A mixture of $\ce{H2}$ and $\ce{He}$ at $\pu{300 K}$ effuses from a small hole in a gas chamber. What is the mole fraction of $\ce{H_2}$ in the original mixture, if $3.0$ times as many $\ce{He}$ atoms as $\ce{H2}$ molecules escape from the hole in one minute?
The answer apparently relies on the fact that $\ce{H2}$ collides with the walls $\displaystyle\frac{1}{\sqrt2}$ as often as $\ce{He}$, and we therefore need $\sqrt{2}$ as many $\ce{He}$ atoms as $\ce{H2}$ molecules to yield an equal amount of substance escaping. Considering that the ratio is in fact 3, the final answer is then that the mole fraction of $\ce{H2}$ is
$$x(\ce{H2}) = \frac{1}{3\sqrt{2} + 1}.$$
Can anybody please explain how they got the $\displaystyle\frac{1}{\sqrt2}$ part?