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 the neutron for which inelastic collision will take place is (assume that mass of proton is nearly equal to the mass of neutron)

Let y = speed of neutron before collision,

v₁ = speed of neutron after collision,
v₂ = speed of proton or hydrogen atom after collision,

and $∆$E = energy of excitation
From conservation of linear momentum,

$mv=m{v}_1+m{v}_2.....\left(i\right)$

From conservation of energy,

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2+\mathrm{\Delta }E.......\left(ii\right)$

From Eq. (i),

$v^2=v_1^2+v_2^2+2v_1{v}_2$

From Eq. (ii),

$$\begin{gathered} v^2=v_1^2+v_2^2+\frac{2\mathrm{\Delta }E}{m} \\ \therefore 2v_1{v}_2=\frac{2\mathrm{\Delta }E}{m} \\ \therefore {\left(v_1-v_2\right)}^2={\left(v_1+v_2\right)}^2-4v_1{v}_2 \\ \Rightarrow {\left(v_1-v_2\right)}^2=v^2-4\frac{\mathrm{\Delta }E}{m} \end{gathered}$$

$\text{ As }{v}_1-v_2\text{ must be real, therefore }$

$v^2-4\frac{\mathrm{\Delta }E}{m}\ge 0$

$\text{ or }\frac{1}{2}m{v}^2\ge 2\mathrm{\Delta }E$

The minimum energy that can be absorbed by hydrogen atom in ground state to go into excited state is 10.2 eV. Therefore,

$\frac{1}{2}m{v}_{min}^2=2\times 10.2\mathrm{eV}=20.4\mathrm{eV}$