Solutions Manual For Alternative Energy Systems 1st Edition By Hodge

Chapter 15

1. Compute the energy resulting from the 92U 235 +0 n1 →56 Ba137 +36 Kr 97 + 2 0 n1
nuclear reaction.
− 27 8 m − 13
Definitions: amu := 1.660539⋅ 10 ⋅ kg c := 3 ⋅ 10 ⋅ MeV := 1.602176⋅ 10 ⋅J
s
Specify the amu of all atoms and particles:
amuU235 := 235.0439223 ⋅ amu
amun1 := 1.008665⋅ amu
amuBa137 := 136.905821⋅ amu
amuKr97 := 96.948560 ⋅ amu
∆mass := amuBa137 + amuKr97 + 2 ⋅ amun1 − amuU235 − amun1
−3
∆mass = −180.876 × 10 ⋅ amu
2
∆E := ∆mass⋅ c
− 12
∆E = −27.032 × 10 J
0
∆E = −168.719 × 10 ⋅ MeV

,2. Verify the energy release of the following fusion reactions:

1 H 2 + 1 H 2 → 1 H 3 + 1 H 1 + 4 . 03 MeV
1 H 2 + 1 H 3 → 2 He 4 + 0 n 1 + 17 .59 MeV
1 H 2 + 2 He 3 → 2 He 4 + 1 H 1 + 18 .35 MeV
1 H 3 + 1 H 3 → 2 He 4 + 2 0 n 1 + 11 .33 MeV

− 27 8 m − 13
Definitions: amu := 1.660539⋅ 10 ⋅ kg c := 3 ⋅ 10 ⋅ MeV := 1.602176⋅ 10 ⋅J
s
Specify the amu of all atoms and particles:
amuH1 := 1.007825⋅ amu
amuH2 := 2.014102⋅ amu
amuH3 := 3.016049⋅ amu
amun1 := 1.008665⋅ amu
amuHe3 := 3.016029⋅ amu
amuHe4 := 4.002603⋅ amu

1 H 2 + 1 H 2 →1 H 3 + 1 H 1 + 4.03 MeV
∆mass := amuHe3 + amuH1 − 2amuH2
−3
∆mass = −4.35 × 10 ⋅ amu
2
∆E := ∆mass⋅ c
− 15
∆E = −650.101 × 10 J
0
∆E = −4.058 × 10 ⋅ MeV

H 2 + 1 H 3 → 2 He 4 + 0 n1 + 17.59 MeV
∆mass := amuHe4 + amun1 − amuH2 − amuH3
−3
∆mass = −18.883 × 10 ⋅ amu
2
∆E := ∆mass⋅ c
− 12
∆E = −2.822 × 10 J
0
∆E = −17.614 × 10 ⋅ MeV

1 H 2 + 2 He 3 → 2 He 4 + 1 H 1 + 18.35 MeV
∆mass := amuHe4 + amuH1 − amuH2 − amuH3
−3
∆mass = −19.723 × 10 ⋅ amu
2
∆E := ∆mass⋅ c
− 12
∆E = −2.948 × 10 J
0
∆E = −18.397 × 10 ⋅ MeV

,1 H 3 + 1 H 3 → 2 He 4 + 2 0 n1 + 11.33 MeV

∆mass := amuHe4 + 2amun1 − 2amuH3
−3
∆mass = −12.165 × 10 ⋅ amu
2
∆E := ∆mass⋅ c
− 12
∆E = −1.818 × 10 J
0
∆E = −11.347 × 10 ⋅ MeV

, 3. In the beta decay reaction, 82 Pb214 →83 Bi 214 + −1e0 + ν , determine the times
required for the number of original atoms to be reduced by 25, 50, and 75 percent.
If the number of nuclides is to be reduced by 75 percent, only 25 percent will be left.
The half life of Pb 214 thalflife := 26.8⋅ min


⎛ 0.75 ⎞ ⎛1⎞
pcatom := ⎜ 0.50 t := ⎜ 1 ⋅ hr
⎜ ⎜
⎝ 0.25 ⎠ ⎝1⎠

Given

pcatom = exp⎛⎜ −0.69315 ⋅ ⎞
t

⎝ thalflife
⎠
t := Find( t)


⎛ 667.378 × 100 ⎞ ⎛ 185.383 × 10− 3 ⎞ ⎛ 11.123 × 100 ⎞
⎜ ⎜ ⎜
t = ⎜ 1.608 × 103 ⎟ s t = ⎜ 446.665 × 10− 3 ⎟ ⋅ hr t = ⎜ 26.8 × 100 ⎟ ⋅ min
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ 3 ⎜ −3 ⎜ 0
⎝ 3.216 × 10 ⎠ ⎝ 893.33 × 10 ⎠ ⎝ 53.6 × 10 ⎠