Question

A neutron moving with velocity $u$, collides elastically with an atom of mass number $A$. If the collision is head on and the kinetic energy of neutron is $E$, then the final kinetic energy of neutron after collision is

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

Answer 1

Answer: (A)

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

Let the final velocity of neutron be v' then
$v^{\prime }\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2$
Here $u_1=u^{}{u}_2=0;m_1=1,m_2=A$
$\therefore v^{\prime }=\left(\frac{1-A}{1+A}\right)u$
For neutron, the final kinetic energy $E^{\prime }=\frac{1}{2}m{\left(v^{\prime }\right)}^2$
$=\frac{1}{2}\left(1\right){\left[\left(\frac{1-A}{1+A}\right)u\right]}^2=\frac{1}{2}{\left(\frac{1-A}{1+A}\right)}^2{u}^2$
given initial kinetic energy $=E$
$\therefore \frac{E^{\prime }}{E}={\left(\frac{1-A}{1+A}\right)}^2$
$\Rightarrow E^{\prime }={\left(\frac{1-A}{1+A}\right)}^{2}E$