For a molecule of an ideal gas $n=3\times {10}^8{\mathrm{cm}}^{-3}$ and mean free path is ${10}^{-2}cm$. Calculate the diameter of the molecule

$\begin{array}{l}\text{ Given number of molecules per unit volume is }\mathrm{n}=3\times {10}^8{\mathrm{cm}}^{-1},mean free path is \lambda ={10}^{-2}\mathrm{cm}; diameter of molecule is d\\ \lambda =\frac{1}{\sqrt{2}\pi n{d}^2}\\ {\mathrm{d}}^2=\frac{1}{\sqrt{2}\pi n\lambda }\\ {\mathrm{d}}^2=\frac{1}{\sqrt{2}\times 3.14\times 3\times {10}^8\times {10}^{-2}}\\ {\mathrm{d}}^2=\frac{{10}^{-6}}{1.414\times 3.14\times 3}\\ \mathrm{d}=\sqrt{7.5\times {10}^{-8}}=\sqrt{7.5}\times {10}^{-4}\\ \mathrm{d}=2.7\times {10}^{-4}\mathrm{cm}\\ \text{ Hence the diameter of the molecule of the gas is }2.7\times {10}^{-4}\mathrm{cm}\end{array}$