Nuclear Fission: The Basic Process — Question 9

Question 9

Missing momentum reveals why two fragments need not be back to back

In a compound-nucleus rest frame, measured fragment momenta are $\mathbf p_1=(100,0)$ MeV/c and $\mathbf p_2=(-96,3)$ MeV/c. Assume the only unobserved momentum is carried by one neutron with $m_nc^2=939$ MeV.

1. Find the neutron momentum vector and direction relative to the positive x axis.

2. Estimate its kinetic energy and check the validity of the nonrelativistic formula.

Question 9 - Solution

Missing momentum reveals why two fragments need not be back to back

(a) Method and calculation

$$\boxed{\mathbf p_n=-(\mathbf p_1+\mathbf p_2)=(-4,-3)\,\mathrm{MeV}/c}.$$$$|\mathbf p_n|=5\,\mathrm{MeV}/c,\quad\boxed{\theta=180^\circ+\tan^{-1}(3/4)=216.87^\circ}.$$

(b) Interpretation and further calculation

$$\boxed{K_n\simeq\frac{(p_nc)^2}{2m_nc^2}=\frac{25}{1878}=.01331\,\mathrm{MeV}}.$$$$\boxed{p_n/(m_nc)=5/939=.005325\ll1}.$$

Physical check

This is an ideal missing-momentum exercise. Additional photons or neutrons would prevent assigning the entire momentum deficit to a single neutron.