$5.3.1.$ At atmospheric pressure and a temperature of $0 \,^{\circ}C$, the mean free path of a hydrogen molecule is $0.1 ~\mu m$. Estimate the diameter of this molecule.
To find the mean free path $\lambda$, it is necessary to find the volume that a particle must travel to meet another.
Each particle has a volume $$ V=\frac{1}{n}\quad(1) $$ In this case, a particle with an effective cross-section $S=\pi(2r)^2=\pi d^2$? cross-section at which it touches at least the edge of another particle, will pass through a volume $$ V = \lambda \pi d^2\quad(2) $$ If we assume that all other particles are at rest, then $$ \lambda = \frac{1}{\pi d^2n}\quad(3) $$ If the beam particle is part of an established equilibrium system with identical particles, then the square of the relative velocity is equal to: $$ \overline{{v}_{\mathrm{relative}}^2}=\overline{({v}_1-{v}_2)^2}=\overline{{v}_1^2+{v}_2^2-2{v}_1\cdot{v}_2}\quad(4) $$ At equilibrium, the values of the velocities ${\displaystyle {\overline { {v} _{1}\cdot {v} _{2}}}=0}$, and the relative velocity is equal at equilibrium, the values of the velocities $$ {\displaystyle v_{\rm {rel}}={\sqrt {\overline {\ {v} _{\rm {relative}}^{2}}}}={\sqrt {\overline {\ {v} _{1}^{2}+\ {v} _{2}^{2}}}}={\sqrt {2}}v.}\quad(5) $$ The mean free path of a particle in this case is then defined as $$ \boxed{\lambda = \frac{1}{\sqrt{2}\pi d^2n}}\quad(6) $$ From the formula for the mean free path we obtain that $$ d=\frac{1}{\sqrt{\sqrt{2}\pi \lambda n}}\quad(7) $$ The concentration of particles can be expressed through the Ideal Gas Law $$ p=nkT_0\Rightarrow \boxed{n=\frac{p}{kT_0}}\quad(8) $$ Substituting $(8)$ into $(7)$, we get $$ \boxed{d=\frac{kT_0}{\sqrt{\sqrt{2}\pi \lambda p}}\approx 0.3 \text{ nm}}\quad(9) $$
$$d=\frac{kT_0}{\sqrt{\sqrt{2}\pi \lambda p}}\approx 0.3 \text{ nm}$$