The temperature at which root mean square velocity of hydrogen molecules equals their escape velocity from the earth is closest to:
[Boltzmann constant ${{k}_{B}}=1.38\times {{10}^{-23}}J/k$
Avogadro number ${{N}_{A}}=6.02\times {{10}^{26}}/kg$
Radius of earth: $6.4\times {{10}^{6}}m$
Gravitational acceleration on Earth = $10m{{s}^{-2}}$ ]
$\begin{align}
  & \text{A}\text{. 650K} \\
 & \text{B}\text{. 3}\times \text{1}{{\text{0}}^{5}}K \\
 & \text{C}\text{. 1}{{\text{0}}^{4}}K \\
 & \text{D}\text{. 800K} \\
\end{align}$

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, $\dfrac{k}{m}=\dfrac{R}{M}$