Nuclear Fission: The Basic Process — Question 6

Question 6

Prompt and delayed energy create different calorimeter readings

A burst of $10^{12}$ fissions deposits 180 MeV promptly per event and an additional 12 MeV per event through a single exponential delayed component with mean time $\tau=30.0$ s. Ignore neutrino energy; all stated energy is absorbed. Use $1\,\mathrm{MeV}=1.602\times10^{-13}$ J.

1. Derive total absorbed energy E(t) for t after the prompt pulse.

2. Find E at 30 s and the delayed thermal power at that time.

Question 6 - Solution

Prompt and delayed energy create different calorimeter readings

(a) Method and calculation

$$E(t)=N\{E_p+E_d[1-e^{-t/\tau}]\}\Rightarrow\boxed{E(t)=28.836+1.9224(1-e^{-t/30})\,\mathrm J}.$$$$E(0^+)=28.836\,\mathrm J,\qquad E(\infty)=30.7584\,\mathrm J.$$

(b) Interpretation and further calculation

$$\boxed{E(30)=30.0512\,\mathrm J}.$$$$P_d=\frac{NE_d}{\tau}e^{-t/\tau}\Rightarrow\boxed{P_d(30)=.023574\,\mathrm W}.$$

Physical check

The delayed power is the derivative of only the delayed energy. A short observation window underestimates the eventual calorimetric yield.