At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from
the earth's atmosphere? 
(Take, mass of oxygen molecule,
${\text{m = 2.76 × 10}}^{-26} \mathrm{kg}$ and Boltzmann's constant,
${\mathrm{k}}_{\mathrm{B}} = 1.38 \times {10}^{-23} \mathrm{J} {\mathrm{K}}^{-1}$)

detailed solution

Correct option is B

The minimum velocity with which the body must be projected vertically upwards, so that it could escape from theearth's atmosphere is its escape velocity (ve ).As,  ve = 2gRSubstituting the value of g (9.8 ms-2) and radius of earth (R = 6.4 × 106m), we get                        νe=2×9.8×6.4×106                              ≅11.2 km s-1=11.2×103 m s-1Let the temperature of molecule be T when it attains ve.According to the question, vrms =ve.where, vrms is the rms speed of the oxygen molecule⇒        3kBTmO2=11.2×103or                 T=112×1032mO23kBSubstituting the given values, i.e.                    kB = 1.38 × 10-23 JK-1and              mO2 = m = 2.76 × 10-26 kg,we get            T=11.2×1032276×10-263×1.38×10-23≃8.326×104 K