Physical Chemistry Third Edition

22.4 The Rotation and Vibration of Polyatomic Molecules 935

Exercise 22.7

Show that the methane molecule is a spherical top. Since any mutually perpendicular axes can

be chosen, the problem is easier if the hydrogen atoms are placed at alternate corners of a cube

that has its faces perpendicular to the coordinate axes and its center at the carbon atom, which is

located at the center of mass.

The total angular momentumLof any rigid object is the vector sum of the angular

momenta of the particles making up the object (see Appendix E for information about

angular momentum). The angular momenta of all particles making up a rigid rotating

object are in the same direction. The classical rotational energy of a rigid object is given

in terms of its angular momentum and its principal moments of inertia by

Eclassical

L^2 A

2 IA

+

L^2 B

2 IB

+

L^2 C

2 IC

(22.4-5)

whereLA,LB, andLCare the components of the vectorLon theA,B, andCaxes.

We can write the quantum mechanical rotational energy eigenvalues by replacing the

classical variables in Eq. (22.4-5) by their quantum mechanical eigenvalues. Consider a

diatomic molecule or linear polyatomic molecule, for whichIAvanishes and for which

IBIC. There can be no component of angular momentum on theAaxis, because all

of the nuclei are on this axis. Equation (22.4-5) becomes

Eclassical

1

2 IB

(

L^2 B+L^2 C

)



L^2

2 IB

(22.4-6)

From Eq. (17.2-28) we have the values thatL^2 can assume in quantum mechanics.

Substitution of this formula into Eq. (22.4-6) gives

EqmEJ

h ̄^2

2 IB

J(J+1) (22.4-7)

where we use the letterJfor the quantum number instead ofl. For a diatomic molecule,

Iμre^2 , so Eq. (22.4-7) agrees with Eq. (22.2-18). Equation (22.4-7) also applies to

a linear polyatomic molecule such as C 2 H 2 or CO 2.

EXAMPLE22.9

Show that for a rigid diatomic moleculeIeμr^2 e.

Solution

Temporarily place the molecule on thezaxis with the center of mass at the origin. Denote

the masses bym 1 andm 2 and theirzcoordinates byz 1 andz 2. Letm 1 m 2 M.

Iem 1 z^21 +m 2 z^22

z 1 

m 2 (z 1 −z 2 )

m 1 +m 2



m 2 (z 1 −z 2 )

M

, z 2 

m 2 (z 2 −z 1 )

m 1 +m 2



m 2 (z 2 −z 1 )

M

Iem 1

(

m 2 (z 1 −z 2 )

M

) 2

+m 2

(

m 1 (z 1 −z 2 )

M

) 2



m 1 m 2 (m 2 +m 1 )(z 1 −z 2 )^2

M^2



m 1 m 2

M

(z 1 −z 2 )^2 μ(z 1 −z 2 )^2