Question

A neutron travelling with a velocity $v$ and kinetic energy KE collides 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

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

Answer 1

The correct option is D ${\left(\frac{A-1}{A+1}\right)}^2$

A neutron collides elastically head on with the nucleus of an atom. Momentum before collision is $u(m_1-m_2)$ and momentum after collision is $v(m_1+m_2)$.
Hence,
$u(m_1-m_2)=v(m_1+m_2)$
$v=\frac{u(m_1-m_2)}{m_1+m_2}$
$v=\frac{u(1-A)}{1+A}$
$\frac{v}{u}=\frac{1-A}{1+A}$ ...............(1)
Now, the fraction of kinetic energy retained is
$\frac{\frac{1}{2}m(v_1{)}^2}{\frac{1}{2}m(u_1{)}^2}=\frac{(v_1{)}^2}{(u_1{)}^2}$
$\therefore \frac{(v_1{)}^2}{(u_1{)}^2}={\left(\frac{1-A}{1+A}\right)}^2$.............from (1)
$\therefore \frac{(v_1{)}^2}{(u_1{)}^2}={\left(\frac{A-1}{1+A}\right)}^2$