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Q. At what temperature, hydrogen molecules will escape from the earth's surface? (Take mass of hydrogen molecule $=0.34\times10^{-26} kg,$ Boltzmann constant $-1.38\times10^{-23} J K^{-1},$ Radius of earth $=6.4\times10^{6} m$ and acceleration due to gravity $=9.8 ms^{-2}$ )

Gravitation

Solution:

The root mean square velocity of gas is $v_{ms}=\sqrt{\frac{3kT}{m}}$ .....(i) Escape velocity of gas molecules is $v_{cs}=\sqrt{2gR_e}$ .....(ii) As the root mean square velocity of gas molecules must be equal to the escape velocity. = From Eqs. (i) and (ii), we get $\sqrt{\frac{3kT}{m}}=\sqrt{2gR_e}$ $\Rightarrow T=\frac{2gR_em}{3k}$ $\Rightarrow T=\frac{2\times9.8\times6.4\times10^6\times0.34\times10^{-26}}{3(1.38\times10^{-23})}$ $=10^4 K$ Therefore, $10^4 K$ is the temperature at which hydrogen molecules will escape from earth's surface.