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. $\frac{2}{11}$
• C. $\frac{48}{121}$
• D. $\frac{48}{169}$

Answer 1

Answer: (D)

$\frac{48}{169}$

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 $v=\frac{(m-M)u}{m+M}$
$\therefore K_f=\frac{1}{2}m{\left(\frac{m-M}{m+M}\right)}^2{u}^2$
$\therefore \frac{K_f}{K_i}={\left(\frac{m-M}{m+M}\right)}^2$
The fraction of kinetic energy transferred from the neutron to the carbon atom is
$f=\frac{4mM}{(m+M{)}^2}$
But for carbon, $M=12m$
$\therefore f=\frac{4m\left(12m\right)}{(m+12m{)}^2}=\frac{48}{169}$.