Question

Estimate the mean free path of a cosmic ray proton in the atmosphere at sea level. Given $\sigma ={10}^{-26}c{m}^2$

• A. ${10}^{4}cm$
• B. ${10}^{-4}cm$
• C. ${10}^{6}cm$
• D. ${10}^{-6}cm$

Answer 1

The correct option is C ${10}^{6}cm$

The mean free path is given by, $\lambda =\frac{1}{\pi \sigma n}$

Here the proton is scattered by the molecules of the atmosphere.

The density of the molecules of the atmosphere is $n=\frac{N}{V}=\frac{P}{kT}=\frac{{10}^6}{(1.38\times {10}^{-16})\left(300\right)}=2.4\times {10}^{19}$ (all units are taken in CGS unit)

So, $\lambda =\frac{1}{\pi \left({10}^{-26}\right)(2.4\times {10}^{19})}=1.3\times {10}^6\sim {10}^6$ $cm$