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

The Correct Option is D

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