thermoelectric, AMTEC and
Stirling converters developed for space applications, also have terrestrial uses.
Rather than use radioisotopes as a heat, a combustion flame can be used. Consider a converter producing 100 W(e) with a 15% thermal to electric conversion
efficiency designed to produce power continuously for one year. (a) What mass
of 238 Pu would be needed? (b) What mass of a petroleum-based fuel, with
density 0.9 g/cm3 and a heat of combustion of 40 MJ/L, would be needed?
Solution:
The thermal power requirement after t = one year is Pt (t) = 100/0.15 = 667 W.
(a) From Table 12.2, 238 Pu has a half life of 87.7 y and a specific thermal
power of P#t = 0.558 W/g. To achieve a thermal power of 667 W after one
year, the initial thermal power Pt (0) would need to be
!
"
ln 2 × 1
Pt (0) = Pt (t)eλt = 667 exp
= 672 W.
87.7
The required initial mass of
m=
238
Pu is then
Pt (0)
672 W
=
= 1.2 kg.
0.558 W/g
P#t
(b) The specific heat of combustion is 40/0.9 = 44.4 MJ/kg. The mass m
required for one year’s operation at 667 W is
m=
667 J/s × 3.156 × 107 s/y
× 1 y = 474 kg.
44.4 × 106 J/kg
12-6
Fusion Reactors and Other Conversion Devices
Chap. 12
12. Discuss the advantages and disadvantages of using RTGs versus solar cells for
the following space missions: (a) communication satellites, (b) lunar surface
experiments, (c) a mars lander, and (d) a planetary mission to Neptune. Consider issues such as launch constraints, safety, reliability, space environment,
mass, size, auxiliary systems.
Solution:
This problem is best suited to classroom discussion.
13. A particular heart pacemaker requires 200 µW of power from its betavoltaic
battery. (a) How many curies of 147 Pm are needed for a betavoltaic battery
that has a conversion efficiency of 0.95%? (b) if the battery is to have a lifetime
of 5 years, what must be the initial loading of 147 Pm?
Solution:
From Table 12.2, 147 Pm has a half life of 2.623 y and delivers P! = 367µW/Ci.
To deliver Pe = 200µW electrical power after t = 5 years at an efficiency of
η = 0.95% requires the following activity.
A=
"
#
Pe /η λt
200/0.0095 µW
ln 2 × 5
e =
exp
= 215 Ci.
367µW/Ci
2.623
P!
14. In a betavoltaic battery, the energy of the beta particles should be below about
0.2 MeV. Explain why betas with significantly greater energy are not as attractive for the battery’s operation.
Solution:
Higher energy beta particles would require excessively thick diodes, adding to
the weight and bulk of the battery without gain in the conversion efficiency.
15. The 147Pm used in betavoltaic batteries, usually contains 146Pm as an impurity.
This impurity is undesirable because it is radioactive (5 y half-life) and emits
energetic gamma rays (0.75 MeV). To avoid bulky shielding around the battery,
it is preferable to eliminating much of this impurity. Suggest how this could
be done. HINT: 146 Pm has a large thermal-neutron absorption cross section.
Solution: The 146 Pm could be eliminated by exposing the Pm mixture to
thermal neutrons in a nuclear reactor. The 146Pm is thus be converted to the
desired 147 Pm.
Fusion Reactors and Other Conversion Devices
12-7
Chap. 12
16. Discuss the relative advantages and disadvantages of using fission reactors versus radioisotopes as thermal energy sources for use in thermal electric converters.
Solution:
This problem is best suited to classroom discussion.
17. Show that 238Pu has (a) a specific activity of 17.1 Ci/g, and (b) a specific
thermal power of 0.558 W/g.
Solution:
(a) For 87.7-y 238 Pu, the decay constant is 2.505 ×10−10 s−1 . One gram of
238
! = 1(g) Na (atoms/mol)/A(g/mol) = 6.022×1023/238 =
Pu consists of N
21
2.53 × 10 atoms. The specific decay rate is thus
! = 6.337 × 1011 Bq/g = 17.1 Ci/g.
