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## Q. 10.2

(a) Given that the spacing between vibrational energy levels of the HCl molecule is 0.36 eV, calculate the effective force constant. (b) Find the classical temperature associated with this difference between vibrational energy levels in HCl.

Strategy (a) Because $\kappa=\mu \omega^{2}$ we first need to find μ and ω. The reduced mass μ is given by  $\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)$. We also know that $\Delta E=h f=\hbar \omega, \text { so } \omega=\Delta E / \hbar$. At the lowest level $n=0 \text { and } E_{\text {vibr }}=\hbar \omega / 2$.

(b) Two degrees of freedom are associated with a onedimensional oscillator, one from the kinetic energy and one from the potential (see Section 9.3). Therefore we can say that

$\Delta E=\hbar \omega=2\left(\frac{k T}{2}\right)=k T$

(where k is Boltzmann’s constant), and so $T=\Delta E / k$.

## Verified Solution

(a) If we assume the most common chlorine-35 isotope, the reduced mass is

\begin{aligned}\mu &=\frac{m_{1} m_{2}}{m_{1}+m_{2}}=\frac{(34.97 u )(1.008 u )}{34.97 u +1.008 u } \\&=0.9798 u =1.63 \times 10^{-27} kg\end{aligned}

Then

$\omega=\frac{\Delta E}{\hbar}=\frac{0.36 eV }{6.58 \times 10^{-16} eV \cdot s }=5.47 \times 10^{14} rad / s$

Putting together the calculated numerical values, the force constant is given by Equation (10.4) as

$\omega=\sqrt{\frac{\kappa}{\mu}}$ (10.4)

\begin{aligned}\kappa &=\mu \omega^{2}=\left(1.63 \times 10^{-27} kg \right)\left(5.47 \times 10^{14} rad / s \right)^{2} \\&=490 N / m\end{aligned}

which is quite close to the value in Table 10.1.

(b) Using the numerical value of $\Delta E$,

$T=\frac{\Delta E}{k}=\frac{0.36 eV }{8.62 \times 10^{-5} eV / K }=4200 K$

In order to excite this vibrational state, we need a temperature of 4200 K. This is why vibrational levels of most diatomic molecules are not thermally excited at ordinary temperatures.

 Table 10.1 Fundamental Vibrational Frequencies and Effective Force Constants for Some Diatomic Molecules Frequency (Hz), Force Constant Molecule n = 0 to n = 1 (N/m) HF $8.72 \times 10^{13}$ 970 HCl $8.66 \times 10^{13}$ 480 HBr $7.68 \times 10^{13}$ 410 HI $6.69 \times 10^{13}$ 320 CO $6.42 \times 10^{13}$ 1860 NO $5.63 \times 10^{13}$ 1530 From G. M. Barrow, The Structure of Molecules, New York: Benjamin (1963).