A neutron travelling with a velocity v and kinetic energy E collides perfectly elastically head on with the nucleus of an atom of mass number A at rest. The fraction of total energy retained by the neutron is:

{(\frac{A - 1}{A + 1})}^{2}

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{(\frac{A + 1}{A - 1})}^{2}

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{(\frac{A - 1}{A})}^{2}

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{(\frac{A + 1}{A})}^{2}

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Solution

The correct option is A ${(\frac{A - 1}{A + 1})}^{2}$
We are given that the neutron strikes the nucleus and undergoes perfectly elastic collision. The velocity ( $v_{1}$ ) of the neutron can be obtained by combining K.E and Momentum laws
Final velocity $v_{1}$ of the first body

After the collision is $\frac{m_{1} - m_{2}}{m_{1} + m_{2}} u_{1} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}}$
But we have
When $u_{2} = 0$ , $v_{1} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} u_{1}$
$\frac{v_{1}}{u_{1}} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}$
But the ratio of $m_{2}$ to $m_{1}$ is mass number A or we have
Thus, $v_{1} = \frac{1 A}{1 A}$
Now, fraction of the kinetic energy retained $= \frac{F i n a l k i n e t i c i n t h e f i s t o b j e c t}{I n i t i a l k i n e t i c e n e r g y}$
$= \frac{\frac{1}{2} m_{1} v_{1} 2}{\frac{1}{2} m_{1} u_{1} 2} = \frac{v_{1} 2}{u_{1} 2}$
$= {(\frac{m_{1} m_{2}}{m_{1} + m_{2}})}^{2}$
Thus, the answer is K.E= ${(\frac{1 A}{1 A})}^{2} = {(\frac{A 1}{A + 1})}^{2}$

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