λN
(b) From Table 12.2, the energy release from the decay is 5.495 MeV = 8.80
×10−13 J. Thus, the specific thermal power is
P! = 8.80 × 10−13 J/decay × 6.337 × 1011 decays/g s = 0.558 W/g.
18. According to Table 11.6 spent fuel from a 1000 MWe PWR contains almost
250 kg of plutonium. The isotopic abundance of the plutonium isotopes is
approximately 53% 239Pu, 25% 240Pu, 15% 241Pu, 5% 242Pu and 2% of 238 Pu.
Why is it not feasible to extract the 238 Pu from this spent fuel?
Solution:
While it is rather easy to chemically separate the plutonium from spent fuel,
such extraction would require a major change in government policy in the
U.S. which todate has endorsed only spent fuel long term storage. But more
important to separate the 238 Pu from the other isotopes is extremely expensive
and difficult.
19. In a thorium fueled power reactor relatively large amounts of
produced. Show the reaction chain that leads to this isotope.
238
Pu can be
Solution:
232
90 Th
β−
+ 10 n −→ 233
90 Th −→
21.8 m
β − 233
233
91 Pa −→ 92 U
6.69 h
β−
1
237
−→ 236
92 U + 0 n −→ 92 U −→
15 %
6.75 d
237
93 Np
1
235
1
+ 10 n −→ 234
92 U + 0 n −→ 92 U + 0 n
10 %
β−
+ 10 n −→ 238
93 Np −→
2.21 d
238
94 Pu
12-8
Fusion Reactors and Other Conversion Devices
Chap. 12
Chapter 13
Nuclear Technology in
Industry and Research
PROBLEMS
1. You are to select a radionuclide to indicate the arrival of a new batch of
petroleum in a 700-km long oil pipeline in which the oil flows with a speed
of 1.4 m/s. Which of the following radioisotopes would you pick and explain
why: 24 Na, 35 S, 60 Co, or 59 Fe?
Solution:
Travel over a distance of 700 km at a speed of 1.4 m/s requires 139 hours. 24 Na
has too short a half life (15 h) and 35 S is difficult to detect because it emits
only beta particles. Either 60 Co or 59 Fe could be used. The latter would be a
better choice, if available, because of its shorter half life (45 d as compared to
5.27 y).
2. A biological tissue contains 131 I (T1/2 = 13.2 h) with an activity concentration
of 3 pCi/g. What is the smallest mass of this tissue for which the 131 I can be
detected? State and explain all assumptions made.
Solution:
Note that 131 I has a half life of 8.04 d, not 13.2 h, but that value is not needed
in the solution of the problem. It may be deduced from Eq. (13.2) that the
minimum detectable activity Amin (Bq) is given by CRmin /!, the ratio of the
minimum acceptable count rate (over background) to the detector efficiency.
Here it is assumed that CRmin = 0.5/s and ! = 0.01. Thus Amin = 50 Bq. If
the activity concentration is 3 pCi/g (0.111 Bq/g), then the minimum tissue
mass is 50/0.111 = 451 g.
13-1
13-2
Nuclear Technology in Industry and Research
Chap. 13
3. What is the atomic concentration (atoms/g) of 60 Co in a tagged 5-gram sample
of iron that produces a photon count rate (1.17 and 1.33 MeV photons) of 335
counts/minute over background. The detector has an efficiency of 0.15 for 60 Co
photons
Solution:
! be the number of 60 Co atoms per gram of tagged material. The decay
Let N
constant for a 5.27 year half life is λ = 4.168 × 10−9 s−1 . Each decay of 60 Co
produces two gamma rays (1.17 and 1.33 MeV). From Eq. (13.1) the count rate
in a detector with efficiency " produced by a sample of mass m is
The number of
60
! m.
CR = 2(photons/decay)"λN
Co atoms per gram of tagged sample is thus
(336 min−1 )(60 s/min)
! = CR =
N
= 8.93 × 108 atoms/g.
2"mλ
(2)(0.15)(5 g)(4.168 × 10−9 s−1 )
4. Iron pistons rings in a motor have 60 Co imbedded as a label with a concentration of 5.0 pCi/kg. The engine is lubricated by recirculation of 8 kg of oil.
