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.

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    <h3 id="back-link"><a href="/#5.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $5.3.1.$ 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.
    </p>

    <h3>Solution</h3>
    <p>
        To find the mean free path $\lambda$, it is necessary to find the volume that a particle must travel to meet another. <p></p>

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
$$\boxed{d=\frac{kT_0}{\sqrt{\sqrt{2}\pi \lambda p}}\approx 0.3 \text{ nm}}\quad(9)$$

    </p>

    <h4>Answer</h4>
    <p>
        $$d=\frac{kT_0}{\sqrt{\sqrt{2}\pi \lambda p}}\approx 0.3 \text{ nm}$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
	  Aliaksandr Melnichenka<br>
	</p>


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