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Physics / Chapter 43: Energy from the Nucleus / Section 43-1

Nuclear Fission: The Basic Process — Question 5

Section 43-1 · Physics pages 7160–7161

Question 5 Mass-energy conversion is a small fraction of the reacting rest mass Use an illustrative fission release of 200 MeV from a system of rest energy 236(931.5) MeV. Let c = 3.00 × 108 m/s and use 1 MeV = 1.602 × 10−13 J. sealed calorimeter rest energy to heat Schematic; labels, not drawing lengths, specify the data. (a) Find the fraction of initial rest mass converted to released energy. (b) A sealed calorimeter retains the released energy. Compare the mass change of the nuclear products alone with that of the entire sealed calorimeter.
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Question 5 - Solution Mass-energy conversion is a small fraction of the reacting rest mass sealed calorimeter rest energy to heat (a) Method and calculation ∆m mi = 200 236(931.5) = 9.0978 × 10−4 . ∆m = Q/c2 = 200(1.602 × 10−13) (3.00 × 108)2 = 3.560 × 10−28 kg . (b) Interpretation and further calculation ∆Mproducts,rest = −Q/c2, ∆Mthermal = +Q/c2. ∆Msealed system = 0 if no energy or matter escapes. Physical check The nuclear rest-mass decrease appears as kinetic and thermal energy. Heating contributes to the total inertial mass of a closed system.

Question 5

Mass-energy conversion is a small fraction of the reacting rest mass

Use an illustrative fission release of 200 MeV from a system of rest energy 236(931.5)236(931.5) MeV. Let c=3.00×108c=3.00\times10^8 m/s and use 1MeV=1.602×10−131\,\mathrm{MeV}=1.602\times10^{-13} J.

Diagram shown in the original SVG worksheet.

  1. Find the fraction of initial rest mass converted to released energy.

  2. A sealed calorimeter retains the released energy. Compare the mass change of the nuclear products alone with that of the entire sealed calorimeter.

Question 5 - Solution

Mass-energy conversion is a small fraction of the reacting rest mass

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Δmmi=200236(931.5)=9.0978×10−4.\boxed{\frac{\Delta m}{m_i}=\frac{200}{236(931.5)}=9.0978\times10^{-4}}.Δm=Q/c2=200(1.602×10−13)(3.00×108)2=3.560×10−28kg.\boxed{\Delta m=Q/c^2=\frac{200(1.602\times10^{-13})}{(3.00\times10^8)^2}=3.560\times10^{-28}\,\mathrm{kg}}.

(b) Interpretation and further calculation

ΔMproducts,rest=−Q/c2,ΔMthermal=+Q/c2.\Delta M_{\mathrm{products,rest}}=-Q/c^2,\qquad\Delta M_{\mathrm{thermal}}=+Q/c^2.ΔMsealedsystem=0if no energy or matter escapes.\boxed{\Delta M_{\mathrm{sealed\ system}}=0}\quad\text{if no energy or matter escapes}.

Physical check

The nuclear rest-mass decrease appears as kinetic and thermal energy. Heating contributes to the total inertial mass of a closed system.

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