Question

for many molecules, only the lowest vibrational energy state is significantly populated at room temperature. In this case, the vibrational partition function is close to

A) 0

B) 1

C) the temperature T

D) infinite

The answer is B.

Answer 1

The overall vibrational partition function is the product of the individual partition functions, and q V = q V(1) q V(2) ..., where q V( K) is the partition function for the Kth normal mode and is calculated by direct summation of the observed spectroscopic levels.

The energy levels of a quantum simple harmonic oscillator of frequency $$\omega$$ are

en = (n+) w n=0,1,2..

so

$\begin{eqnarray*} Z_1\!\!\!&=&\!\!\!\sum_{n=0}^\infty = e^{-\varepsilon_n\beta}=... ...\Bigl(2 \sinh({\textstyle \frac 1 2}\hbar\omega\beta)\Bigr)^{-1} \end{eqnarray*}$

where we have used the expression for the sum of a geometric series, $$\sum_n x^n=(1-x)^{-1}$$, with .

e-huß

From this we obtain

In Z1 0B2

The low temperature limit of this ( $$k_{\scriptscriptstyle B}T \ll\hbar\omega$$; $$\hbar\omega\beta\to \infty$$) is $${\textstyle \frac 1 2}\hbar \omega$$, which is what we expect if only the ground state is populated.