Given: Mass of nitrogen molecule, $m = 4.65 \times 10^{-26} \text{ kg}$ Boltzmann constant, $k = 1.38 \times 10^{-23} \text{ J K}^{-1}$ Temperature, $T = 0^\circ \text{C} = 273 \text{ K}$ Radius of Earth, $R_e = 6.4 \times 10^6 \text{ m}$ Mass of Earth, $M_e = 6 \times 10^{24} \text{ kg}$ Gravitational constant, $G = 6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ The root mean square velocity $v_{rms}$ is: $$ \begin{aligned} v_{rms} &= \sqrt{\frac{3kT}{m}} \\ &= \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{4.65 \times 10^{-26}}} \\ &= 493.01 \text{ m s}^{-1} \end{aligned} $$ Using the law of conservation of energy, the kinetic energy at the surface equals the change in gravitational potential energy at height $h$: $$ \begin{aligned} \frac{1}{2}mv^2 &= G M_e m \left( \frac{1}{R_e} - \frac{1}{R_e + h} \right) \\ \frac{1}{2}v^2 &= G M_e \left( \frac{h}{R_e(R_e + h)} \right) \end{aligned} $$ Substituting the values: $$ \begin{aligned} \frac{1}{2}(493.01)^2 &= \frac{6.673 \times 10^{-11} \times 6 \times 10^{24} \times h}{6.4 \times 10^6 (6.4 \times 10^6 + h)} \\ 121532.4 &= \frac{4.0038 \times 10^{14} \times h}{4.096 \times 10^{13} + 6.4 \times 10^6 h} \end{aligned} $$ Solving for $h$: $$ \begin{aligned} 4.9798 \times 10^{18} + 777807.36 h &= 4.0038 \times 10^{14} h \\ h &\approx 12457 \text{ m} = 12.457 \text{ km} \end{aligned} $$ The molecule will rise to a height of $12.457 \text{ km}$.