Nuclear Energy, Eighth Edition Solutions to Exercises Raymond L. Murray/ Complete Solved Questions to all chapters, A+ Graded.

Nuclear Energy, Eighth Edition

Solutions to Exercises

Raymond L. Murray
North Carolina State University

Keith E. Holbert
Arizona State University
Tempe, AZ 85287-5706



Please do not make these solutions publicly accessible, online or otherwise. The author
welcomes corrections and suggestions for improvements, additions, and deletions. Since
Microsoft deleted the classic Equation Editor in January 2018, the MathType font must be
installed to view some equations correctly, (https://www.dessci.com/en/dl/fonts/getfont.asp).


Chapter 1 – Energy

1.1 m = 75 kg; v = 8 m/s; EK = ½mv2 = (0.5)(75 kg)(8 m/s)2 = 2400 J

1.2 C = (5/9)(F – 32), F = (9/5)C + 32
(a) C = (5/9)(68 °F – 32) = 20 °C
(b) C = (5/9)(500 °F – 32) = 260 °C
(c) F = (9/5)(–273 °C) + 32 = –459.4 °F
(d) F = (9/5)(1000 °C) + 32 = 1832 °F
(e) F = (9/5)(–40 °C) + 32 = –40 °F
(f) C = (5/9)(212 °F – 32) = 100 °C

1.3 cp = 450 J/(kg·°C), m = 0.5 kg, ΔT = 100 °C
Q = m cp ΔT = (0.5 kg)(450 J/(kg·°C))(100 °C) = 22,500 J = 22.5 kJ

1.4 Molecular weight 28, m = (28 amu)(1.66 × 10–27 kg/amu) = 4.65 × 10–26 kg
At 20 °C, E = 6.07 × 10–21 J, from Example 1.3
v  2 E / m  2 (6.07  10 21 J ) (4.65  10 26 kg )  511 m/s

1.5 Conversion factor (Appendix): 1 hp = 745.7 W
P = (200 hp)(745.7 W/hp) = 149,140 W  149 kW
E = (149 kW)(4 h) = 596 kWh

1.6 (a) ν = c/λ = (3.0 × 108 m/s) / (1.5 × 10–12 m) = 2 × 1020 Hz
(b) ν = c/λ = (3.0 × 108 m/s) / (0.10 × 10–9 m) = 3 × 1018 Hz

1.7 (a) m/m0 = 1/[1  (v/c)2]1/2


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,Nuclear Energy Chapter 21 – Reactor Safety and Security


series expansion: (1 + x)n = 1 + nx + ...
Let n = –1/2, x =  (v/c)2
m/m0 = 1/[1  (v/c)2]1/2  1 + 1/2 (v/c)2
EK = (m  m0)c2 = m0(m/m0  1)c2 = (1/2)m0v2 q.e.d.
(b) m0 = 1000 kg, v = 20 m/s, c = 3 × 108 m/s
Δm = EK /c2  (1/2)(m0)(v/c)2 = (1/2)(1000 kg)[(20 m/s)/(3 × 108 m/s)]2
Δm = 2.22 × 10–12 kg = 2.22 × 10–9 g

1.8 E = (190 × 106 eV)(1.60 × 10–19 J/eV) = 3.04 × 10–11 J

1.9 m = E/c2 = (3.04 × 10–11 J) / (3 × 108 m/s)2 = 3.38 × 10–28 kg

1.10 m = (235 u)(1.66 × 10–27 kg/u) = 3.90 × 10–25 kg
E = mc2 = (3.90 × 10–25 kg)(3 × 108 m/s)2 = 3.51 × 10–8 J

1.11 kilogram fraction: (3.38 × 10–28 kg) / (3.90 × 10–25 kg) = 8.67 × 10–4
joule fraction: (3.04 × 10–11 J) / (3.51 × 10–8 J) = 8.66 × 10–4
MeV fraction: (190 MeV) / [(235)(931.5 MeV)] = 8.68 × 10–4
(differences due to rounding)

1.12 [F (fissions/s)](190 MeV/fission)(1.60 × 10–13 J/MeV) = 1 W
F = (1 J/s)/(3.04 × 10–11 J/fission) = 3.29 × 1010 fissions/s  3.3 × 1010 /s

1.13 (a) E0 = mpc2 = (1.673 × 10–27 kg)(2.998 × 108 m/s)2/(1.602 × 10–13 J/MeV) = 938.6 MeV
(b) E0 = mnc2 = (1.675 × 10–27 kg)(2.998 × 108 m/s)2/(1.602 × 10–13 J/MeV) = 939.8 MeV
These values differ from Table A.2 in a manner proportionate to the number of
significant digits used.

1.14 E = mc2 = (1.6605389 × 10–27 kg)(299792458 m/s)2(1 MeV/1.602176565 × 10–13 J)
= 931.49 MeV

1.15 (a) E = mc2, E0 = m0c2, ΔE = (m  m0)c2
ΔE/E0 = m/m0  1 = [1  (v/c)2]–1/2  1
(1 + ΔE/E0)2 = [1 (v/c)2]–1
(v/c)2 = 1  1/(1 + ΔE/E0)2
v/c = [1  (1 + ΔE/E0)–2]1/2
(b) ΔE/E0 = 0.01, v/c = [1  (1.01)–2]1/2 = 0.140
ΔE/E0 = 0.1, v/c = [1  (1.1)–2]1/2 = 0.417
ΔE/E0 = 1, v/c = [1  (2)–2]1/2 = 0.866

