1. Tardigrade
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  3. Physics
  4. 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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Q. 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

NTA AbhyasNTA Abhyas 2020

Solution:

Let us assume the mass of the neutron is $m_{n}=m$ and mass of the nucleus is $m_{A}=Am$
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,
$v_{f}=\frac{\left(m - A m\right)}{\left(\right. m + A m \left.\right)}v_{i}=\left(\frac{1 - A}{1 + A}\right)v_{i}$
$\Rightarrow \, \, \, \frac{v_{f}}{v_{i}}=\left(\frac{1 - A}{1 + A}\right)$
$\Rightarrow \frac{K_{f i n a l}}{K_{i n i t i a l}}=\left(\frac{1 - A}{1 + A}\right)^{2}$