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
$K_i=\frac{1}{2}m{u}^2$ and $K_f=\frac{1}{2}m{v}^2$, 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 $M=12m$
$\therefore f=\frac{4m\left(12m\right)}{{\left(m+12m\right)}^2}=\frac{48}{169}$