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Physics Question on Nuclei

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

1112\frac{11}{12}

B

211\frac{2}{11}

C

48121\frac{48}{121}

D

48169\frac{48}{169}

Answer

48169\frac{48}{169}

Explanation

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 Ki=12mu2andKf=12mv2,K_{i}=\frac{1}{2}mu^{2} and K_{f} =\frac{1}{2}mv^{2}, But v=(m−M)um+Mv=\frac{\left(m-M\right)u}{m+M} ∴Kf=12m(m−Mm+M)2 u2\therefore\quad K_{f} =\frac{1}{2}m\left(\frac{m-M}{m+M}\right)^{2}\,u^{2} ∴KfKi=(m−Mm+M)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 f=4mM(m+M)2\quad\quad f=\frac{4mM}{\left(m+M\right)^{2}} But for carbon, M = 12m ∴f=4m(12m)(m+12m)2=48169\therefore\quad f =\frac{4m\left(12m\right)}{\left(m+12m\right)^{2}}=\frac{48}{169}