(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 κ = μω^2 we first need to find μ and

Question

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

(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

[latex]\Delta E=\hbar \omega=2\left(\frac{k T}{2} ight)=k T[/latex]

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

Accepted Answer

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

[latex]\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 imes 10^{-27} kg\end{aligned}[/latex]

Then

[latex]\omega=\frac{\Delta E}{\hbar}=\frac{0.36 eV }{6.58 imes 10^{-16} eV \cdot s }=5.47 imes 10^{14} rad / s[/latex]

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

[latex]\omega=\sqrt{\frac{\kappa}{\mu}}[/latex] (10.4)

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

which is quite close to the value in Table 10.1.

(b) Using the numerical value of [latex]\Delta E[/latex],

[latex]T=\frac{\Delta E}{k}=\frac{0.36 eV }{8.62 imes 10^{-5} eV / K }=4200 K[/latex]

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	[latex]8.72 imes 10^{13}[/latex]	970
HCl	[latex]8.66 imes 10^{13}[/latex]	480
HBr	[latex]7.68 imes 10^{13}[/latex]	410
HI	[latex]6.69 imes 10^{13}[/latex]	320
CO	[latex]6.42 imes 10^{13}[/latex]	1860
NO	[latex]5.63 imes 10^{13}[/latex]	1530
From G. M. Barrow, The Structure of Molecules, New York: Benjamin (1963).

Question 10.2: (a) Given that the spacing between vibrational energy levels...

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