# The temperature, at which the root mean square velocity of hydrogen molecules

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 $k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$

Avogadro Number $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{26} / \mathrm{kg}$

Radius of Earth : $6.4 \times 10^{6} \mathrm{~m}$

Gravitational acceleration on Earth $=10 \mathrm{~ms}^{-2}$ ]

1. (1) $800 \mathrm{~K}$

2. (2) $3 \times 10^{5} \mathrm{~K}$

3. (3) $10^{4} \mathrm{~K}$

4. (4) $650 \mathrm{~K}$

Correct Option: , 3

Solution:

(3) $v_{\text {rms }}=v_{e}$

$\sqrt{\frac{3 R T}{M}}=11.2 \times 10^{3}$

or $\sqrt{\frac{3 k T}{m}}=11.2 \times 10^{3}$

or $\sqrt{\frac{3 \times 1.38 \times 10^{-23} T}{2 \times 10^{-3}}}=11.2 \times 10^{3} \quad \therefore v=10^{4} \mathrm{~K}$