Question

A neutron moving at a speed 'v' undergoes a head-on elastic collision with a nucleus of mass number 'A' at rest. The ratio of the kinetic energies of the neutron after and before collision is

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

Answer 1

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

Mass of neutron $\left(m_1\right)=1unit$. Mass of nucleus $\left(m_2\right)=Aunits$.
The velocity of the neutron after the collision is
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u=\left(\frac{1-A}{1+A}\right)u$
KE of neutron after collision = $\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}\times 1\times {\left(\frac{1-A}{1+A}\right)}^2{u}^2$
KE of neutron before collision = $\frac{1}{2}m{u}^2=\frac{1}{2}\times 1\times u^2=\frac{1}{2}{u}^2$.
Their ratio is ${\left(\frac{1-A}{1+A}\right)}^2$, which is choice (a)