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 neutron for which inelastic collision will take place is

Let $v=$ speed of neutron before collision $v_1=$ speed of neutron after collision $v_2=$ speed of hydrogen atom after collision and $\Delta E=$ energy of the excitation From the conservation of linear momentum, $m v=m v_1+m v_2$ ...(i) From the conservation of energy, $\frac{1}{2} m v^2=\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2+\Delta E$ ...(ii) from Eq. (i), $v^2=v_1^2+v_2^2+2 v_1 v_2$ From Eq. (ii), $\begin{array}{rlrl} & & v^2 & =v_1^2+v_2^2+\frac{2 \Delta E}{m} \\ & \therefore & 2 v_1 v_2 & =\frac{2 \Delta E}{m} \\ & \therefore & \left(v_1-v_2\right)^2 & =\left(v_1+v_2\right)^2-4 v_1 v_2 \\ & & =v^2-\frac{4 \Delta E}{m}\end{array}$ As, $v_1-v_2$ must be real $\Rightarrow \quad v^2-\frac{4 \Delta E}{m} \geq 0$ or $\frac{1}{2} m v^2 \geq 2 \Delta E$ The maximum energy that can be absorbed by hydrogen atom in ground state to go into excited state is 10.2 eV . Therefore, the minimum kinetic energy required is $\frac{1}{2} m v_{\min }^2=2 \times 10.2 \mathrm{eV}=20.4 \mathrm{eV}$