Kinetic Theory

JEE Main level, single correct, medium

61 of 168

Question 61 of 168

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape their escape velocity from the earth, is closest to:
[[ Boltzmann Constant kB=1.38×10−23J/K{k}_{B}=1.38\times {10}^{-23}J/K
Avogadro number
NA=6.02×1026/kg{N}_{A}=6.02\times {10}^{26}/kg
Radius of Earth:
6.4×106m6.4\times {10}^{6}m
Gravitational acceleration on Earth
=10ms−2]=10m{s}^{-2}]

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Answer: (D) 104K{10}^{4}K

3RTm=11.2km/s\sqrt{\dfrac{3RT}{m}}=11.2km/s
3×253×T2×10−3=(11.2×103)2\dfrac{3\times \dfrac{25}{3}\times T}{2\times {10}^{-3}}={\left(11.2\times {10}^{3}\right)}^{2}
T=10035.2K≈104KT=10035.2K\approx {10}^{4}K

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