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The temperature, at which the 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 = 10ms^-2]

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9727^oC

527^oC

650K

6.4 x 10⁶ K

answer is A.

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Detailed Solution

The escape speed of the molecule, $v_{e} = \sqrt{2 gR}$ root mean square velocity, $v_{rms} = \sqrt{\frac{3 R T}{m}}$
So, for $v_{e} = v_{rms} \Rightarrow 2 gR = \frac{3 R T}{m}$
$\Rightarrow T = \frac{2 gRm}{3 R} = 10240 K = 9967^{o} C$

Nearest option is (a)

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