At what temperature root mean square speed of h2 gas will become escape velocity of the Earth?

I believe your mistake is with units, and it is the following:

$$T=\dfrac{M[c_{rms}]^2}{3R} = \dfrac{\left( 1 \text{amu} \right) [11.2 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} $$

This doesn't even cancel out because you're left with a $\text{mol}$ unit. Add avogadro's number.

$$T = \dfrac{\left( 1 \text{amu} \right) [11.2 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} \left( 6.022 \times 10^{23} \frac{1}{\text{mol}} \right) = 5,028 K $$

This is the case for Earth. For the Moon:

$$T = \dfrac{\left( 1 \text{amu} \right) [2.4 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} \left( 6.022 \times 10^{23} \frac{1}{\text{mol}} \right) = 230 K $$

This is negative in Celsius units. This is not a problem. It is merely saying that even freezing temperatures are enough for a lone Hydrogen atom to escape the gravity of the moon with. All we required was that this number be less than the temperature of the sun, which it is. Any surface that faces away from the sun will again see those photons within a month, unless it's in a crater on one of the poles, which we know can have ice near the surface. So that makes sense.

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Answer

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Hint: To solve this question, first obtain the mathematical expressions for the escape velocity from earth and the root mean square velocity. Escape velocity is the velocity required by a particle to escape the earth’s gravitational attraction and rms velocity is the square root of the average of the square of the velocity. Equate these expressions and put the given values to find the answer to this question.

Complete step by step answer:

The escape velocity of a particle from earth is given by,${{V}_{e}}=\sqrt{2g{{R}_{e}}}$ Where g is the acceleration due to gravity and ${{R}_{e}}$ is the radius of earth.Again, the root mean square velocity of hydrogen molecule is given by the mathematical expression,${{V}_{rms}}=\sqrt{\dfrac{3RT}{M}}$ Where, R is the gas constant, T is the temperature and M is the mass of the hydrogen molecule.The value of the gas constant R is,$R=8.314Jmo{{l}^{-1}}{{K}^{-1}}$ Mass of the hydrogen molecule in kg is,$M=2\times {{10}^{-3}}kg$ The acceleration due to gravity is, $g=9.8m{{s}^{-2}}$ Radius of earth is given as, ${{R}_{e}}=6.4\times {{10}^{6}}m$ Now, the root mean square velocity and the escape velocity are equal to each other.So, we can write,$\begin{align} & \sqrt{\dfrac{3RT}{M}}=\sqrt{2g{{R}_{e}}} \\ & \dfrac{3RT}{M}=2g{{R}_{e}} \\ & T=\dfrac{2Mg{{R}_{e}}}{3R} \\ \end{align}$ Putting the given values on the above equation, we get that,$\begin{align} & T=\dfrac{2\times 2\times {{10}^{-3}}\times 9.8\times 6.4\times {{10}^{6}}}{3\times 8.314} \\ & T=10.05\times {{10}^{3}}K \\ & T\approx {{10}^{4}}K \\ \end{align}$ So, the temperature at which the root mean square velocity of a hydrogen molecule is equal to the escape velocity from earth is ${{10}^{4}}K$.The correct option is (C).

Note:

We can also express the root mean square velocity in terms of the Boltzmann constant as, $\sqrt{\dfrac{3kT}{m}}$, where, k is the Boltzmann constant and m is the mass of the particle.Here, $k=\dfrac{R}{{{N}_{A}}}$ and $m=\dfrac{M}{{{N}_{A}}}$.So, we can write, $\dfrac{k}{m}=\dfrac{R}{M}$