After 1,000 hours of operation, the oil is found to have a 60 Co concentration of
2.7 fCi/g. How much iron has been worn from the piston rings?
Solution:
NOTE: The
pCi/kg.
60
Co label concentration in the rings should be 5.0 nCi/g, not
The total activity in the oil is (2.7 × 10−15 Ci/g)(8000 g) = 2.16 × 10−11 Ci.
With the corrected specific activity of the piston rings of 5 × 10−9 Ci/g(Fe),
the grams of iron eroded into the oil is (2.16 ×10−11 Ci)/(5 ×10−9 Ci/g(Fe)) =
0.0043 g = 4.3 mg.
5. A radioisotope with a half-life of 30 days is injected into the center of a river at
a steady rate of 10 µCi/min. At 10 km downstream the water activity is found
to be 7.3 nCi/g. What is the river flow rate in m3 /h? What assumptions did
you make?
Solution:
Assume that there is uniform mixing of the radioisotope in the river, that
there is no radioactive decay in passage, and that the water density is ρ =
1000 kg/m3 . Let Q (m3 /h) be the river flow rate and R = 10−5 (Ci/min) ×
60(min/h) = 6 × 10−4 Ci/h be the rate of injection of the radionuclide into
the river. The rate of activity injected upstream must equal the rate at
which activity flows past the measurement site downstream, i.e., R(Ci/h) =
!w (Ci/m3 ) ρw (g/m3 ), from which the flow rate Q is found as
Q(m3 /h) A
Q=
R
6 × 10−4 (Ci/h)
=
= 0.082 m3 /h,
6
!w
(10 g/m3 )(7.3 × 10−9 Ci/g)
ρA
a very sluggish river, indeed.
Nuclear Technology in Industry and Research
13-3
Chap. 13
NOTE: This problem would be more realistic if the downstream water specific
activity were 7.3 pCi/kg. In this case, the volumetric flow rate would be 106
times larger.
6. Five grams of water containing a radionuclide with a concentration of 107 Bq/L
and a half life of 1.3 d are injected into a small pond without an outlet. After
10 days, during which the radioisotope is uniformly mixed with the pond water,
the concentration of the water is observed to be 1.75 µBq/cm3 . What is the
volume of water in the pond?
Solution:
Let V = the unknown pond volume (m3 ) and Ao the activity injected (50,000
Bq). The decay constant of the radionuclide is λ = ln 2/T1/2 = 0.5332 d−1 and
the decay time is t = 10 d. Given is C, the concentration in the pond after
time t = 10 d, namely, 1.75 µBq/cm3 = 1.75 Bq/m3 . Since
V C = Ao e−λt ,
it follows that
V =
50, 000 Bq × e−5.332
3
1.75 Bq/m
= 138 m3 ,
a truly small pond.
7. A radioactive tracer, with a half life of 10 days, is injected into an underground
aquifer at t = 0. Eighty-five days later the radioisotope is first observed in a
monitoring well 500 m from the injection point. What is the speed of water
flowing in the aquifer between the two wells?
Solution:
The speed is 500 m in 85 days, or 5.82 m/d.
8. The sizes of objects imbedded in a homogeneous material are to be determined
by radiography. Explain what physical properties the objects and the homogeneous material should have for (a) x-ray radiography, (b) neutron radiography,
and (c) beta-radiography to be the preferred choice.
Solution:
X ray images have greater contrast when the objects have greater mass thickness (density times distance along the x-ray beam) than the displaced homogeneous medium. Contrast is further enhanced when the object is made
of materials of greater atomic mass than the homogeneous matrix. Neutron
radiography requires objects of low atomic mass with high scattering cross sections, for example, water or plastics. Beta radiography, like x radiography, is
sensitive to mass thickness; however the entire subject must be thinner than
the range of the beta particles used.
13-4
Nuclear Technology in Industry and Research
Chap. 13
9. What type of a radioactive source should be used, and how should it be used
in thickness gauging processes if it is to be used for (a) gauging paper thickness, (b) the control of sheet metal thickness between 0.1 and 1 cm, and (c)
controlling the coating thickness of adhesive on a cloth substrate?
