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 $$N_A = 6.02 \times 10^{26}$$/kg, Radius of Earth: $$6.4 \times 10^{6}$$ m, Gravitational acceleration on Earth = 10 ms$$^{-2}$$]

We want the root mean square (r.m.s.) speed of a hydrogen molecule to be exactly equal to the escape speed from the surface of the earth. The two speeds are:

$$v_{\text{rms}}=\sqrt{\dfrac{3k_B T}{m}} \qquad\text{and}\qquad v_{\text{esc}}=\sqrt{2gR}$$

Here $$k_B$$ is the Boltzmann constant, $$T$$ the absolute temperature, $$m$$ the mass of one hydrogen molecule, $$g$$ the acceleration due to gravity and $$R$$ the radius of the earth.

We set $$v_{\text{rms}}=v_{\text{esc}}.$$ Doing so and then squaring both sides gives

$$\dfrac{3k_B T}{m}=2gR.$$

Now we isolate $$T$$:

$$T=\dfrac{2gR\,m}{3k_B}.$$

Next we evaluate each quantity. The escape-speed part $$2gR$$ is calculated first. Substituting $$g=10\ \text{m s}^{-2}$$ and $$R=6.4\times10^{6}\ \text{m}$$ we obtain

$$2gR = 2\times10\times6.4\times10^{6} = 1.28\times10^{8}\ \text{m}^2\text{s}^{-2}.$$

For the molecular mass $$m$$ we remember that one hydrogen atom has mass $$\dfrac{1}{N_A}\ \text{kg},$$ because $$N_A$$ particles make up $$1\ \text{kg}$$ (the statement of the Avogadro number in the data). A hydrogen molecule (H$$_2$$) consists of two atoms, so

$$m = \dfrac{2}{N_A} = \dfrac{2}{6.02\times10^{26}} = 3.32\times10^{-27}\ \text{kg}.$$

The denominator in the temperature formula is $$3k_B$$:

$$3k_B = 3\times1.38\times10^{-23} = 4.14\times10^{-23}\ \text{J K}^{-1}.$$

We now substitute every value into $$T=\dfrac{2gR\,m}{3k_B}:$$

$$T =\dfrac{(1.28\times10^{8})\;(3.32\times10^{-27})} {4.14\times10^{-23}} =\dfrac{4.25\times10^{-19}}{4.14\times10^{-23}}.$$

Dividing the numerical coefficients and combining powers of ten step by step, we first handle the coefficients:

$$\dfrac{4.25}{4.14}\approx1.03.$$

Then we combine the exponents:

$$10^{-19}\div10^{-23}=10^{\,4}.$$

So

$$T\approx1.03\times10^{4}\ \text{K}.$$

The temperature that makes the r.m.s. velocity of hydrogen molecules equal to the earth’s escape velocity is therefore about $$10^{4}$$ kelvin, which matches the fourth option.

Hence, the correct answer is Option D.