The mean free path of a gas molecule at STP is $2.1\times {10}^{-7 }m$ . The diameter (in $\stackrel{\circ }{\mathrm{A}}$) of the molecule is (Given : Boltzmann constant $1.4 \times {10}^{-23} J/K$))

Mean free path, $\lambda =\frac{1}{\sqrt{2}{\mathrm{\pi d}}^2\mathrm{n}}$

where, $n$is the number density and $d$is the diameter of the molecule.

As $PV = N{k}_{B}T$         $\therefore n=\frac{N}{V}=\frac{P}{k_{B}T}$

Thus , $\lambda =\frac{1}{\sqrt{2}{\mathrm{\pi d}}^2\left({\displaystyle \frac{\mathrm{P}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}\right)}=\frac{k_{B}T}{\sqrt{2}{\mathrm{\pi d}}^{2}P}$ or $d^2=\frac{k_{B}T}{\sqrt{2}\mathrm{\pi P\lambda }}$

Here , $k_B=1.4\times {10}^{-23} J/K, \lambda = 2.1\times {10}^{-7} m$

At STP, $T=0^{°}C=273K, P=1 atm=1.01\times {10}^5 N/m^2$

$$\begin{gathered} \therefore d^2=\frac{1.4\times {10}^{-23}\times 273}{\sqrt{2}\times 3.14\times 1.01\times {10}^5\times 2.1\times {10}^{-7}} \\ =4.06\times {10}^{-20} m^2 \mathrm{or} \mathrm{d}=2.01\times {10}^{-10} \mathrm{m}\approx 2.01\stackrel{\circ }{\mathrm{A}} \end{gathered}$$