A neutron moving with speed v makes a head on collision with a...

Let speed of neutron before collision $=\mathrm{V}$ Speed of neutron after collision $=\mathrm{V}_{1}$ Speed of proton or hydrogen atom after collision $=\mathrm{V}_{2}$ Energy of excitation $=\Delta \mathrm{E}$ From the law of conservation of linear momentum, $\mathrm{mv}=\mathrm{mv}_{1}+\mathrm{mv}_{2}$ And for law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\frac{1}{2} \mathrm{mv}_{2}^{2}$ From squaring eq. (i), we get v^{2}=v_{1}^{2}+v_{2}^{2}+2 v_{1} v_{2} From squaring eq. (ii), we get v^{2}=v_{1}^{2}+v_{2}^{2}+$\frac{2 \Delta E}{m}$ From eqn (3) \& (4) $$\begin{aligned} \therefore2 \mathrm{v}_{1} \mathrm{v}_{2}&=\frac{2 \Delta \mathrm{E}}{\mathrm{m}}\\ \therefore\left(\mathrm{v}_{1}-\mathrm{v}_{2}\right)^{2}&=\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)^{2}-4 \mathrm{v}_{1} \mathrm{v}_{2}=\mathrmv^{2}-\frac{4 \Delta \mathrm{E}}{\mathrm{m}} \end{aligned}$$ As, $v_{1}-v_{2}$ must be real, $v^{2}-\frac{4 \Delta E}{m} \geq 0$ \Rightarrow \quad $\frac{1}{2} \mathrm{mv}^{2} \geq2$\Delta $\mathrm{E}$ The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is $10.2 \mathrm{eV}$. Therefore, the maximum kinetic energy needed is $\frac{1}{2}$ m v_{\min }^{2}=2 $\times10.2=20.4 \mathrm{eV}$