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Physics / Chapter 43: Energy from the Nucleus / Section 43-3

The Nuclear Reactor — Question 2

Section 43-3 · Physics pages 7194–7195

Question 2 Thermal power measures the number of fission events per second A teaching model assigns 190 MeV of recoverable heat per fission. A thermal source produces 1.00 MW steadily for one hour. Use 1 MeV = 1.602 × 10−13 J. fission rate Rf ϵf thermal power Schematic; labels, not drawing lengths, specify the data. (a) Infer the fission rate and the event count in one hour. (b) If the assumed recoverable energy is revised upward by 5 percent with measured power unchanged, find the factor by which the inferred event rate changes.
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Question 2 - Solution Thermal power measures the number of fission events per second fission rate Rf ϵf thermal power (a) Method and calculation ϵf = 190(1.602 × 10−13) = 3.0438 × 10−11 J. Rf = P/ϵf = 3.28537 × 1016 s−1 . (b) Interpretation and further calculation N = Rf(3600) = 1.18273 × 1020 . Rf′/Rf = ϵf/(1.05ϵf) = .952381 a 4.762 percent decrease. Physical check Only recoverable heat belongs in this calorimetric inference. Escaping neutrino energy is excluded from the specified heat per event.

Question 2

Thermal power measures the number of fission events per second

A teaching model assigns 190 MeV of recoverable heat per fission. A thermal source produces 1.00 MW steadily for one hour. Use 1MeV=1.602×10−131\,\mathrm{MeV}=1.602\times10^{-13} J.

Diagram shown in the original SVG worksheet.

  1. Infer the fission rate and the event count in one hour.

  2. If the assumed recoverable energy is revised upward by 5 percent with measured power unchanged, find the factor by which the inferred event rate changes.

Question 2 - Solution

Thermal power measures the number of fission events per second

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ϵf=190(1.602×10−13)=3.0438×10−11J.\epsilon_f=190(1.602\times10^{-13})=3.0438\times10^{-11}\,\mathrm J.Rf=P/ϵf=3.28537×1016s−1.\boxed{R_f=P/\epsilon_f=3.28537\times10^{16}\,\mathrm{s^{-1}}}.

(b) Interpretation and further calculation

N=Rf(3600)=1.18273×1020.\boxed{N=R_f(3600)=1.18273\times10^{20}}.Rf′/Rf=ϵf/(1.05ϵf)=.952381a 4.762 percent decrease.\boxed{R_f\prime/R_f=\epsilon_f/(1.05\epsilon_f)=.952381}\quad\text{a 4.762 percent decrease}.

Physical check

Only recoverable heat belongs in this calorimetric inference. Escaping neutrino energy is excluded from the specified heat per event.

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