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Question: A neutron travelling with a velocity v and K.E. E collides perfectly elastically head on with the nu...

A neutron travelling with a velocity v and K.E. 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 neutron is

A

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

B

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

C

(A−1A)2\left( \frac{A - 1}{A} \right)^{2}

D

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

Answer

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

Explanation

Solution

Fraction of kinetic energy retained by projectile

ΔKK=(m1−m2m1+m2)2\frac{\Delta K}{K} = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)^{2}

Mass of neutron (m1) = 1 and Mass of atom (m2) = A

∴ ΔKK=(1−A1+A)2\frac{\Delta K}{K} = \left( \frac{1 - A}{1 + A} \right)^{2} or (A−1A+1)2\left( \frac{A - 1}{A + 1} \right)^{2}.