Question:
The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to :
[Boltzmann Constant \(k_B = 1.38\times10^{-23}\ \mathrm{J/K}\)
Avogadro Number \(N_A = 6.02\times10^{26}\ \mathrm{/kg}\)
Radius of Earth : \(6.4\times10^{6}\ \mathrm{m}\)
Gravitational acceleration on Earth = \(10\ \mathrm{ms^{-2}}\)]
The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to :
[Boltzmann Constant \(k_B = 1.38\times10^{-23}\ \mathrm{J/K}\)
Avogadro Number \(N_A = 6.02\times10^{26}\ \mathrm{/kg}\)
Radius of Earth : \(6.4\times10^{6}\ \mathrm{m}\)
Gravitational acceleration on Earth = \(10\ \mathrm{ms^{-2}}\)]
[Boltzmann Constant \(k_B = 1.38\times10^{-23}\ \mathrm{J/K}\)
Avogadro Number \(N_A = 6.02\times10^{26}\ \mathrm{/kg}\)
Radius of Earth : \(6.4\times10^{6}\ \mathrm{m}\)
Gravitational acceleration on Earth = \(10\ \mathrm{ms^{-2}}\)]
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Set \(\sqrt{3k_BT/m} = \sqrt{2gR}\); the mass of one H\(_2\) molecule is 2 kg per \(N_A\) (in the given per-kg units).
Updated On: Nov 5, 2026
- \(3\times10^{5}\ \mathrm{K}\)
- \(10^{4}\ \mathrm{K}\)
- \(650\ \mathrm{K}\)
- \(800\ \mathrm{K}\)
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The Correct Option is B
Solution and Explanation
Escape velocity: \(v_e = \sqrt{2gR} = \sqrt{2\times10\times6.4\times10^{6}} = \sqrt{1.28\times10^{8}}\ \mathrm{m/s}\).
RMS speed: \(v_{rms} = \sqrt{\frac{3k_BT}{m}}\).
Mass of an H\(_2\) molecule: \(m = \frac{2\ \mathrm{kg\ per\ kmol}}{N_A} = \frac{2}{6.02\times10^{26}} = 3.32\times10^{-27}\ \mathrm{kg}\).
Equate: \(\frac{3k_BT}{m} = 2gR\), so \(T = \frac{2gR\,m}{3k_B} = \frac{1.28\times10^{8}\times3.32\times10^{-27}}{3\times1.38\times10^{-23}}\).
\(T = \frac{4.25\times10^{-19}}{4.14\times10^{-23}} \approx 1.03\times10^{4}\ \mathrm{K}\), closest to \(10^4\ \mathrm{K}\).
Answer: (2) \(10^{4}\ \mathrm{K}\)
RMS speed: \(v_{rms} = \sqrt{\frac{3k_BT}{m}}\).
Mass of an H\(_2\) molecule: \(m = \frac{2\ \mathrm{kg\ per\ kmol}}{N_A} = \frac{2}{6.02\times10^{26}} = 3.32\times10^{-27}\ \mathrm{kg}\).
Equate: \(\frac{3k_BT}{m} = 2gR\), so \(T = \frac{2gR\,m}{3k_B} = \frac{1.28\times10^{8}\times3.32\times10^{-27}}{3\times1.38\times10^{-23}}\).
\(T = \frac{4.25\times10^{-19}}{4.14\times10^{-23}} \approx 1.03\times10^{4}\ \mathrm{K}\), closest to \(10^4\ \mathrm{K}\).
Answer: (2) \(10^{4}\ \mathrm{K}\)
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