Solution:
As described in Sec. 13.4.2, gauging paper thickness is well suited to betaparticle sources. The greater mass thickness of sheet metal requires x ray or
gamma ray sources. Controlling coating thickness is a special challenge because
the coating is often similar to the substrate in density and atomic mass. In
that case, flourescence thickness gauging is called for.
10. It is proposed to a use a 90 Sr source (half-life 29.1 y) in a thickness gauge for
aluminum sheets. This radionuclide source is in secular equilibrium with its
90
Y daughter (half-life 64 h). Both radioisotopes emit beta particles with no
accompanying gamma rays. The average energies of the 90 Sr and 90 Y beta
particles are 195 keV and 602 keV, respectively. What is the maximum thickness of aluminum for which gauging by 90 Sr is practical? Ranges of electrons
in aluminum are shown in Fig. 7.16.
Solution:
The mass thickness CSDA range for 90 Y beta particles of mean energy is about
0.2 g/cm2 (Fig. 7.16). For aluminum of density 2.7 g/cm3 , the range is about
0.074 cm. This is approximately the greatest thickness of aluminum suitable
for gauging using beta particles from 90 Y.
11. It has been suggested that Napoleon died as a result of arsenic poisoning since
the growing end of his hair has been found, through NAA analysis, to have
abnormally high levels of arsenic. What hair sample mass (in grams) is needed
to detect arsenic with a concentration of 0.001 ppm?
Solution:
From Table 13.3, the minimum mass of arsenic detectable using neutron activation analysis is 0.0002 µg. For a mass fraction of 10−9 (0.001 ppm) of arsenic
in hair, the required hair sample mass is 2 × 10−10 /10−9 = 0.2 g.
Nuclear Technology in Industry and Research
Chap. 13
13-5
12. Mercury pollution in water is of concern since fish often concentrate this element in their tissues. To measure such mercury contamination by neutron
activation analysis, a fish sample is irradiated in a reactor in a thermal neutron
flux φ = 1.5 × 1012 cm−2 s−1 . The stable mercury isotope 196 Hg has a neutron
absorption cross section for thermal neutrons of 3.2 kb. The resulting 197 Hg has
a half life of 2.67 d and can be detected in a fish sample at a minimum activity
of 15 Bq. What irradiation time is needed to detect mercury contamination at
a level of 30 µg/g (30 ppm) in a 10-g fish sample?
Solution:
Note: A more realistic problem is for mercury contamination at a level of
30 ng/g (30 ppb). We assume this value in the following solution.
Let No196 be the number of atoms of 196Hg present in the fish sample, assumed
to be constant during irradiation. The isotopic abundance of 196 Hg is found
from Ap. A to be 0.0015. Thus
No196 = 0.0015N Hg
=
10 g(sample) × 30 × 10−9 g(Hg)/g(sample) × Na atoms/mol
196 g/mol
= 1.383 × 1012 atoms.
Let N 197 (t) be the number of atoms and A197 (t) the activity (Bq) of 197 Hg at
time t (s) after the start of neutron irradiation. It is given that A197 (t) = 15
Bq, and the problem is to solve for the required time t. The decay constant
for 197Hg is λ197 = ln 2/T1/2 = 3.005 × 10−6 s−1 . The number of 197 Hg atoms
N 197 (t) may be determined by solving the governing radioactive decay and
buildup equation
dN 197(t)
= −λN 197 (t) + No196 σ 196 φ,
dt
to obtain
N 197(t) =
The
197
No196σ 196 φ
(1 − e−λ197 t ).
λ197
Hg activity is λ197 N 197(t) or
A197 (t) = No196 σ 196 φ (1 − e−λ197 t ),
in which σ 196 = 3.2 × 10−21 cm2 /atom and φ = 1.5 × 1012 cm−2 s−1 . Solving
this equation for the irradiation time to produce A196 (t) = 15 s−1 , we find
!
"
1
A197(t)
t=−
ln 1 − 196 196
= 753 s = 12.5 min.
