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

The root mean square speed is given by
$v_{r m s} = \frac{\sqrt{3 k T}}{m}$
The escape velocity is given by $v_{e} = \sqrt{2 g R}$
For $v_{r m s} = v_{e}$ we require
$\sqrt{\frac{3 k T}{m}} = \sqrt{2 g R} o r T = \frac{2 m g R}{3 k},$ which is choice (d).

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