The temperature, at which the root mean square velocity of hydrogen… | JEE Main 2019 (8 Apr, Shift 2)

Q80JEE Main 2019 · 8 APR · SHIFT 2Difficult

Molecular Speeds & Maxwell Distribution

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to:

[Boltzmann Constant kB = 1.38 × 10–23 J/K Avogadro Number NA = 6.02 × 1026 /kg Radius of Earth: 6.4 × 106 m Gravitational acceleration on Earth = 10ms–2]

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Answer: (b)

Set $\sqrt{\dfrac{3kT}{m}} = v_e = \sqrt{2gR}$ , so $T = \dfrac{m\,(2gR)}{3k}$ .
$2gR = 2 \times 10 \times 6.4\times10^{6} = 1.28\times10^{8}\text{ m}^2\text{s}^{-2}$
Mass of H $_2$ molecule: $m = \dfrac{2}{6.02\times10^{26}} = 3.32\times10^{-27}$ kg
$T = \dfrac{3.32\times10^{-27}\times1.28\times10^{8}}{3\times1.38\times10^{-23}} \approx 1.0\times10^{4}$ K
So the answer is (b) $10^4$ K.

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