Physics - Modern Physics Question with Solution | TestHub

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 $$v^2=v_1^2+v_2^2+2v_1{v}_2$$ From squaring eq. (ii), we get $$v^2=v_1^2+v_2^2+\frac{2\Delta E}{m}$$ From eqn (3) \& (4) $$\begin{array}{rl}\therefore 2{\mathrm{v}}_1{\mathrm{v}}_2& =\frac{2\Delta \mathrm{E}}{\mathrm{m}}\\ \therefore {\left({\mathrm{v}}_1-{\mathrm{v}}_2\right)}^2& ={\left({\mathrm{v}}_1+{\mathrm{v}}_2\right)}^2-4{\mathrm{v}}_1{\mathrm{v}}_2={\mathrm{v}}^2-\frac{4\Delta \mathrm{E}}{\mathrm{m}}\end{array}$$ As, $v_1-v_2$ must be real, $v^2-\frac{4\Delta E}{m}\ge 0$ $$\Rightarrow \frac{1}{2}{\mathrm{mv}}^2\ge 2\Delta \mathrm{E}$$ 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 $$\frac{1}{2}m{v}_{\min }^2=2\times 10.2=20.4\mathrm{eV}$$