Answer
(a) $3.3\times 10^{12}\frac{1}{m^3}$
(b) $1.71\times 10^{6}m$
Work Step by Step
(a) We can determine the required number density as follows:
$\frac{N}{V}=\frac{\rho gh}{k_b T}$
We plug in the known values to obtain:
$\frac{N}{V}=\frac{13600\times 9.8\times 1\times 10^{-10}}{1.38\times 10^{-23}\times 293}$
This simplifies to:
$\frac{N}{V}=3.3\times 10^{12}\frac{1}{m^3}$
(b) We can determine the mean free path as follows:
$\lambda=\frac{1}{4\sqrt{2}\pi \frac{N}{V}r^2}$
We plug in the known values to obtain:
$\lambda=\frac{1}{4\sqrt{2}\pi \times 3.3\times 10^{12}(10^{-10})^2}$
$\implies \lambda=1.71\times 10^{6}m$
