1. Use the escape velocity from Earth

The escape velocity is

$v_e = \sqrt{2gR}$

Given:

• $g = 10\,\text{m s}^{-2}$
• $R = 6.4\times 10^6\,\text{m}$

So,

$v_e = \sqrt{2\times 10\times 6.4\times 10^6}$

$v_e = \sqrt{1.28\times 10^8}$

$v_e \approx 1.13\times 10^4\,\text{m s}^{-1}$

2. RMS speed of hydrogen molecules

For a gas,

$v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}$

We are told that the RMS speed of hydrogen molecules equals escape velocity:

$\sqrt{\frac{3k_B T}{m}} = v_e$

Squaring both sides,

$\frac{3k_B T}{m} = v_e^2$

$T = \frac{m v_e^2}{3k_B}$

3. Mass of one hydrogen molecule

Hydrogen is $H_2$, so its molar mass is

$M = 2\times 10^{-3}\,\text{kg mol}^{-1}$

Mass of one molecule:

$m = \frac{M}{N_A} = \frac{2\times 10^{-3}}{6.02\times 10^{23}}$

$m \approx 3.32\times 10^{-27}\,\text{kg}$

4. Substitute values

Using

$v_e^2 = 1.28\times 10^8$

we get

$T = \frac{(3.32\times 10^{-27})(1.28\times 10^8)}{3(1.38\times 10^{-23})}$

First calculate numerator:

$3.32\times 1.28 \approx 4.25$

$\Rightarrow (3.32\times 10^{-27})(1.28\times 10^8) \approx 4.25\times 10^{-19}$

Denominator:

$3\times 1.38\times 10^{-23} = 4.14\times 10^{-23}$

Thus,

$T \approx \frac{4.25\times 10^{-19}}{4.14\times 10^{-23}}$

$T \approx 1.03\times 10^4\,\text{K}$

5. Choose the closest option

$T \approx 10^4\,\text{K}$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.