Download Rigid and non-rigid Rotors

Rigid and non-rigid Rotors
N. Chandrakumar
We consider free rotation of a rigid diatomic molecule around an axis which
passes through its center of mass, and is perpendicular to the molecular axis.
r1
r2
m2
m1
r
The diatomic molecule comprises two atoms, numbered 1 and 2, with
respective atomic masses m1 and m2, a distance r apart. The center of mass (which is
at a distance r1 from atom 1 and r2 from atom 2, such that r1+r2 = r) is located by
recognizing that the moments of the masses should be equal around this point, ie,
m1r1=m2r2. We thus have:
r1  m1  m2   r1m1  r1m2  r2 m2  r1m2   r1  r2  m2  rm2
 r1 
rm2
m
 1  m2 
By an analogous calculation, we find:
r2   m1rm m1 2 
For the moment of inertia, we find in terms of the reduced mass eff:
I   mi ri 2  m1r12  m2 r22  r 2
i

I  eff r 2
m1m22
 m1  m2 
2
 r2
 eff   mm11mm22 
m2 m12
 m1  m2 
2
 r2
m1m2
 m1  m2 

The free rigid rotor is not subject to a force, ie, sees a constant potential that
may be set to zero, so that the total energy comprises the kinetic energy alone. For the
kinetic energy of the rotor, we have:
2
1
i i
2
i
It is more convenient to express the kinetic energy in terms of the angular velocity
which characterizes the rotation, rather than in terms of the linear velocity v which
varies for different particles of the rotor, by employing the standard relation viri.
[Recall that the length of an arc of a circle is directly proportional to the angle
subtended by it: l = r. This clearly results in: dl/dt = r (d/dt), ie, v = r]
mv
T
We find, in terms of the moment of inertia I:
T  2
2
m r
i i
i
2
 12 I  2
We also recognize that the angular momentum of the rotating object is given by:

J   mi ri vi    mi ri 2  I 
i
i
Quantum mechanically, however the magnitude of the angular momentum is given by
[J(J+1)]1/2.
It may be noted that rotational angular velocities are quantized accordingly:
  J  J  1 I
We thus find the following expression for the energy of the free rigid rotor:
EJ  T 
1  I 
2
I
2

1
2I
J  J  1  2
The allowed rotational quantum numbers are: J = 0,1,2,3,… .
Expressed in units of wavenumbers, rotational energies are given as:
EJ 
EJ
hc
 J  J  1 8h2cI  BJ  J  1
The eigenfunctions of the rigid free rotor are the spherical harmonics, YJ,M, M
being the eigenvalue of the z component of the rotational angular momentum, ie, Jz.
Selection rules for rotational transitions may be found by considering conditions
under which the transition dipole moment integral fi vanishes. The required integral
is given in terms of the molecular permanent electric dipole moment by:
2 
 Y

J f ,M f

μYJ i , M i sin  d d
0 0
  Mf

 2  iM f  iM i 
Mi
ˆ
e d 
k   PJ f  cos   cos  PJ i  cos   sin  d   e

 0

 0



2




Mf


iM
M

iM
f
i

i
~ 0  ˆi   PJ f  cos   sin  PJ i  cos   sin  d   e
cos  e d  
 0
 0
 

  Mf

  2  iM f 
Mi
iM i
ˆ
j
P
cos
sin
P
cos
sin
d





e
sin

e
d





   Jf 
 

Ji
 0
 
  0
Because the spherical harmonics – which form a complete set of orthonormal
functions in and may be factored into the associated Legendre polynomials
(which are polynomials in [cos multiplied by oscillatory complex functions of ,
we infer the following selection rules for rotational transitions: J = 1,  .
Further, we notice that , the permanent electric dipole moment of the molecule
needs to be non-zero, for a rotational transition to be allowed.
The spacing between two successive rotational transitions, one between the
levels J and J+1 and the other between levels J+1 and J+2, is clearly 2B.
The intensities of rotational transitions are, in addition, a function of the population of
the rotational energy levels. We have:
NJ ~ gJ e
 EJ kT
  2 J  1 e
 J  J 1 2 2 IkT
Clearly, population is maximal at a value of J which may be found by treating it as a
continuous variable, with respect to which NJ is differentiated:
dN J
dJ
 0  2e
  2 J  1
 2J 1 
J max 
IkT

2
 J  J 1  2 2 IkT
2
2 IkT
2

  2 J  1
2
2
2 IkT
e
 J  J 1  2 2 IkT
2
IkT
 
1
2

2 kT h 8 2 Bc
h

 
1
2
kT
2 Bch

1
2
 12
Moments of Inertia
Moments of inertia may be defined along three principal molecular axes A, B and C
that are mutually orthogonal.
Spherical tops: All three moments of inertia are equal (eg. CH4).
Symmetric tops: Two components are equal, the third one being different. There are
two sub-categories here, viz., prolates, for which the unique IA < IB = IC (eg. CH3CN),
and oblates, for which the unique IA > IB = IC (eg. benzene).
Linear molecules: Two components of the moments of inertia, around axes
perpendicular to the molecular axis and to each other are equal; while the third
component along the molecular axis vanishes.
Asymmetric tops: All three moments of inertia are unequal (eg. CH2Cl2, O3, SO2).
Non-rigid rotors
Centrifugal forces due to rotation, which are significant especially at high
rotational angular velocities, are balanced by restoring forces which keep the
molecule from flying apart.
The centrifugal force may be readily calculated from the centrifugal
acceleration, which in turn is the rate of change of radial velocity of the rotating
system. For a system with linear velocity v and rotational velocity , the centrifugal
acceleration is readily seen to be:
v [cos ((d)]/dt = v (sin d/dt) = v (ddt) = v  = r .
The centrifugal force is thus given by eff r.
We thus have:
eff r 2  k  r  r0 


 r k  eff  2  kr0
r
k
k 
eff 

2

r0 
r0

 eff  2 
1 k 



 r0 1  2
2
vib

 r 1  r01 1  2 ; r 2  r02 1  2
2
vib
2
vib


1
2


 r02 1  22 ; r  r0  r0 2
2
vib
2
vib
   vib 

In the foregoing, we have taken into account the fact that rotational velocities are
typically a tiny fraction of the angular frequencies of vibration, the square of their
ratio being normally in the range 103 to 107.
For the energy of the non-rigid rotor, we need to take into account now the
rotational kinetic energy, as well as the potential energy corresponding to the
restoring force.
We have:
T  J  J  1 2 I  J  J  1 2 
2
2
eff r0
 J  J  1 2 
2
2
eff r0
 2 J  J  1 2 
T  J  J  1 2 I  2 J 2  J  1
2
V  12 k  r  r0   k2 r02 4 
2
4
vib
eff
2
1  
2 2
2
2
vib
J  J 1
2
2 2
r
I2

eff 0 vib
2
4
2
2 I 3vib
r02 J 2  J  1
EJ  T  V  J  J  1 2 I  J 2  J  1
2
2
2
2
4
2 4
vib
I
 J 2  J  1
2
4
2
2 I 3vib
4
2
2 I 3vib
In units of wavenumbers, we have for the energy levels of the non-rigid rotor:
EJ  BJ  J  1  4 B2 J 2  J  1  BJ  J  1  DJ 2  J 1
3
2
vib

[Note that vib is the vibrational frequency expressed in rad s1, while  vib is the
vibrational frequency expressed in wavenumbers.]
2