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A neutron makes a head-on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is
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The correct answer is:
$8 / 9$

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
$$
E_d=\frac{1}{2} \times(2 m) v_2^2=m \times\left(\frac{2}{3} u\right)^2=\frac{4}{9} m u^2 .
$$
$E_d$ is the amount of energy lost by the neutron. Therefore the fractional loss of energy of the neutron is
$$
\frac{E_d}{E_1}=\frac{(4 / 9) m u^2}{(1 / 2) m u^2}=\frac{8}{9}
$$
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