# Why is the mean free path divided by $\sqrt{2}$?

In the equation in the picture, the mean free path $\lambda$ is described as the volume occupied by a molecule, divided by the volume of the molecule times root two. I do not exactly grasp the purpose of dividing by $\sqrt{2}$, could someone explain it?

• Note that mean free path is not a volume. The diagram's wording "Mean free path can be estimated as" would be more properly written "Mean free path can be estimates from". Mar 27, 2017 at 21:54

In reality, all of the other molecules are moving, so the typical relative speed between molecules will be larger. To see this, consider the one-dimensional case where the velocities of two particles, $v_x$ and $v_y$, are normally distributed with zero mean and standard deviation $v_0$. Then the relative velocity $v_x - v_y$ is normally distributed with standard deviation $$\sqrt{v_0^2 + v_0^2} = \sqrt{2} v_0$$ since variances of independent variables add. Since the typical relative speed is about $\sqrt{2} v_0$, we can imitate having all molecules moving (a 'dynamic system') by just giving a single moving molecule an effective speed $\sqrt{2}$ times as big.