Explanation

(d) As we know gas molecules cannot escape from earth’s atmosphere because their root mean square velocity is less than escape velocity at earth’s surface. If we fill this requirement, then gas molecules can escape from earth’s atmosphere.
$ie,\ \ \ \ \ \ \ \ \ \ \ \ v_{rms} = v_{es}$
$or\ \ \ \ \ \ \ \ \ \sqrt{\frac{3RT}{M}} = \sqrt{2gR_{e}}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = \frac{2MgR_{e}}{3R}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
$Given,\ \ \ \ \ M = 2 \times 10^{- 3}kg,\ \ g = 9.8\ {ms}^{- 2}$
$R_{e} = 6.4 \times 10^{6}\ m,\ R = 8.31\ J{mol}^{- 1}–K^{- 1}$
Substituting in Eq. (i), we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = \frac{2 \times 2 \times 10^{- 3} \times 9.8 \times 6.4 \times 10^{6}}{3 \times 8.31}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10^{4}\ K$