Question

A neutron with velocity $v$ suffers head on elastic collision with the nucleus of an atom of mass number A at rest. The fraction of energy retained by the neutron is :

• A. $(\frac{A+1}{A}{)}^2$
• B. $(\frac{A+1}{A-1}{)}^2$
• C. $(\frac{A-1}{A+1}{)}^2$
• D. $(\frac{A-1}{A}{)}^2$

Answer 1

Answer: (C)

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

Let $M_1$ be mass of neutron and its initial velocity be $U_1$

Let $m$ be mass of nucleus and its velocity $u$ is $0$, since at rest

According to law of conservation of energy,

$V_1=\frac{\left(\right(M-m)U_1+2mu)}{(M+m)}$

Since $u$ is zero

$V_1=\frac{(M-m)U_1}{(M+m)}$

$\frac{V_1}{U_1}=\frac{(M-m)}{(M+m)}$

It can also be written as:

$\frac{V_1}{U_1}=\frac{(1-m/M)}{(1+m/M)}=\frac{(1-A)}{(1+A)}$

We know that, Kinetic energy $K.E=\frac{1}{2}\times m{v}^2$

Therefore, $\frac{V_1^2}{U_1^2}=\frac{(1-A{)}^2}{(1+A{)}^2}$

$⟹(\frac{A-1)}{A+1}{)}^2$ is the fraction of the total energy retained.