A neutron makes a head-on elastic collision with a stationary...

One deuteron consists of one proton and one neutron. As the mass of a proton and a neutron is approximately same $\left(\approx 1.67 \times 10^{-27} \mathrm{~kg}\right)$, we assume that the mass of the neutron be $m$ and the mass of the deuteron be $2 m$ as the electron mass is negligibly small compared to that of proton and neutron. Let the initial velocity of the neutron be $u$. As the deuteron is initially at rest, the final velocity of the neutron is v_1=$\left(\frac{m-2 m}{m+2 m}\right)$u=-$\frac{m u}{3 m}=-\frac{u}{3} .$and the velocity of the deuteron is v_2=$\frac{2 m}{m+2 m}$u=$\frac{2}{3}$u $\text{. }$Total energy before collision is $E_1=\frac{1}{2} m u^2$ After collision, the energy gained by the deuteron is E_d=$\frac{1}{2} \times($2 m) v_2^2=m \times\left(\frac{2}{3}u\right)^2=$\frac{4}{9}$m u² . $E_d$ is the amount of energy lost by the neutron. Therefore the fractional loss of energy of the neutron is $\frac{E_d}{E_1}=\frac{(4 / 9) m u^2}{(1 / 2) m u^2}=\frac{8}{9}$