91Ó°ÊÓ

We have given that the translational motion of a nitrogen molecule will freeze out, in a box of width $1cm$

We need to find that the temperature.

The transitional motion of the nitrogen molecule freezes out when $kT$is less than or comparable to the spacing between the lowest energy levels. The energy levels inside a box width of $L$is given by

$E_n=\frac{n^2{h}^2}{8m{L}^2}$

the energies of the first two levels are given by

$E_1=\frac{h^2}{8m{L}^3}{E}_2=\frac{4h^2}{8m{L}^2}$

therefore the spacing between the two levels is:

$E_2-E_1=\frac{3h^2}{8m{L}^2}$

the freeze-out temperature is, therefore:

$$\begin{gathered} T=\frac{E_2-E_1}{k} \\ T=\frac{3h^2}{8mk{L}^2} \end{gathered}$$

Consider nitrogen at the room temperature $T=300k$inside a box with a side width of $1cm$, the mass of the nitrogen gas $N_2$is $28u$, where $u=1.66×{10}^{-27}kg$ , substitute with the givens to get (not that $h=6.636×{10}^{-34}j.s$and $k=1.38×{10}^{-23}j/k$):

$$\begin{gathered} T=\frac{3{(6.626×{10}^{-34}J.s)}^2}{8(28×1.66×{10}^{-27}kg)(1.38×{10}^{-23}J/k){(1.0×{10}^{-2}m)}^2} \\ =2.567×{10}^{-15}k \\ T=2.567×{10}^{-15}k \end{gathered}$$