Nuclear Fission: The Basic Process — Question 10

Question 10

Calorimetry and event counts give an experimental energy per fission

A detector infers $N=(2.00\pm.04)\times10^9$ events in a pulse. A separate calorimeter measures deposited energy $E=.0600\pm.0018$ J. The stated errors are independent standard uncertainties. Use $1\,\mathrm{MeV}=1.602\times10^{-13}$ J.

1. Infer the deposited energy per fission and its propagated standard uncertainty.

2. A model predicts 190 MeV deposited per fission with negligible uncertainty. Quantify the discrepancy in standard deviations.

Question 10 - Solution

Calorimetry and event counts give an experimental energy per fission

(a) Method and calculation

$$\boxed{q=E/N=3.00\times10^{-11}\,\mathrm J=187.266\,\mathrm{MeV}}.$$$$\frac{\sigma_q}{q}=\sqrt{(.0018/.0600)^2+(.04/2.00)^2}=.036056.$$

(b) Interpretation and further calculation

$$\boxed{\sigma_q=6.752\,\mathrm{MeV},\quad q=(187.3\pm6.8)\,\mathrm{MeV}}.$$$$\boxed{z=\frac{187.266-190}{6.752}=-.405}\quad\text{consistent at this precision}.$$

Physical check

Shared calibration errors would require a covariance term. A measurement of deposited energy need not equal the full nuclear Q value.