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, $N a = 6.02 \times 10^{26} / k g$
Radius of Earth : $6.4 \times 10^{6} m$
Gravitational acceleration on Earth $= 10 m s^{- 2}$ ]

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$10^{4} K$

$3 \times 10^{5} K$

$800 K$

$650 K$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The escape speed of the molecule, $v_{e} = \sqrt{2 g R}$

Root mean square velocity, $v_{r m s} = \sqrt{\frac{3 (k_{B} N) T}{m}}$

So, for $v_{e} = v_{r m s}; 2 g R = \frac{3 k_{B} N T}{m}$

$\Rightarrow T = \frac{2 g R m}{3 k_{B} N} = \frac{2 \times 10 \times 6.4 \times 10^{6} \times 2 \times 10^{- 3}}{3 \times 1.38 \times 10^{- 23} \times 6.02 \times 10^{23}} = 10^{4} K$

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!