Question

At what temperature hydrogen molecules will escape from the earth’s surface? (Take mass of hydrogen molecule ${m_H} = 0.34 \times {10^{ - 26}}$ $kg$, Boltzmann constant ${k_B} = 1.38 \times {10^{ - 23}}$ $J{K^{ - 1}}$, radius of the earth $R = 6.4 \times {10^6}$ $m$ and acceleration due to gravity $g = 9.8$ $m{s^{ - 2}}$)
(A) $10$ $K$
(B) ${10^2}$ $K$
(C) ${10^3}$ $K$
(D) ${10^4}$ $K$

Answer 1

All the matter that exists inside the atmosphere of the earth, experiences a force of gravitation due to the gravitational field of earth. If the matter escapes from the influence of the earth’s gravitational field, the matter will escape from the earth’s surface. For an object of mass $m$ to escape from the earth’s surface, the object needs to gain the escape velocity.

Formula used:
The escape velocity for an object on earth is given as ${v_e} = \sqrt {\dfrac{{2GM}}{R}}$

Complete step by step answer:
Consider a hydrogen molecule of mass ${m_H}$. The rms velocity of the molecule will be ${v_{rms}} = \sqrt {\dfrac{{3{k_B}T}}{{{m_H}}}}$where $T$ the temperature.
Now, in order to let the hydrogen molecule escape from the earth’s surface, the temperature should have a value such that the rms velocity of the molecule equals the escape velocity.
Let ${T_e}$ be the temperature at which the rms velocity equals the escape velocity,
$\implies {v_{rms}} = {v_e} \\\ \implies \sqrt {\dfrac{{3{k_B}T}}{{{m_H}}}} = \sqrt {\dfrac{{2GM}}{R}} \\\$
${v_e}$ can be expressed as follows
$\sqrt {\dfrac{{2GM}}{R}} = \sqrt {\dfrac{{2GM}}{{{R^2}}}R} = \sqrt {2gR}$
So,

$$\sqrt {\dfrac{{3{k_B}T}}{{{m_H}}}} = \sqrt {2gR} \\\ \implies \dfrac{{3{k_B}T}}{{{m_H}}} = 2gR \\\ \implies T = \dfrac{{2gR{m_H}}}{{3{k_B}}} \\\$$

$T = \dfrac{{(2)(9.8)(6.4 \times {{10}^6})(0.34 \times {{10}^{ - 26}})}}{{(3)(1.38 \times {{10}^{ - 23}})}}$
$T = 10301.38$ $K$
The temperature is nearly equal to $10000$ $K$.
Therefore, the temperature at which the hydrogen molecules will escape from the earth’s surface is ${10^4}$ $K$.

So, the correct answer is “Option D”.

Additional Information:
The velocity required by an object to escape the gravitational influence of the earth is called the ‘escape velocity’.
Consider a system with earth and an object on it. Let the object be projected up with an initial velocity $u$. Now, by conservation of energy, we have
$K.{E_{initial}} + P.{E_{initial}} = K.{E_{final}} + P.{E_{final}} \\\ \dfrac{1}{2}m{u^2} + \left( { - \dfrac{{GMm}}{R}} \right) = \dfrac{1}{2}m{v^2} + \left( { - \dfrac{{GMm}}{r}} \right) \\\ \implies \dfrac{1}{2}m{v^2} = \left( {\dfrac{1}{2}m{u^2} - \dfrac{{GMm}}{R}} \right) + \dfrac{{GMm}}{r} \\\$
If the quantity inside the bracket is positive, then $v$ will never become zero and the particle will never stop no matter what the value of $r$ will be. Hence, we get
$u \geqslant \sqrt {\dfrac{{2GM}}{R}}$, $u$ is the minimum velocity required to escape the field.

Note:
Remember the escape velocity for earth required by any object or matter and also the rms velocity of a molecule. The rms velocity can also be given in terms of the universal gas constant $R$. We have $R = {k_B} \times {N_A}$ , where ${N_A}$ is the Avogadro number. The molecular mass of gas is given as $M = {N_A} \times m$, where $m$ is the mass of a single molecule.
${v_{rms}} = \sqrt {\dfrac{{3{k_B}T}}{m}} \\\ \implies {v_{rms}} = \sqrt {\dfrac{{3\left( {\dfrac{R}{{{N_A}}}} \right)T}}{{\dfrac{M}{{{N_A}}}}}} \\\ \implies {v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \\\$
This information can be used in case if the molecular mass of the gas is given.