Question

A neutron moving with a speed $'v'$ 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

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,
$mv=mv_{1}+mv_{2}$
$\Rightarrow v^{2}=\left(v_{1}\right)^{2}+\left(v_{2}\right)^{2}+2v_{1}v_{2}...\left(1\right)$
And for the law of conservation of energy,
$\frac{1}{2}mv^{2}=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}mv_{2}^{2}+\Delta E$
$\Rightarrow v^{2}=\left(v_{1}\right)^{2}+\left(v_{2}\right)^{2}+\frac{2 \Delta E}{m}...\left(2\right)$
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,
$\Rightarrow \left(v_{1} - v_{2}\right)^{2}=\left(v_{1} + v_{2}\right)^{2}-4v_{1}v_{2}$
$\Rightarrow v^{2}-\frac{4 \Delta E}{m}\geq 0$
$\Rightarrow \frac{1}{2}mv^{2}\geq 2\Delta E$
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
$\frac{1}{2}mv^{2}_{min}=2\times 10.2=20.4\approx20eV$