Question

A neutron moving with a speed $v$ makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of neutron for which inelastic collision will take place is

• A. $10.2eV$
• B. $20.4eV$
• C. $12.1eV$
• D. $16.8eV$

Answer 1

The correct option is B $20.4eV$

Let $v=$ speed of neutron before collision
$v_1=$ speed of neutron after collision
$v_2=$ speed of hydrogen atom after collision and
$\Delta E=$ energy of the excitation
From the conservation of linear momentum,
$mv=m{v}_1+m{v}_2$ ....(i)
From the conservation of energy,
$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2+\Delta E$ ...(ii)
From equation (i),
$v^2=v_1^2+v_2^2+2v_1{v}_2$
From equation (ii),
$v^2=v_1^2+v_2^2+\frac{2\Delta E}{m}$
$\therefore 2v_1{v}_2=\frac{2\Delta E}{m}$
$\therefore {(v_1-v_2)}^2={(v_1+v_2)}^2-4v_1{v}_2$
$=v^2-\frac{4\Delta E}{m}$
As, $v_1-v_2$ must be real
$\Rightarrow v^2-\frac{4\Delta E}{m}\ge 0$
or $\frac{1}{2}m{v}^2\ge 2\Delta E$
The maximum energy that can be absorbed by hydrogen atom in ground state to go into excited state is $10.2eV$. Therefore, the minimum kinetic energy required is
$\frac{1}{2}m{v}_{min}^2=2\times 10.2eV=20.4eV$