Q100 / 180 · AIIMS 2017 · Physics

A neutron in a nuclear reactor collides head on elastically with the nucleus of a carbon atom initially at rest. The fraction of kinetic energy transferred from the neutron to the carbon atom is

  • A
    $\frac{11}{12}$
  • B
    $\frac{2}{11}$
  • C
    $\frac{2}{11}$
  • Correct answer
    D
    $\frac{48}{169}$

Solution

Let $m$ and $M$ be the masses of neutron and carbon nucleus (at rest) respectively. If $u$ and $v$ are the velocity of neutron before and after collision, then $K_i=\frac{1}{2} m u^2$ and $K_f=\frac{1}{2} m v^2$, But $v=\frac{(m-M) u}{m+M}$ $\begin{aligned} & \therefore \quad K_f=\frac{1}{2} m\left(\frac{m-M}{m+M}\right)^2 u^2 \\ & \therefore \quad \frac{K_f}{K_i}=\left(\frac{m-M}{m+M}\right)^2\end{aligned}$ The fraction of kinetic energy transferred from the neutron to the carbon atom is $f=\frac{4 m M}{(m+M)^2}$ But for carbon, $M=12 \mathrm{~m}$ $\therefore \quad f=\frac{4 m(12 m)}{(m+12 m)^2}=\frac{48}{169}$

Hence, the correct answer is option: 'D'.

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