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ADVANCED QUANTUM MECHANICS
1. MORSE potential
It is a good approximation to an intermolecular potential energy curve.
First let x = l – l 0
so that I write.
D
e
is dissociation energy (measured from the minimum of V(l)),
β is a measure of the
curvature of V(l). QUESTION: Derive relation between the force constant and
D
e
, β .
SOLUTION: expand V(x) about x = 0:
(1)
First term: a constant (depends upon where I choose the zero of energy)
Second term: l = l 0
is minimum of potential energy curve, dV/dl vanishes there; no linear term
in the displacement.
dV/dl vanishes at l = l 0
means that the force acting between the nuclei is zero at this point; l =
l 0
is called the equilibrium bond length.
With l – l 0
by x , ( d
2
V/dl
2
) l=l
by k , ( d
3
V/dl
3
) l=l
by
γ
3
equation (1) becomes to
V(0) = 0
dV
dx
x= 0
[
2 D
e
β ( e
− βx
−e
− 2 βx
]
x= 0
(
d
2
V
d x
2
)
x= 0
[
− 2 D
e
β ( β e
−βx
− 2 β e
− 2 βx
]
x= 0
= 2 D
e
β
2
Therefore
V
x
=D
e
β
2
x
2
Comparing this with
V ( x )=
k
x
2
gives
k = 2 D
e
β
2
- VIBRATIONAL FREQUENCY
A diatomic molecule can make a transition from one vibrational energy state to another by
absorbing or emitting electromagnetic radiation whose observed frequency satisfies BOHR
frequency condition
∆ E=h ν
obs
The infrared spectrum of Br
19
F
❑
75
consists of an intense line at 380 cm
− 1
.
Calculate: force constant of Br
19
F
❑
75
SOLUTION: force constant: k =( 2 πc
ω)
2
μ
(2)
The reduced mass:
μ=
( 75.0 amu) ( 19.0 amu)
( 75.0+19.0 ) amu
( 1.661 x 10
− 27
kg∗amu
− 1
)=2.52 x 10
− 26
kg
and so
k = [
2 π (2.998 x 10
8
m∗s
− 1
) ( 380 cm
− 1
) ( 100 cm∗m
− 1
]
2
(2.52 x 10
− 26
kg) = 129 kg∗s
− 2
= 129 N∗m
− 1
k = 129 N∗m
− 1
Why equation (2)?
With BOHR frequency condition
∆ E=h ν
obs
harmonic- oscillator allows transitions only between adjacent states → condition: ∆ ν=± 1 ,
this is called selction rule.
For absorption to occur,
∆ ν=± 1
and so
∆ E=E
ν+ 1
−E
ν
=hν
with the vibrational energy
(3)
where
(3), (4) is a solution of SCHRÖDINGER equation for a one-dimensional harmonic oscillator
with
- BOND POTENTIAL
of a diatomic molecule has the form
bondlength of this molecule is 2.8 Ängström
and dissociations energy is 8 eV.
Calculate A and B.
SOLUTION:
The bond length is the separation R where the bond potential has a minimum.
Dissociation energy is the bond potential evaluated at R 0
.
Check this by evaluating
If the ground state potential of HCL molecule in the vicinity of the minimum is approximated
by E
pot
k
R−R
e
2
, one obtains a vibration frequency of ν
0
= 9 x 10
13
s
− 1
.
Calculate constant k.
What is the classical oscillation amplitude for an oscillation with the same energy as the ν= 1
state?
m H
= 1u, m Cl
= 35.45u, u = 1.66 x 10
kg
SOLUTION
Vibrational frequency is
ν
0
= 9 x 10
13
Hz=
2 π
√
k /m
r
where k is spring constant and m
r
1 /m
a
b
− 1
is the reduced mass. The spring
constant for a molecule HCL is 521 N/m. The energy for ν = 1 is
hf
ν +
=8.945 x 10
− 20
J.
Classically, this energy is k A
2
where A is the amplitude.
A=
√
2 x 8.945 x 10
− 20
=1.853 x 10
− 11
m
For the normal modes of a molecule, the following energies were found: 0.18678 eV and
1.05344 x 10
− 19
J.
Calculate the corresponding frequencies and wavelengths
SOLUTION
For 0.18678 eV the frequency is 4.516 x 10
13
Hz and the wavelength is 6.
μm .
For 1.05344E-19J the frequency is 1.590 x 10
14
Hz and the wavelength is 1.886 μm.
Figure shows a simple term schema of the vibration-rotation transitions in HCl.
You get two series of transitions, which differ by the respective ∆J.
For the energies one obtains:
P-branch (∆J = −1):
∆E = E( ν + 1, J – 1) – E( ν, J) = −2B J + E
0
vib
R-branch (∆J = +1):
∆E = E( ν + 1, J + 1) − E(v, J) = 2B(J + 1) +E
0
vib
The transition | J = 0, ν = 0 ⟩ → | J = 0, v = 1 ⟩ is prohibited. For this reason, a gap in the
spectrum arises, which is centered exactly in the transition energy of the vibration transition.
With 1/λ = 2885.90cm
results in
E
0
vib
= ℏ ω
0
hc
λ
=0.359 eV
(b)Calculate the mean nuclear distance R e
of the HCl molecule
Solution: The core distance can be determined from the energy interval between the
rotational peaks, which is given by ∆ = 21cm
∆ E
rot
J + 1
−∆ E
rot
J
2
I
2
M R
e
2
you get R
e
2
M ∆ ( hc)
2 πM ∆ c
With the following values you finally get the core distance:
m 1H
= 1.67 · 10
kg
m 35Cl
= 5.85 · 10
kg
M = 1.62 · 10
kg
R e
= 1.3 Ängström
On closer inspection, one will find that the distance between the rotational peaks is not
strictly constant. This is because the core distance is different in the two virtualization states.
(c) On closer inspection, it is noticeable that the absorption peaks have a substructure
(double peak). How do you explain this?
Solution:
Chlorine occurs in nature in two isotopes:
35
Cl (75.5%) and
37
Cl (24.5%).
This leads to slightly different reduced masses M and thus to shifted absorption peaks in the
rotation spectrum.
- Infared spectrum of H
79
Br
Here is an intense line 2559 cm
- . Calculate the force constant of H
79
Br and the period of
vibration of H
79
Br.
Solution:
ν
obs
2 πc
k
μ
1 / 2
k =( 2 πc
ω)
2
μ
[
2 π ( 2.998 x 10
10
cm∗s
− 1
) ( 2.559 cm
− 1
]
2
x
[
( 1.008 amu) ( 78.92 amu)
79.93 amu
]
(1.661 x 10
− 27
kg∗amu
− 1
) = 384 N∗m
− 1
k = 384 N∗m
− 1
The period of vibration is
T =
ν
c
ν
=1.30 x 10
− 14
s
- Bond length of H
35
Cl
To a good approximation, the microwave spectrum of H
35
Cl consists of a series of equally
spaced lines, separated by 6.26 x 10
11
Hz.
Calculate the bond length of H
35
Cl.
Solution:
With ν obs
= 2 B(J + 1 ) J = 0, 1, 2, ⋯ where
B=
h
8 π
2
I
(Hz) the spacing of the lines in the
microwave spectrum of H
35
Cl is given by
2 B=
h
4 π
2
I
= 6.26 x 10
11
Hz
Solving the equation for I, we have
The reduced mass :
