Escape velocity for hydrogen molecules The question:
At what temperature is the RMS speed of Hydrogen molecules equal to the escape speed from the earth's surface? Values of radius of earth($r$) and gas constant $R$ has been supplied only.
We know that the escape velocity of a body on earth is given by $(2rg)^{1/2}$. Putting in the values of $r$ and $g$ we get $11.2\ \text{km/s}$ as the answer.
We also know that for a mole of ideal gas $PV=RT=\dfrac13M[c_{rms}]^2$,where $M$ is molecular weight of the gas. 
So substituting for $T$ we get $T=\dfrac{M[c_{rms}]^2}{3R}$
I have put $11.2\ \text{km/s}$ in place of $c_{rms}$ and $M=2$ and put the value of $R$ and I have got $904.8\ \text{K}$ as my answer, which is equal to $631.8\ ^o\text{C}$.
For moon the answer came as $98\ ^o\text{C}$.
My question is that moon certainly has a temperature was lower than $98\ ^o\text{C}$ so then shouldn't it have an atmosphere of hydrogen? But the moon has no atmosphere!!
I cannot understand where i have gone wrong.
 A: 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.
