Answer
2820.5 m
Work Step by Step
Here we use the principle of conservation of mechanical energy to find the solution.
Initial mechanical energy = Final mechanical energy
$\frac{1}{2}mv_{f}^{2}+mgh_{f}=\frac{1}{2}mv_{0}^{2}+mgh_{0}$
We know that,
$\frac{1}{2}mv_{0}^{2}=KE_{Avg}=\frac{3}{2}kT$ and $\frac{1}{2}mv_{f}^{2}=0$ (Since the atom comes to a halt at the top of its trajectory)
Let's take the height at the earth's surface to be $h_{0}=0\space m$. So we can write,
$$mgh_{f}=\frac{3}{2}kT=>h_{f}=\frac{3kT}{2mg}-(1)$$
We know that, $m=\frac{M}{N_{A}}-(2)$ ; Where M-molecular mass, $N_{A}$-Avagadro's number
(2)=>(1),
$h_{f}=\frac{3kN_{A}T}{2Mg}=\frac{3RT}{2Mg}$ ; Let's plug known values into this equation.
$h_{f}=\frac{3(8.31\space J/mol\space K)(291\space K)}{2(131.29\times10^{-3}kg/mol)(9.8\space m/s^{2})}=2820.5\space m$