91Ó°ÊÓ

$\begin{array}{c}\frac{N}{V}=1\frac{\text{molecule}}{{\text{cm}}^3}\\ =1×{10}^6\frac{\text{molecule}}{{\text{m}}^3}\end{array}$

$\text{h}=2500\text{ k³¾}$

$\begin{array}{c}d=2.0×{10}^{−8}\text{cm}\\ =2.0×{10}^{−10}\text{m}\end{array}$

The expression for the mean free path is given by,

$\mathrm{λ}=\frac{1}{\sqrt{2}\mathrm{π}{\text{d}}^2\left(\frac{\text{N}}{\text{V}}\right)}$

Here $\mathrm{λ}$ is the mean free path, $\frac{N}{V}$is the molecule density in $molecule/c{m}^3$.

The mean free path is given by

$\mathrm{λ}=\frac{1}{\sqrt{2}\mathrm{π}{d}^2\left(\frac{N}{V}\right)}$

Substitute $2.0×{10}^{−10}\text{m}$ for d and $1×{10}^6\frac{\text{molecule}}{{\text{m}}^3}$ for $\frac{N}{V}$ into the above equation,

$\begin{array}{c}\mathrm{λ}=\frac{1}{\sqrt{2}\left(3.142\right){(2.0×{10}^{−10})}^2(1×{10}^6)}\\ =5.62×{10}^{12}\\ ≈6×{10}^{12}\text{m}\end{array}$

Therefore, the mean free path is $6×{10}^{12}\text{m}.$

At height $h$ the atmospheric density is very low. From the above formula, we can note that the mean free path is inversely proportional to the density. Hence the mean free path is very large. It implies that at the given height, air collision occurs rarely. So the predicted value of the mean free path is not meaningful