1. The document discusses the rotational energy levels of a carbon monoxide (CO) molecule. 2. It is found that the characteristic rotational energy (E0r) of CO is 0.239 meV and an energy level diagram is drawn for rotational levels from l=0 to l=5. 3. Transitions between these levels that obey the selection rule Δl=-1 are indicated and the energies of the photons emitted during these transitions are calculated. 4. The wavelengths of the photons are then found, and it is determined that they fall within the microwave region of the electromagnetic spectrum.

1. Chapter 37
Molecules
Conceptual Problems
1 • Would you expect NaCl to be polar or nonpolar?
Determine the Concept Yes. Because the center of charge of the positive Na ion
does not coincide with the center of charge for the negative Cl ion, the NaCl
molecule has a permanent dipole moment. Hence, it is a polar molecule.
5 •• The elements on the far right column of the periodic table are
sometimes called noble gases, both because they are gasses under a wide range of
conditions, and because atoms of these elements almost never react with other
atoms to form molecules or ionic compounds. However, atoms of noble gases
can react if the resulting molecule is formed in an electronic excited state. An
example is ArF. When it is formed in the excited state, it is written ArF* and is
called an excimer (for excited dimer). Refer to Figure 37-13 and discuss how a
diagram for the electronic, vibrational, and rotation energy levels of ArF and
ArF* would look in which the ArF ground state is unstable and the ArF* excited
state is stable. Remark: Excimers are used in certain kinds of lasers.
Determine the Concept The diagram would consist of a non-bonding ground
state with no vibrational or rotational states for ArF (similar to the upper curve in
Figure 37-4) but for ArF* there should be a bonding excited state with a definite
minimum with respect to inter-nuclear separation and several vibrational states as
in the excited state curve of Figure 37-13.
Energy Levels of Spectra of Diatomic Molecules
23 • The separation of the two oxygen atoms in a molecule of O2 is actually
slightly greater than the 0.100 nm used in Example 37-3. Furthermore, the
characteristic energy of rotation E0r for O2 is 1.78 × 10–4
eV rather than the result
obtained in that example. Use this value to calculate the separation distance of the
two oxygen atoms.
Picture the Problem We can relate the characteristic rotational energy E0r to the
moment of inertia of the molecule and model the moment of inertia of the O2
molecule as two point objects separated by a distance r.
The characteristic rotational energy
of a molecule is given by: I
E r
2
2
0
h
=
24

2. Molecules 25
The moment of inertia of the
molecule is given by:
2
O2
1
2
O
2
2 rM
r
MI =⎟
⎠
⎞
⎜
⎝
⎛
=
Substitute for I to obtain:
( ) 2
p
2
2
O
2
2
O2
1
2
0
162 rmrMrM
E r
hhh
===
Solving for r gives:
p0
1
4 mE
r
r
h
=
Substitute numerical values and evaluate r:
( )( )( )
nm121.0
kg10673.1J/eV10602.1eV1078.1
1
4
sJ10055.1
27194
34
=
×××⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅×
= −−−
−
r
27 •• The equilibrium separation between the atoms of a LiH molecule is
0.16 nm. Determine the energy separation between the l = 3 and l = 2 rotational
levels of this diatomic molecule.
Picture the Problem We can use the expression for the rotational energy levels
of the diatomic molecule to express the energy separation ΔE between the
and rotational levels and model the moment of inertia of the LiH
3=l
2=l molecule
as two point objects separated by a distance r0.
The energy separation between the
and rotational levels of
this diatomic molecule is given by:
3=l 2=l
23 == −=Δ ll EEE
Express the rotational energy levels
El = 3 and El = 2 in terms of E0r:
( ) rr EEE 003 12133 =+==l
and
( ) rr EEE 002 6122 =+==l
Substitute for El = 3 and El = 2 to
obtain:
rrr EEEE 000 6612 =−=Δ
The characteristic rotational energy
of a molecule is given by
I
E r
2
2
0
h
= .
Hence:
II
E
22
3
2
6
hh
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=Δ

