An H -atom moving with speed v makes a head-on collision with...

For {v}_\min collision should be completely inelastic.

According to energy conservation principle,

∴ \dfrac{1}{2}m{v}_\min^{2} = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}mv^{2} + ∆ E ...(i))

According to momentum conservation principle.

m{v}_\min = mv + mv

∴ v = \dfrac{{m}_\min}{2} ...(ii)

After solving eq. (i) an d(ii), we get

\dfrac{1}{2}m{v}_\min^{2} = 2 ∆ E

∴ {v}_\min = \sqrt{\left(\dfrac{4 ∆ E}{m}\right)}

Here, $∆ E =$ minimum excitation energy

$= 10.2eV$

$m = 1.67 \times {10}^{- 27}kg$

∴ {v}_\min = \sqrt{\left(\dfrac{4 \times 10.2 \times 1.6 \times {10}^{- 19}}{1.67 \times {10}^{- 27}}\right)}

$= \text{6.25} \times {10}^{4}ms^{- 1}$