A hydrogen atom travelling at a speed v collides with a stationary...

Let the mass of the hydrogen atom be $m$ and its initial speed be $v$.

For the maximum loss of kinetic energy, the collision must be perfectly inelastic. By conservation of linear momentum, the final common velocity $v'$ of the two atoms is given by:

$mv = (m + m)v' \Rightarrow v' = \dfrac{v}{2}$

The maximum loss in kinetic energy $\Delta K$ is:

$\Delta K = \dfrac{1}{2}mv^2 - \dfrac{1}{2}(2m)\left(\dfrac{v}{2}\right)^2 = \dfrac{1}{4}mv^2$

For one of the hydrogen atoms to be ionized, this loss in kinetic energy must be at least equal to the ionization energy of a hydrogen atom, which is $13.6 \text{ eV}$.

$\dfrac{1}{4}mv^2 \ge 13.6 \times 1.6 \times 10^{-19} \text{ J}$

$v \ge \sqrt{\dfrac{4 \times 13.6 \times 1.6 \times 10^{-19}}{m}}$

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

$v \ge \sqrt{\dfrac{4 \times 21.76 \times 10^{-19}}{1.67 \times 10^{-27}}}$

$v \ge \sqrt{52.12 \times 10^8}$

$v \ge 7.2 \times 10^4 \text{ m/s}$

The minimum value of $v$ is $7.2 \times 10^4 \text{ m/s}$.