1.16 (a) (34.18 kcal/g)(454 g/lb)/(0.252 kcal/Btu) = 6.16 × 104 Btu/lb
(b) (34.18 kcal/g)(1000 cal/kcal)(4.184 J/cal) = 1.43 ×105 J/g
(c) MH = 2.016 g/mol, molecular weight of H2
NA = 6.022 × 1023 molecules/mol, Avogadro’s number
NA/MH = 2.987 × 1023, number of H2 molecules per gram
H = heat of combustion per molecule


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,Nuclear Energy Chapter 21 – Reactor Safety and Security


= (1.43 × 105 J/g) / [(1.60 × 10–19 J/eV)(2.987 × 1023 molecules/g)]
H  3.0 eV/molecule

m0 c 2  1 
2 2
1.17 E K  ET E 0  m c m0 c  m0 c 2  m0 c 2  1
1 (v 2 c 2 )  2 2
 1 (v c )  
2
EK 1 v  1 1
1 2
  1    2
 v  c 1 2
m0 c 1 (v 2 c 2 ) c   E K   EK 

1  1 
 2   2 
 m0 c   m0 c 

1.18 Graph of error in using classic rather than relativistic EK equation
Error = [EK(classic) – EK(relat.)] (100%) / EK(relat.)
0
Error Using Classic Expression (%)




-10
-20
-30
-40
-50
-60
-70
-80
-90
-100
0 0.2 0.4 0.6 0.8 1
Fraction of the Speed of Light, v/c


1.A Sampling of ALBERT results:
(a) Electron at 0.5c: Ratio of mass to rest mass is 1.1547
(b) Proton of 1000 MeV total energy: Kinetic energy is 9.8899e-012 J, or 61.728 MeV
(c) Neutron of 0.025 eV kinetic energy: Momentum is 3.663e-024 kg-m/sec
(d) Deuteron with m/m0 = 1.01: Velocity is 42082095.5242 m/sec
(e) Alpha with momentum 10–19 kg·m/s: Relativistic mass is 6.653e-027 kilograms


Chapter 2 – Atoms and Nuclei

2.1 Density of graphite ρC = 1.65 g/cm3, molecular weight MC = 12.011
NC = ρC NA/MC = (1.65 g/cm3)(6.022×1023 atoms/mol)/(12.011 g/mol) = 8.273 × 1022/cm3
(a) NC-12 = γC-12 NC = (0.988922)(8.273 × 1022/cm3) = 8.181 × 1022/cm3
(b) NC-13 = γC-13 NC = (0.011078)(8.273 × 1022/cm3) = 9.165 × 1020/cm3

2.2 N = ρ NA/M = (19.3 g/cm3)(6.022 × 1023 atoms/mol)/(197 g/mol) = 0.0590 × 1024/cm3
Side of cube s such that s3 = 1, so that
s = 1/N1/3 = 1/(59.0 × 1021)1/3 = 2.57 × 10–8 cm
Equal volumes means s3 = (4/3)πr3 or r = [3/(4π)]1/3s


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, Nuclear Energy Chapter 21 – Reactor Safety and Security


r = (0.620)(2.57 × 10–8 cm) = 1.59 × 10–8cm
V = s3 = (2.57 × 10–8 cm)3 = 1.70 × 10–23 cm3

2kT 2 (1.38  10 23 J/K ) (293 K )
2.3 (a) v p    2200 m/s
mn 1.675  10 27 kg
2kT 2 (1.38  1023 J/K) (500  273K)
(b) v p    3570 m/s
mn 1.675  1027 kg

2.4 Average energy per molecule (3/2)kT
Total energy for N molecules E = (3/2)NkT
Add energy ΔE = (3/2)NkΔT which is also ΔE = cVNmΔT from basic heat concepts,
where cV = specific heat and m = mass of molecule. Equating, (3/2)k = cVm or
cV = (3/2)(k/m) q.e.d.

2.5 For elemental helium, which is almost entirely comprised of He-4:
3k (3)(1.3806  10 23 J/K )
cV    3116 J/(kg·K)
2 m (2)(4.0026 amu)(1.6605  10 27 kg/amu)

2.6 (a) E = h ν = h c / λ = (4.1357×10–15 eV·s)(2.998×108 m/s) / (280×10–9 m) = 4.43 eV
(b) E = h ν = h c / λ = (4.1357×10–15 eV·s)(2.998×108 m/s) / (1.5×10–12 m) = 8.27×105 eV
(c) E = h ν = h c / λ = (4.1357×10–15 eV·s)(2.998×108 m/s) / (0.1×10–9 m) = 1.24×104 eV

2.7 Energy E1 = (13.6 eV)(1.602 × 10–19 J/eV) = 2.18 × 10–18 J
Frequency ν = E/h = (2.18 × 10–18 J) / (6.63 × 10–34 J·s) = 3.29 × 1015 Hz

2.8 E3 = E1 /(3)2 = –13.6/9 = –1.51 eV
R3 = (3)2 R1 = 9 (0.53 × 10–10 m) = 4.77 × 10–10 m = 4.77 × 10–8 cm
ΔE = E3  E1 = –1.5  (–13.6) = (12.1 eV)(1.60 × 10–19 J/eV) = 1.94 × 10–18 J
ν = ΔE/h = (1.94 × 10–18 J) / (6.63 × 10–34 J·s) = 2.92 × 1015 Hz

2.9 Carbon-14: Z = 6, A = 14
Numbers: electrons 6, protons 6, neutrons 8
Electrons in orbits: inner 2, outer 4




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