Nuclear Fission: The Basic Process — Question 4

Question 4

Unequal fission fragments share recoil energy inversely with mass

In the rest frame of a compound nucleus, a simplified two-fragment split releases 170 MeV entirely as kinetic energy. Fragment masses are $m_1=100\,\mathrm u$ and $m_2=140\,\mathrm u$. Ignore emitted neutrons and use nonrelativistic motion.

1. Derive each fragment kinetic energy using momentum conservation.

2. Find the speed ratio and check that the lighter fragment receives more energy.

Question 4 - Solution

Unequal fission fragments share recoil energy inversely with mass

(a) Method and calculation

$$\mathbf p_1=-\mathbf p_2,\quad K_i=p^2/(2m_i)\Rightarrow K_1/K_2=m_2/m_1.$$$$\boxed{K_1=170\frac{140}{240}=99.17\,\mathrm{MeV},\quad K_2=70.83\,\mathrm{MeV}}.$$

(b) Interpretation and further calculation

$$m_1v_1=m_2v_2\Rightarrow\boxed{v_1/v_2=140/100=1.40}.$$$$\boxed{K_1+K_2=170\,\mathrm{MeV},\quad K_1/K_2=1.40>1}.$$

Physical check

Equal and opposite momenta do not imply equal speeds or energies. Actual neutron and photon emissions perturb this two-body partition.