Nuclear Fission: The Basic Process — Question 2

Question 2

Atomic masses determine energy released in a specified split

For one supplied channel, initial neutral-atom-plus-neutron mass is 236.052600 u and final neutral-atom-plus-neutrons mass is 235.847600 u. Electron counts balance. Use $1\,\mathrm u c^2=931.5\,\mathrm{MeV}$ and $1\,\mathrm{MeV}=1.602\times10^{-13}$ J.

1. Calculate the Q value and energy per event in joules.

2. If neutrinos carry away 8.00 MeV per event after the full decay sequence, find the maximum energy retained locally and its fraction of Q.

Question 2 - Solution

Atomic masses determine energy released in a specified split

(a) Method and calculation

$$\Delta m=236.052600-235.847600=.205000\,\mathrm u.$$$$\boxed{Q=.205000(931.5)=190.9575\,\mathrm{MeV}=3.05914\times10^{-11}\,\mathrm J}.$$

(b) Interpretation and further calculation

$$\boxed{E_{\mathrm{retained}}=Q-8.00=182.9575\,\mathrm{MeV}}.$$$$\boxed{f_{\mathrm{retained}}=1-8.00/190.9575=.95811}.$$

Physical check

Balanced atomic electron counts permit the atomic-mass subtraction. Local retention also presumes that the other particles and photons deposit their energy.