λ197
No σ φ
For the stated 30 ppm concentration, the irradiation time is only 0.725 s.
13-6
Nuclear Technology in Industry and Research
Chap. 13
13. Neutron transmutation doping is a process in which high purity silicon is placed
in a nuclear reactor where some of the silicon atoms are transmuted to stable
phosphorus atoms through the reactions
30
14 Si
+ 10 n −→
31
14 Si
β−
−→
2.62 h
31
15 P.
After annealing to remove crystalline defects caused by the fast neutrons, the
resulting semiconductor can be used in devices such as power thyristors which
require a large area of a semiconductor with uniform resistivity. The thermalneutron capture cross section in 30 Si is 0.107 b. What thermal flux is needed
to produce a phosphorus impurity concentration of 8 parts per billion following
an 20-hour irradiation?
Solution:
The phosphorus impurity concentration is evaluated after decay of all 31 Si.
Thus, the number of atoms N (t) of 31 Si after irradiation for time t is required
to be the fraction 8 × 10−9 of the number of atoms No of 30 Si. As in Problem
13.12,
N (t)
σφ
=
(1 − e−λt ) = 8 × 10−9 ,
No
λ
in which σ = 0.107 × 10−24 cm2 , t = 20 h = 72,000 s, and λ = 7.349 × 10−5
s−1 . Solving this equation for the required thermal neutron flux yields
φ=
λ N (t)
1
= 5.5 × 1012 cm−2 s−1 .
σ No (1 − e−λt )
14. If potatoes receive gamma-ray doses between 60 and 150 Gy, premature sprouting is inhibited. Such irradiation can be done in an irradiator with a large 60 Co
source. The maximum dose the potatoes receive from a given source in a given
time can be estimated by assuming all the gamma rays are absorbed in the
potatoes. What is the minimum activity of 60 Co needed in an irradiator to
deliver such a dose of 250 kGy to 100,000 kg of potatoes in 8 hours? How
realistic is it to suppose all the gamma-ray energy is deposited in the potatoes?
Solution:
First, let it be said that the supposition that all gamma-ray energy is deposited
in the potatoes is very unrealistic, and yields a very conservative (lower) estimate of the minimum activity required in an irradiator. For irradiation by
60
Co, the gamma-ray energy released per decay, and carried by two photons,
is Eγ = 2.5 MeV (Appendix D). The energy absorption rate in the potatoes is
Eabs =
2.5 × 105 J/kg × 105 kg
= 5.42 × 1018 MeV/s.
8 × 3600 s × 1.602 × 10−13 J/MeV
The energy release rate of a 60 Co source with activity A is Erel = A Bq ×
2.5 MeV. Equating Eabs = Erel , we find A = 2.17 × 1018 Bq = 5.86 × 107 Ci.
Nuclear Technology in Industry and Research
Chap. 13
13-7
15. Bacteria such as salmonella can be killed by radiation doses of gamma rays of
several kGy. Estimate the irradiation time needed to sterilize a small delicate
medical instrument sealed in plastic if the instrument is placed 30 cm from a
small 60 Co source with an activity of 10 kCi.
Solution:
In this problem it is assumed that, per decay, 60 Co releases two 1.25-MeV
gamma rays and that attenuation in 30 cm of air may be neglected. From
Eq. (7.25), the gamma-ray flux density is
φ=
Sp
7.4 × 1014 cm−2 s−1
=
= 6.54 × 1010 cm−2 s−1 .
4πr 2
4π(30 cm)2
From Sec. 13.5.1, the sterilization dose is assumed to be 20 kGy, evaluated as
kerma in water. From Eq. (9.5), the kerma is given by
K(Gy) = 1.602 × 10−10E MeV ×
µtr cm2
× φ cm−2 s−1 t s = 2 × 104 Gy.
ρ
g
From Table C.3, µtr /ρ = 0.02974 cm2 / g. Solving for the irradiation time t
yields
K Gy
t=
= 5.14 × 104 s = 14.3 h.
