91Ó°ÊÓ

A water molecule can vibrate in various ways, but the easiest type of vibration to excite is the "flexing' mode in which the hydrogen atoms move toward and away from each other but the HO bonds do not stretch. The oscillations of this mode are approximately harmonic, with a frequency of 4.8 x 10¹³Hz. As for any quantum harmonic oscillator, the energy levels are $\frac{1}{2}hf,\frac{3}{2}hf,\frac{5}{2}hf$, and so on. None of these levels are degenerate.

(a) The partition function is: because none of the levels are degenerate.

$Z=\underset{s}{∑}{e}^{-E\left(s\right)/kT}$

Where,

$E\left(s\right)=\left(s+\frac{1}{2}\right)hfs=0,1,2,…$

So,

$Z=e^{-hf/2kT}+e^{-3hf/2kT}+e^{-5hf/2kT}+…$

Let $x=hf/kT$, so

$Z=e^{-x/2}+e^{-3x/2}+e^{-5x/2}+…\left(1\right)$

As a result, the value of z is (at T = 300 K, we substitute the Boltzmann constant in eV, k = 8.617x 10^-5 eV/K, and the Planck constant in eV, h = 4.136 x 10^-15 eV s):

$x=\frac{\left(4.136×{10}^{-15}\mathrm{eV}·\mathrm{s}\right)\left(4.8×{10}^{13}\mathrm{Hz}\right)}{\left(8.617×{10}^{-5}\mathrm{eV}\right)(300\mathrm{K})}=7.68$

Substitute x into (1)

$$\begin{gathered} Z=e^{-(7.68)/2}+e^{-3(7.68)/2}+e^{-5(7.68)/2} \\ Z=0.0215 \end{gathered}$$

Probability of first state is:

$P=\frac{1}{Z}{e}^{-x/2}$

Substitute with x and z:

role="math" localid="1647366550518" $$\begin{gathered} P_1=\left(\frac{1}{0.0215}\right)e^{-7.68/2} \\ {P}_1=0.9997 \end{gathered}$$

Probability of second state is:

$P_2=\frac{1}{Z}{e}^{-3x/2}$

Substitute x and z:

role="math" localid="1647366573852" $$\begin{gathered} P_2=\left(\frac{1}{0.0215}\right)e^{-3(7.68)/2} \\ {P}_2=4.618×{10}^{-4} \end{gathered}$$

Probability of third state:

$P_3=\frac{1}{Z}{e}^{-3x/2}$

Substitute x and z:

role="math" localid="1647366591263" $$\begin{gathered} P_3=\left(\frac{1}{0.0215}\right)e^{-5(7.68)/2} \\ {P}_3=2.133×{10}^{-7} \end{gathered}$$

We have temperature value, T=300k, so

$x=\frac{\left(4.136×{10}^{-15}\mathrm{eV}·\mathrm{s}\right)\left(4.8×{10}^{13}\mathrm{Hz}\right)}{\left(8.617×{10}^{-5}\mathrm{eV}\right)(700\mathrm{K})}=3.2913$

Substitute x into (1):

role="math" localid="1647366281815" $$\begin{gathered} Z=e^{-(3.2913)/2}+e^{-3(3.2913)/2}+e^{-5(3.2913)/2} \\ Z=0.20033 \end{gathered}$$

Substitute the value of x and z to get the probabilities:

$$\begin{gathered} P_1=\left(\frac{1}{0.20033}\right)e^{-3.2913/2} \\ {P}_1=0.962847 \\ {P}_2=\left(\frac{1}{0.20033}\right)e^{-3(3.2913)/2} \\ {P}_2=0.035823 \\ {P}_3=\left(\frac{1}{0.20033}\right)e^{-5(3.2913)/2} \\ {P}_3=0.001333 \end{gathered}$$