The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to [Boltzmann constant $k_B = 1.38\times10^{-23}\ \text{J/K}$, Avogadro number $N_A = 6.02\times10^{26}$ /kg, radius of Earth $6.4\times10^6\ \text{m}$, gravitational acceleration on Earth $= 10\ \text{m s}^{-2}$]
Answer: (D) $10^4\ \text{K}$
Escape velocity: $v_e^2 = 2gR = 2\times10\times6.4\times10^6 = 1.28\times10^8\ \text{m}^2\text{s}^{-2}$.
Mass of an $\text{H}_2$ molecule ($N_A$ is given per kg-mole): $m = \dfrac{2}{6.02\times10^{26}} = 3.32\times10^{-27}\ \text{kg}$.
Setting $\dfrac{3k_BT}{m} = v_e^2$:
$$T = \frac{mv_e^2}{3k_B} = \frac{3.32\times10^{-27}\times1.28\times10^8}{3\times1.38\times10^{-23}} \approx 1.0\times10^4\ \text{K}$$
Solution by Sreeraj P, M.Sc Physics