3. Chapter 3726
The moment of inertia of the
molecule is given by:
2
0rI μ=
where μ is the reduced mass of the
molecule.
Substituting for I yields:
( )
2
0HLi
HLi
2
2
0
HLi
HLi
2
2
0
2
3
33
rmm
mm
r
mm
mmr
E
+
=
+
==Δ
h
hh
μ
Substitute numerical values and evaluate ΔE:
( ) ( )
( )( )( ) ( )( )
meV6.5
kg/u10661.1J/eV10602.1nm16.0u1u94.6
u1u94.6sJ10055.13
27192
234
=
××
+⋅×
=Δ −−
−
E
31 •• The equilibrium separation between the atoms of a CO molecule is
0.113 nm. For a molecule, such as CO, that has a permanent electric dipole
moment, radiative transitions obeying the selection rule Δl = ±1 between two
rotational energy levels of the same vibrational level are allowed. (That is, the
selection rule Δν = ±1 does not hold.) (a) Find the moment of inertia of CO and
calculate the characteristic rotational energy E0r (in eV). (b) Make an energy-level
diagram for the rotational levels from l = 0 to l = 5 for some vibrational level.
Label the energies in electron volts, starting with E = 0 for 0=l . Indicate on your
diagram the transitions that obey Δl = −1, and calculate the energies of the
photons emitted. (c) Find the wavelength of the photons emitted during each
transition in (b). In what region of the electromagnetic spectrum are these
photons?
Picture the Problem We can find the reduced mass of CO and the moment of
inertia of a CO molecule from their definitions. The energy level diagram for the
rotational levels from to0=l 5=l can be found using Finally,
we can find the wavelength of the photons emitted for each transition using
.2 01, rEE lll =Δ −
rE
hc
E
hc
01,
1,
2 Δ
=
Δ
=
−
−
lll
llλ .
(a) Express the moment of inertia
of CO:
2
0rI μ=
where μ is the reduced mass of the CO
molecule.

4. Molecules 27
The reduced mass of the CO
molecule is given by: OC
OC
mm
mm
+
=μ
Substituting for μ gives:
OC
2
0OC
mm
rmm
I
+
=
Substitute numerical values (r0 = 0.113 nm) and evaluate I:
( )( ) ( )( )
246
246227
mkg1045.1
mkg10454.1nm113.0kg/u10661.1
u16u12
u16u12
⋅×=
⋅×=×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
=
−
−−
I
The characteristic rotational energy
E0r is given by: I
E r
2
2
0
h
=
Substitute numerical values and evaluate E0r:
( ) ( )
( ) meV0.239meV0.2385
mkg10454.12
J/eV10602.1seV1058.6
246
19216
0 ==
⋅×
×⋅×
= −
−−
rE
(b) The energy level diagram is
shown to the right. Note that ΔEl,l−1,
the energy difference between
adjacent levels for Δl = −1, is
.2 01, rEE lll =Δ −
meV76.4,4 == El
meV86.2,3 == El
meV43.1,2 == El
meV476.0,1 == El
0,0 == El
meV14.7,5 == El

5. Chapter 3728
The energies of the photons emitted
in each transition are given by
. Hence:1, −Δ llE
meV38.2
meV76.4meV14.74,5
=
−=ΔE
meV.901
meV86.2meV76.43,4
=
−=ΔE
meV.431
meV43.1meV86.22,3
=
−=ΔE
meV.9540
meV476.0meV43.11,2
=
−=ΔE
meV476.0
0meV476.00,1
=
−=ΔE
(c) Express the energy difference
between energy levels in
terms of the frequency of the emitted
radiation:
1. −Δ llE
1,1, −− =Δ llll hfE
Because :1,1, −−= llll λfc
rE
hc
E
hc
01,
1,
2 Δ
=
Δ
=
−
−
lll
llλ
Substitute numerical values to obtain:
( )( )
( ) ll
ll
m2600
meV2385.02
m/s10998.2seV10136.4 815
1,
μ
λ =
×⋅×
=
−
−
For l = 1:
m2600
1
m2600
0,1 μ
μ
λ ==
For l = 2:
m1300
2
m2600
1,2 μ
μ
λ ==
For l = 3:
m867
3
m2600
2,3 μ
μ
λ ==
For l = 4:
m650
4
m2600
3,4 μ
μ
λ ==

6. Molecules 29
For l = 5:
m520
5
m2600
4,5 μ
μ
λ ==
These wavelengths fall in the microwave region of the electromagnetic spectrum.