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 the moon is \[R\], \[g\] is the acceleration due to gravity on the moon's surface, \[m\] is the mass of the hydrogen molecule and is the Boltzmann constant.
A.\[\dfrac{{mgR}}{{2K}}\]
B. \[\dfrac{{2mgR}}{K}\]
C. \[\dfrac{{3mgR}}{{2K}}\]
D. \[\dfrac{{2mgR}}{{3K}}\]

Hint:Recall the formula for the root mean square speed and the escape velocity of an object. Rearrange these equations using the condition given in the question.

Formulae used:
The root mean square velocity is given by
\[\Rightarrow{v_{rms}} = \sqrt {\dfrac{{3KT}}{m}} \]
Here, \[{v_{rms}}\] is the root mean square velocity, \[K\] is Boltzmann constant, \[T\] is the temperature in Kelvin and \[m\] is the mass of one mole of gas in kilogram.
The escape velocity of an object from a planet is
\[\Rightarrow{v_{esc}} = \sqrt {2gR} \]
Here, \[{v_{esc}}\] is the escape velocity of the object, \[g\] is the acceleration due to gravity and \[R\] is the radius of the planet.

Complete step by step answer:
Rewrite equation (2) for the root mean square speed of the hydrogen molecule.
\[\Rightarrow{v_{rms}} = \sqrt {\dfrac{{3KT}}{m}} \]
Here, \[{v_{rms}}\] is the root mean square speed of a hydrogen molecule and \[m\] is the mass of one mole of hydrogen in kilogram.
Rewrite equation (2) for the escape velocity of hydrogen molecule from the moon is
\[\Rightarrow{v_{esc}} = \sqrt {2gR} \]
Here, \[{v_{esc}}\] is the escape velocity of the hydrogen molecule from the moon and \[R\] is the radius of the moon.
The root mean square velocity \[{v_{rms}}\] of the hydrogen molecule equals the escape velocity \[{v_{esc}}\] of the hydrogen molecule from the moon.
\[\Rightarrow{v_{rms}} = {v_{esc}}\]
Substitute \[\sqrt {\dfrac{{3KT}}{m}} \] for \[{v_{rms}}\] and \[\sqrt {2gR} \] for \[{v_{rms}}\] in the above equation.
\[\Rightarrow\sqrt {\dfrac{{3KT}}{m}} = \sqrt {2gR} \]
Take square on both the sides of the above equation.
\[\Rightarrow\dfrac{{3KT}}{m} = 2gR\]
Rearrange the above equation for the temperature \[T\].
\[\Rightarrow T = \dfrac{{2mgR}}{{3K}}\]
Hence, the absolute temperature at which the root mean square speed of the hydrogen molecule is equal to the escape velocity of the hydrogen molecule from the moon is \[\dfrac{{2mgR}}{{3K}}\].

Hence, the correct option is D.

Note:The quantity in the root mean square speed formula is the mass of one mole of gas in kilogram and not the molar mass of the gas.And also remember that root mean square velocity is the square root of the mean of squares of the velocity of individual gas molecules while average velocity is the arithmetic mean of the velocities of different molecules of a gas at a given temperature.So don’t get confuse among both of them.