Physical Chemistry Third Edition

22.2 The Nonelectronic States of Diatomic Molecules 921

The Rigid Rotor

The rigid rotor is a model system that consists of two nuclei with fixed internuclear

distance equal tore. No vibration is possible. Becauseris fixed at the valuere, the

derivatives with respect torin Eq. (22.2-13) do not occur, and becauseV depends

only onr, the potential energy has a constant value that we set equal to zero:

VV(re)constant 0 (22.2-14)

Equation (22.2-13) now becomes

1

2 μr^2 e

̂L^2 YErelY (22.2-15)

This equation is the same as Eq. (17.2-6) except for the factor 1/ 2 μr^2 e. It is satisfied

by the spherical harmonic functionYJMof Section 17.2, which we now denote as the

rotational wave functionψrot:

ψrotYJM(θ,φ)ΘJM(θ)ΦM(φ) (22.2-16)

It is customary to use the lettersJandMfor the quantum numbers instead oflandm

for rotational states:

J0, 1, 2,... (22.2-17a)

MJ,J−1,...,−J+1,−J (22.2-17b)

From the angular momentum eigenvalues in Eq. (17.2-27),

EEJ

h ̄^2

2 μr^2 e

J(J+ 1 )

(

rigid rotor

)

(22.2-18)

The degeneracy of the energy level for a given value ofJis equal to the number of

possible values ofM(Jpositive values,Jnegative values, plus one zero value):

gJ(2J+1) (22.2-19)

Figure 22.1 shows the first few energy levels. The degeneracies are depicted by placing

a line segment for each value ofM.

20

12

6

2

0

J 4

J 3

J 2

J 1

J 0

Energy/

(^2) h 2 r
(^2) e
Figure 22.1 Energy Levels of
the Rigid Rotor.

EXAMPLE22.3

Assume that the carbon monoxide molecule is a rigid rotor withre 1. 128 × 10 −^10 m. Find

the rotational energy forJ0 andJ1.

Solution

μ

(12.000 amu)(15.9949 amu)

12 .000 amu+ 15 .9949 amu

(

1kg

6. 02214 × 1026 amu

)

 1. 1385 × 10 −^26 kg

E 0  0

E 1 

h^2

8 πμr^2 e

(1)(2) 2. 412 × 10 −^22 J