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

$\frac{11}{12}$

$\frac{2}{11}$

$\frac{48}{121}$

$\frac{48}{169}$

Solution

$\frac{48}{169}$

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 = $\frac{1}{2}$ mu²

and K_f = $\frac{1}{2}$ mv²

But v = $\frac{(m - M) u}{m + M}$

∴ K_f = $\frac{1}{2} m {(\frac{m - M}{m + M})}^{2}$ u²

∴ $\frac{K_{f}}{K_{i}} = {(\frac{m - M}{m + M})}^{2}$

The fraction of kinetic energy transferred from the neutron to the carbon atom is

f = $\frac{4 m (12 m)}{{(m + 12 m)}^{2}}$

f = $\frac{48}{169}$

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