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 $k_B=1.38\times {10}^{-23}\text{J/K}$
Avogadro Number $N_A=6.02\times {10}^{26}\text{/kg}$
Radius of Earth$:6.4\times {10}^6\text{m}$
Gravitational acceleration on Earth, $g=10{\text{ms}}^{-2}]$

• A. $3\times {10}^5\text{K}$
• B. ${10}^4\text{K}$
• C. $800\text{K}$
• D. $650\text{K}$

Answer 1

The correct option is B ${10}^4\text{K}$

Here, $v_{\text{rms}}=v_e$

$\Rightarrow \sqrt{\frac{3k_B{N}_{A}T}{M}}=\sqrt{2gR}$

$\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 6.02\times {10}^{26}\times T}{2\times {10}^{-3}}}=\sqrt{2\times 10\times 6.4\times {10}^6}$

$1.246\times {10}^4\times T=12.8\times {10}^7$

$\therefore T=1.027\times {10}^4\text{K}$

Hence, $\left(C\right)$ is the correct answer.