Physical Chemistry Third Edition

23.3 Rotational and Vibrational Spectra of Diatomic Molecules 967

It looks like the rotational spectrum except that these lines start from the band center

instead of from zero reciprocal wavelength. In the same approximation the reciprocal

wavelengths of the lines of thePbranch are given by

̃νP

1

λP

(E1,J− 1 −E0,J)/hc ̃νe− 2 ̃BeJ (J1, 2, 3,...) (23.3-11)

whereJis the value ofJin the lower (v0) state, which must be at least as large as

unity since the transition is to the next lower value ofJ. ThePbranch consists of a set

of equally spaced lines with spacing 2 ̃Beas does theRbranch, but the lines are on the

other side of the band center. The splitting between the first line of thePbranch and

the first line of theRbranch is 4 ̃Be.

EXAMPLE23.6

a.Using the values of the parameters for CO in Table A.22 in Appendix A, find the reciprocal

wavelength at the band center and the splitting between the lines of the fundamental band

in the rigid rotor–harmonic oscillator approximation.

b.Find the value of the force constantkfor the CO molecule.

c.Assuming that all transition dipole moments are equal and that the temperature is 298 K,

find the most intense line in thePbranch and in theRbranch of the fundamental band.

Solution

a.The reciprocal wavelength of the band center is

̃vBC

1

λBC

 ̃ve 2169 .81 cm−^1

The splitting is

∆ ̃ν∆(1/λ) 2 ̃Be2(1.93127 cm−^1 ) 3 .86254 cm−^1

b.From Example 23.4 the reduced mass of CO equals 1. 1385 × 10 −^26 kg. From

Eq. (22.2-30):

ν ̃νec

1

2 π

√

k

μ

or k 4 π^2 μ ̃νe^2 c^2

k 4 π^2 (1. 1385 × 10 −^26 kg)(2169.81 cm−^1 )^2 (2. 9979 × 1010 cm s−^1 )^2

 1901 .8kgs−^2  1901 .8Nm−^1  1901 .8Jm−^2

The value of the force constant of the CO molecule is fairly large, because CO has a triple

bond, which is quite stiff compared to most single and double bonds.

c.From Example 22.12, the most populated level is that forJ7. This corresponds to the

eighth line from the band center in theRbranch and the seventh line from the band center

in thePbranch.

Figure 23.9 shows the fundamental band of the HCl molecule. The double lines

are due to the presence of two isotopes of chlorine. The lines are not equally spaced,

mostly due to the effect of theαterm in the energy level expression of Eq. (22.2-45).