A neutron collides elastically with the stationary nucleus of an atom of mass number A . If the collision is perfectly elastic, then after the collision the fraction of the initial kinetic energy retained by the neutron, is

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Aicharya · Accepted Answer

Let us assume the mass of the neutron is $m_{n} = m$ and mass of the nucleus is $m_{A} = A m$

Also, let us assume that the initial and the final velocities of the neutron are $v_{i}$ and $v_{f}$ respectively.

Using the formula of final velocity for a perfectly elastic collision we get,

$v_{f} = \frac{(m - A m)}{(m + A m)} v_{i} = (\frac{1 - A}{1 + A}) v_{i}$

$\Rightarrow \frac{v_{f}}{v_{i}} = (\frac{1 - A}{1 + A})$

$\Rightarrow \frac{K_{f i n a l}}{K_{i n i t i a l}} = {(\frac{1 - A}{1 + A})}^{2}$

A neutron collides elastically with the stationary nucleus of an atom of mass number A . If the collision is…

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