Question

A neutron moving with a speed ${}^{\prime }{v}^{\prime }$ makes a head on collision with a stationary hydrogen atom in the ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place in $eV$ (closest to nearest integer) is:

Answer 1

Answer: 20

Let the speed of neutron before the collision $=v$
Speed of neutron after collision $=v_1$
Speed of proton or hydrogen atom after collision $=v_2$
Energy of excitation $=\Delta E$
From the law of conservation of linear momentum,
$mv=m{v}_1+m{v}_2$
$\Rightarrow v^2={\left(v_1\right)}^2+{\left(v_2\right)}^2+2v_1{v}_2...\left(1\right)$
And for the law of conservation of energy,
$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2+\Delta E$
$\Rightarrow v^2={\left(v_1\right)}^2+{\left(v_2\right)}^2+\frac{2\Delta E}{m}...\left(2\right)$
Now squaring equation, $\left(1\right)$ ,
$v^2={\left(v_1\right)}^2+{\left(v_2\right)}^2+2v_1{v}_2$
$\Rightarrow 2v_1{v}_2=\frac{2\Delta E}{m}$
form $v_1-v_2$ , must be real,
$\Rightarrow {\left(v_1-v_2\right)}^2={\left(v_1+v_2\right)}^2-4v_1{v}_2$
$\Rightarrow v^2-\frac{4\Delta E}{m}\ge 0$
$\Rightarrow \frac{1}{2}m{v}^2\ge 2\Delta E$
The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is $10.2eV$ . Therefore, the maximum kinetic energy needed is
$\frac{1}{2}m{v}_{min}^2=2\times 10.2=20.4\approx 20eV$