At what absolute temperature T , is the root mean square speed of a hydrogen molecule equal to its escape velocity from the surface of the moon? The radius of moon is R, g is the acceleration due to gravity on moon's surface, m is the mass of hydrogen molecule and "k" is the Boltzmann constant

\frac{m g R}{2 k}

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\frac{2 m g R}{k}

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

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

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Solution

The correct option is D $\frac{2 m g R}{3 k}$
$V_{r m s} = \sqrt{\frac{3 k T}{m}}$
$V_{C} = \sqrt{2 g R}$
$\sqrt{\frac{3 k T}{m}} = \sqrt{2 g R}$
$T = \frac{2 m g R}{3 k}$

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At what absolute temperature $T$ is the root mean square speed of a hydrogen molecule equal to its Escape velocity from the surface of the moon? The radius of moon $R = g$ is the acceleration due to gravity on moon's surface $m$ is the mass of a hydrogen molecule and $k$ is the Boltzmann constant