Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 10

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.

Step-by-Step

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).