28Jan2011
Chemistry 21b – Spectroscopy
Lecture # 11 – VibrationRotation Spectra of Diatomic Molecules
What happens to the rotation and vibration spectra of diatomic molecules if more
realistic potentials are used to describe the interatomic interaction? So for we have just
assumed that
R
=
R
e
, and that no change in the bond length is possible.
Clearly,
for a realistic interatomic potential, such as those outlined in Figure 4.1, considerable
changes in bond length with occur as one moves up in energy from the ground vibrational
state. As a onedimensional equation, the radial nuclear motion described by Eq. (6.4)
can be solved numerically quite simply, and there are in fact computer routines which
invert experimental data to yield diatomic potentials
V
(
R
) =
E
el
(
R
) with spectroscopic
precision. Considerable physical insight, however, can be gained by using simple, analytical
potentials for
V
(
R
), and analyzing the resulting energy level expressions.
One of the
simplest and most useful potentials was developed by Morse in the early part of this
century, which we’ll proceed to examine now.
11.a. The Morse Potential
Clearly, an appropriate potential must go to small values as
R
→ ∞
, have a minimum
at
R
=
R
e
, and go to very large values as
R
→
0. A simple potential which does this, the
Morse potential, is given by
V
(
R
)
=
D
e
(1
−
e
−
a
(
R
−
R
e
)
)
2
,
(11
.
1)
where the dissociation energy
D
e
, the equilibrium bond length
r
e
, and the curvature of
the potential near the minimum
a
are three adjustable parameters. The major qualitative
problem with the Morse potential is that is does not go to
∞
as it should as
R
→
0.
However, we know from our examination of the harmonic oscillator that the wavefunctions
tail away rapidly to zero in the classically forbidden regions, and so as long as the potential
is sufficiently large at small R this should not pose a significant problem.
If the Morse potential is placed into the radial nuclear motion equation, Eq. (6.4) on
page 47 of the notes, we find:
d
2
S
(
R
)
dR
2
+
braceleftbigg
−
J
(
J
+ 1)
R
2
+
2
μ
¯
h
2
[
E
−
D
e
−
D
e
e
−
2
a
(
R
−
R
e
)
+ 2
D
e
e
−
a
(
R
−
R
e
)
]
bracerightbigg
S
(
R
)
=
0
,
(11
.
2)
where the substitution
F
(
R
) =
1
R
S
(
R
) has been made.
Like all such problems in quantum mechanics, the challenge is in finding the right
substitution of variables that allows Eq. (11.2) to be recast in a form that has wellknown
solutions.
We’ll only briefly outline the procedure here.
The first set of substitutions
involve the equations
y
=
e
−
a
(
R
−
R
e
)
and
A
=
J
(
J
+ 1)
¯
h
2
2
μR
2
e
.
(11
.
3)
86
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If these are subsituted into Eq. (11.2), we find
d
2
S
dy
2
+
1
y
dS
dy
+
2
μ
a
2
¯
h
2
parenleftbigg
E
−
D
e
y
2
+
2
D
e
y
−
D
e
−
AR
2
e
y
2
R
2
parenrightbigg
=
0
(11
.
4)
For
A
negationslash
= 0, the last (
R
2
e
/R
2
) term must be expanded in order to generate an equation
containing only
y
. The first three terms of the Taylor expansion are given by
R
e
R
2
=
1
[1
−
(ln
y/aR
e
)]
2
= 1 +
2
aR
e
(
y
−
1) +
parenleftbigg
−
1
aR
e
+
3
a
2
R
2
e
parenrightbigg
(
y
−
1)
2
+
...
(11
.
5)
Retaining these first three terms and regrouping yields
d
2
S
dy
2
+
1
y
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 Atom, Mole, dipole moment, heteronuclear diatomic molecule

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