At what temperature will the $r m s$ speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given : Mass of oxygen molecule ( $m$ ) $= 2.76 \times 10^{- 26} k g$ , Boltzmann's constant $k_{B} = 1.38 \times 10^{- 23} J K^{- 1}$ )

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$8.360 \times 10^{4} K$

$2.508 \times 10^{4} K$

$5.016 \times 10^{4} K$

$1.254 \times 10^{4} K$

answer is B.

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Detailed Solution

Escape velocity from the Earth's surface is

$ν_{e s c a p e} = 11200 m s^{- 1}$

Say at temperature $T$ , oxygen molecule attains escape velocity.

So, $v_{e s c a p e} = \sqrt{\frac{3 k_{B} T}{m_{O_{2}}}} \Rightarrow 11200 = \sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times T}{2.76 \times 10^{- 26}}}$

On solving, $T = 8.360 \times 10^{4} K .$

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