(1.602 × 10−10 E)(µtr /ρ)φ
Note: In the remaining problems you may assume that the proton, deuteron,
and helium nucleus have masses of, respectively, 1.6726×10−27, 3.3436×10−27,
and 6.6447×10−27 kg. The unit of charge is 1.6022×10−19 C. Thus, q/m values
are, respectively, 9.5788×107, 4.7918×107, and 4.8225×107 C/kg.
16. How long does it take for a helium nucleus to make one revolution in a magnetic
field of strength 1.5 Wb/m2 ? If the speed of the particle is 2 × 107 m/s what
is the radius of the circular trajectory?
Solution:
From Eq. 13.5, the radius is given by
R=
v
2 × 107
=
= 0.2765 m.
Bq/m
1.5 × 4.8225 × 107
The period of revolution is the circumference of the trajectory divided by the
velocity, namely,
t=
2πR
= 0.0869 µs.
v
13-8
Nuclear Technology in Industry and Research
Chap. 13
17. Consider a fixed-frequency cyclotron with a radius of 30 cm, a radiofrequency
voltage supply of 107 cycles/s, and a magnetic field of 1.3 Wb/m2 (telsa). What
is the energy of deuterons than can be produced in such a machine?
Solution:
From Eqs. (13.5) and (13.7), The energy is given by
E=
!
"2
1 2
1 rBq
mv =
.
2
2 m
Substitution of r = 0.3 m, and B = 1.3 T = 1.3 Wb/m2 yields a velocity of
1.868 ×107 m/s and an energy of 5.839 ×10−13 J = 3.644 MeV.
18. If the frequency of the applied voltage to the dees of a cyclotron is 1.5×107 s−1 ,
what are the magnetic field strengths needed to accelerator protons, deuterons
and helium nuclei? If the radius at ejection is 40 cm, what is the energy of
each of the particles?
Solution:
From Eqs. (13.6) and (13.7),
v
q
1
m
= 2πf = B
and E = mv2 = r 2 B 2 (q/m)2 .
r
m
2
2
Substitution of r = 0.4 m and f = 1.5 ×107 s−1 , yields the following results:
For all particles, velocities are 3.77 ×107 m/s. For protons, deuterons, and He
nuclei, respectively, B = 0.984, 1.97, and 1.95 T (Wb/m2 ; E = 7.42, 14.83,
and 29.47 MeV.
19. What is the mass (compared to its rest mass) of a 1 GeV electron? At what
fraction of the speed of light are such electrons traveling?
Solution:
From Chapter 2, Section 2.1,
2
m = mo + T /c and β =
#
!
T
1− 1+
mo c2
"−2 $1/2
#1−
!
"−2
1
T
1+
.
2
mo c2
With the electron kinetic energy T equal to 1000 MeV = 1.602×10−10 J, mo =
9.109 × 10−31 kg, and c = 2.998 × 10−8 m/s, it follows that m = 1.782 × 10−27
kg and 1 − β # 1.31 × 10−7 .
Nuclear Technology in Industry and Research
Chap. 13
13-9
20. How many orbits per second do the 1 TeV protons make in the Tevatron?
Solution:
From Eq. 13.10,
"
#−1
q !
q
T
2
ω = 2πf = B
1−β = B
1+
.
mo
mo
mo c2
Given that T = 106 MeV, mo c2 = 931.5 MeV, and, from Section 13.6.6, B =
3.64 tesla, it follows that f = 5.16 × 104 s−1 .
21. Consider a betatron with the following properties: maximum field strength at
orbit of 0.4 Wb/m2, operational frequency of 60 cycles/s, and a stable orbit
radius of 84 cm. Show that the energy gain per orbit is about 400 eV and the
final energy is about 100 MeV.
13-10
Nuclear Technology in Industry and Research
Chap. 13
Chapter 14
Medical Applications of
Nuclear Technology
PROBLEMS
1. Relate a personal experience with a diagnostic-radiology or nuclear-medicine
procedure. Were risks and benefits of the procedure explained to you? Were
you given information about radiation doses associated with the procedure? In
reading this chapter, have you gained a better understanding of the procedure?
What insights have been gained and what new questions have come to mind?
Solution:
This question is best suited to classroom discussion.
