A hydrogen atom in ground state, moving with speed v collides with another hydrogen atom in ground state at rest. If  $v<v_0,$  then the collision is elastic. If the value of $v_0=x\times {10}^{4}m/s,$ find the value of x to nearest integer $\frac{m{e}^4}{2h^2}=E_0=13.6eV/atom=2.18\times {10}^{-18}J/atom\Rightarrow$
 ionisation energy of H-atom $\Rightarrow m_H=1.67\times {10}^{-27}kg\Rightarrow$  Mass of hydrogen atom.

According to Bohr’s model
$\Delta E=E_0(1-\frac{1}{n^2})\Rightarrow \Delta E_{\min }=\frac{3E_0}{4}$
During inelastic collision, a part of kinetic energy of colliding particles is converted into internal energy. The internal energy of the system of two hydrogen atoms, considered in the problem cannot be changed less than $\Delta E_{\min }$. It means if the change in kinetic energy of system in grour frame is less than  $\Delta E_{\min }$ (or if the kinetic energy of colliding atoms with respect to their centre mass is less than $\Delta E_{\min }$), then collision must be an elastic one. Hence considering the critical case
$$\begin{gathered} \frac{1}{2}{m}_H{\left(\frac{v_0}{2}\right)}^2\times 2=\frac{3E_0}{4}\Rightarrow v_0=\sqrt{\frac{3E_0}{m_H}} \\ \sqrt{\frac{3\times 2.18\times {10}^{-18}}{1.67\times {10}^{-27}}}=\sqrt{39.1617}\times {10}^4=6.257\times {10}^{4}m/s\approx 6.26\times {10}^{4}m/s \\ \frac{a\times b\times c}{k\times n}=\frac{6\times 2\times 6}{3\times 4}=6 \end{gathered}$$