[PoM] Average rotational energy

[BRN]

Homework Statement

Determines the average energy of rotation (per molecule) of a rarefied gas of HF at T = 50 K, knowing that the wave number of rotational absorption for the transition L = 2 → 3 worth 121.5 cm-1. [Hint: it performs the calculation by limiting the number of levels taken into account, making sure to get the numerical result by 5%.]

Solution:$E_{rot}=5.493*10^{-22} [J]$

The Attempt at a Solution

In the classical limit, the average rotational energy per molecule is:

$\frac{E_{rot}}{N}=\frac{1}{2}I\omega^2$

For transition L = 2 → 3, I have:

$\nu=\frac{E_{rot}}{h}=\frac{\hbar^2l(l+1)}{2hI} \Rightarrow I=\frac{\hbar^2l(l+1)}{\nu 2h}$

then,

$\omega^2=(2\pi \nu)^2=\frac{2k_BT}{I}=3.3318*10^{17} [rad/s]$

so:

$E_{rot}=\frac{1}{2}I\omega^2=k_BT=6.903*10^{-22} [J]$#

But, in what way I have to limit the number of levels taken into account?

 

[BRN]

Nobody?