Question

At some higher altitude, density of free electrons is around
${10}^{12}$ per cubic meter and due to low density of air the mean path of an electron is about 0.1m. If the mean speed of electrons is ${10}^{5}m/s$, then the conductivity of atmosphere is
(answer upto 2 decimal points)

Answer 1

$\sigma =\frac{N{e}^2\tau }{m}$
=$\frac{{10}^{12}\times {\left(1.6\times {10}^{-19}\right)}^2}{9.1\times {10}^{-31}}\times \frac{0.1}{{10}^5}$ $\sigma =0.281\times {10}^{-1}=0.0281$