Question

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. $\frac{2}{12}$
• C. $\frac{48}{121}$
• D. $\frac{48}{169}$

Answer 1

Answer: (D)

Let $m$ and $M$ be the masses of neutron and carbon nucleus (at rest) respectively.
If $u$ and $v_1$ are the velocity of neutron before and after collision, then
$K_{i} = \frac{1}{2}mu^{2}$ and $K_{f} = \frac{1}{2}mv^{2}$, But $v_{1} = \frac{\left(m-M\right)u}{m+M}$
$\therefore K_{f} = \frac{1}{2}m\left(\frac{m-M}{m+M}\right)^{2}\,u^{2}$
$\therefore \frac{K_{f}}{K_{i}} = \left(\frac{m-M}{m+M}\right)^{2}$
The fraction of kinetic energy transferred from the neutron to the carbon atom is
$f = \frac{4mM}{\left(m+M\right)^{2}}$
But for carbon $M = 12m$
$\therefore f = \frac{4m\left(12m\right)}{\left(m+12m\right)^{2}} = \frac{48}{169}$