2. Consider the two x-ray spectra depicted in Fig. 14.2 produced by the same
x-ray machine. Explain why the spectra are shifted in energy.
Solution:
As discussed in Section 7.3.1, the cross section for the photoelectric effect increases strongly as the photon energy decreases. An aluminum filter thus is
more effective in attenuating photons of lower energy than those of higher energy. This explains why the bremsstrahlung spectra in Fig. 14.2 are heavily
attenuated at low energy rather than increasing without limit as the energy
approaches zero. It also explains why a thicker filter is more effective than a
thinner filter in attenuating lower energy photons and why the energy spectrum
of photons passing through the filter is ”hardened” to a greater extent with the
thicker filter.
14-1
14-2
Medical Applications
Chap. 14
3. Consider an x-ray examination of bone, approximated as a 2-cm diameter cylinder. In the image, what is the bone’s range of subject contrast for 100-keV x
rays? Explain why the image contrast of the bone is less than the subject
contrast. What can be done to improve the image contrast?
Solution:
According to Eq. (14.2), the object contrast is given approximately by
Cs ≡
Ib − If
If
=1−
= 1 − exp[−(µf − µb )x].
Ib
Ib
From the data for 100-keV x rays given beneath Eq. (14.2), the attenuation
coefficient µf for tissue is 0.164 × 1.0 = 0.164 cm−1 and that for bone, µb , is
0.175×1.85 = 0.324 cm−1 . Using the above equation it is found that the object
contrast varies from 0 at x = 0 to 0.27 at x = 2 cm.
As discussed in Sec. 14.1.1, image contrast is less than subject contrast for several reasons. One is the contribution of scattered x rays to the “background.”
At low exposures, statistical variations in the x-ray fluence at the image result
in “quantum mottle” in the image, which degrades the image contrast.
To increase the contrast, lower energy x rays would produce a larger difference
(µf − µb ) and thus a larger value of Cs .
4. Many source and detector configurations can be used in x-ray tomography.
Explain how the machine design is influenced by (a) minimizing costs, (b)
increasing image resolution, (c) minimizing patient exposure, (d) minimizing
the exposure time, (e) and increasing image contrast.
Solution:
Cost is governed by the usual industrial considerations of purchasing, marketing, mass production, quality control, and customer support. As x-ray tomography has progressed from generation 1 (single pencil beam of x rays with a
single receiver rotated synchronously) to generation 4 (rotating x-ray fan beam
with a stationary circular receiver array) the increased costs due to complexity were balanced by improvements in receiver efficiency and vastly increased
computer power. These improvements all increased image resolution and minimized exposure time and patient dose. Improvements in receiver efficiency also
greatly improved image resolution and contrast.
Medical Applications
Chap. 14
14-3
5. Why is 99m Tc useful in diagnostic nuclear medicine but not therapeutic? Describe how decay characteristics such as half-life, atomic number, type of radiation, and energy of radiation affect the choice of a radionuclide for a particular
nuclear-medicine procedure.
Solution:
As indicated in Fig. 5.9, 99m Tc decays by gamma-ray emission and internal
conversion with a half-life of six hours. Appendix D lists the 140.5 keV photon
as the principal radiation emitted, with other emissions consisting of very low
energy internal-conversion and Auger electrons. The gamma ray energy is high
enough to penetrate tissues of the body and low enough to interact efficiently
with radiation detectors such as scintillators in gamma-ray cameras. Therapy
requires the cell-killing effects of charged particles such as alpha and beta particles. The choice of a radionuclide for a procedure depends on many factors,
among which are:
(a) The feasibility of production. For example, ideally one would have available
an isotopically pure precursor (target) nuclide and reaction products that
can be purified, leading to a radionuclide free of contaminants and in an
appropriate chemical form.
(b) The half-life. For example, the radionuclide would have a radiological halflife long enough for practical transport from the point of production to the
point of use. The chemical form of the radionuclide would have an effective
half-life (biological and radiological) in the body short enough to minimize
exposure of the patient, care givers, and family. The radiopharmaceutical
99m
Tc itself has a radiological half-life of six hours, which is very appropriate for medical procedures. It is extracted from the parent 99 Mo which
has a half-life of 66 hours, which is adequate for transport from production
site (nuclear reactor) to hospital site.
