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The absolute temperature at which the rms speed of a hydrogen molecule is equal to its escape speed from the Moon’s surface is (radius of Moon is R, g is acceleration duet to gravity on Moon’s surface, m is mass of hydrogen molecule and k is Boltzmann constant)

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$\frac{m g R}{2 k}$

$\frac{2 m g R}{k}$

$\frac{3 m g R}{2 k}$

$\frac{2 m g R}{3 k}$

answer is D.

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Detailed Solution

$V_{rms} \geq V_{esc} \Rightarrow \sqrt{\frac{3 RT}{M}} \geq \sqrt{2 gR}$

$T \geq \frac{2 MgR}{3 K}$

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