(c) The radiations emitted. For imaging purposes, a radionuclide such as
99m
Tc is desired because its radiations penetrate the body with minimal
radiation exposure and interact efficiently with radiation detectors outside
the body. For positron emission tomography, of course, a positron emitter
such as 18 F is required. For therapy purposes, a beta emitter such as 90 Y
is required so that the radiation energy is deposited close to the particles
point of origin, as discussed in Sec. 14.5.3,
6. Explain why positron-emitting isotopes are not usually produced in nuclear
reactors.
Solution:
Positron emitters are ”proton-rich” isotopes, whose decay is described in Sec. 5.3.4.
They are commonly produced in accelerators by bombarding stable isotopes
with beams of hydrogen or deuterium nuclei. A good example is the reaction
18
O(p,n)18 F. In contrast, fission products and isotopes produced in nuclear reactors by neutron capture or reactions such as (n,γ), (n,p), etc. are ”neutronrich” and typically decay by beta-particle emission.
14-4
Medical Applications
Chap. 14
7. To treat thyroid cancer 131 I is injected in the patient where it rapidly accumulates in the thyroid. 131I with a half-life of 8.0 d emits beta particles with
an average energy of 182 keV/decay and gamma rays with an average energy
of 382 keV/decay. In addition, 131I has a biological half-life in the thyroid of
4.1 d. (a) What is the effective half-life of 131 I in the thyroid. (b) How many
millicuries of this radioisotope should be injected to deliver a 250 Gy dose to
the 20-gram thyroid?
Solution:
(a) The effective decay constant is the sum of the radiological and biological
decay constants, namely,
!
"
1
1
λeff = λr + λb = ln 2
+
= 0.2557 d−1 = 2.96 × 10−6 s−1 .
r
b
T1/2
T1/2
The effective half-life is thus
eff
T1/2
= ln 2/λeff = 2.71 d.
(b) Because of the small physical size of the thyroid, the approximation may
be made that all of the beta-particle energy and none of the gamma-ray is
absorbed. For the purpose of this problem, it may also be assumed that
all of the injected radioiodine is immediately taken up by the thyroid.1
Thus, for a 250 Gy dose, the energy absorbed in the thyroid is 250 J/kg
× 0.02 kg = 5.0 J. The number of decays required to deliver this energy is
No =
5.0 J
= 1.72 × 1014 .
182 keV/decay × 1.602 × 10−16 J/keV
If Ao is the injected activity (Bq), the time dependent activity in the
thyroid is A(t) = Ao e−λeff t and the time dependent number of atoms
present in the thyroid is
N (t) = A(t)/λr .
The number of decays in the thyroid in all time is
# ∞
No =
dt A(t) = Ao /λeff .
0
Thus, initial activity that must be injected is
Ao = No λeff = 5.1 × 108 Bq = 1.4 mCi.
1 In
fact, according to ICRP Publication 30, only 30 percent of the injected radioiodine is deposited
in the thyroid, where it is retained with a biological half-life of 80 days.
Medical Applications
Chap. 14
14-5
8. Explain why SPECT must use a physical collimator, while PET has no need of
such a collimator. Also explain what determines the image resolution in both
procedures.
Solution:
In SPECT, single gamma or x rays emitted in an object are detected by a
plane receiver. Collimation is required to identify the coordinates of a line
in the object, perpendicular to the receiver, along which the emission occurs.
In PET, by contrast, two annihilation photons, travelling simultaneously in
opposite directions are detected by two small receivers. The orientation of the
receivers identify a line within the object along which the positron decay and
photon emission occurs.
9. Explain why nuclear wastes per unit activity is of less societal and physical
concern for medical applications than those produced in nuclear power plants.
Solution:
Radionuclides used in medical applications must be short lived and, in most
cases are photon or electron emitters. In contrast, high level wastes from
nuclear power plants contain long-lived radionuclides, contain alpha particle
emitters, and contain many nuclides taken up and retained in bone.
æ