CHEM 301: Homework assignment #12
Solutions
1. Let’s practice converting between wavelengths, frequencies, and wavenumbers. (10%)
• Express a wavelength of 442 nm as a frequency and as a wavenumber.
What is the photon energy for this wavelength?
The frequency ν of a photon is related to its wavelength λ in vacuum
as follows:
2.998 · 108 m/s
c
=
= 6.78 · 1014 s,
λ
442 · 10−9 m
where c is the speed of light in vacuum. The energy of the photon is:
ν=
E = hν = 6.626 · 10−34 J · s · 6.78 · 1014 s = 4.49 · 10−19 J = 2.79 eV.
• What are the wavenumber and the wavelength of the radiation used
by an FM radio transmitter broadcasting at 88.0 MHz?
The wavelength is related to the frequency as follows:
λ=
c
2.998 · 108 m/s
=
= 3.41 m.
ν
88.0 · 106 Hz
The wavenumber is
νe =
ν
88.0 · 109 Hz
=
= 2.94 m−1 = 2.94 · 10−3 cm−1 .
c
2.998 · 108 m/s
2. Suppose you were seeking the presence of (planar) SO3 molecules in the
microwave spectra of interstellar gas clouds.
32 16
• Calculate
the rotational constants
A and B for S O3 . Assume
32
16
m S = 31.972 a.m.u., m O = 15.995 a.m.u., and an SO bond
length of 143 pm. (15%)
SO3 is a trigonal planar molecule. Ik is perpendicular to the plane of
the molecule and along the vertical axis passes through the S atom.
I⊥ is shown in the figure. The bond angles are 120◦ .
Ik = 2m0 R2 (1 − cos θ) ,
1
where θ = 120◦ ,
2
Ik = 215.995·1.661·10−27 kg· 143 · 10−12 m ·(1 − cos 120◦ ) = 1.629·10−45 kg · m2 .
The second component of the moment of inertia
I⊥ = m0 R2 (1 − cos θ) = 12 Ik = 12 ·1.629·10−45 kg · m2 = 8.149·10−46 kg · m2 .
The rotational constants are
A=
1.0546 · 10−34 J · s
h̄
=
= 5.152 · 109 Hz
4πIk
4 · 3.1416 · 1.629 · 10−45 kg · m2
and
B=
h̄
1.0546 · 10−34 J · s
= 1.030 · 1010 Hz.
=
4πI⊥
4 · 3.1416 · 8.149 · 10−46 kg · m2
• Could you use microwave spectroscopy to distinguish the relative
abundances of 32 S16 O3 and 33 S16 O3 ? (10%)
The S atom is located at the center of mass of the molecule and does
not affect the moments of inertia, and hense the rotational constants.
Consequently, microwave spectroscopy cannot be used to distinguish
between 32 S16 O3 and 33 S16 O3 .
3. How many rotational states have an energy equal to hBJ (J + 1) with
J = 8 for (a) a methane molecule and (b) a chloromethane molecule
described by the quantum numbers J, MJ , and K? (10%)
A methane molecule is a spherical rotor; hence Ik = I⊥ and A = B and
the rotational energy levels only depend on J, not on K or MJ :
EJ = hBJ (J + 1) .
There are 2J + 1 possible values of K and 2J + 1 possible values of MJ
for each value of J. Thus, for J = 8 there are 17 possible K values and 17
possible MJ values. All levels with the same J and any K and MJ values
have the same energy, so the level is 172 = 289-fold degenerate.
If methane is replaced by chloromethane (CH3 Cl) then the molecule is no
longer a spherical rotor (it is a symmetric rotor). Thus, the rotational
energy levels depend both on J and K:
EJ,K = hBJ (J + 1) + h (A − B) K 2 .
The energy is equal to hBJ (J + 1) only for K = 0, and for J = 8 are 17
possible MJ values, so this energy level is 17-fold degenerate.
4. Microwave spectroscopy can be used to determine the bond length of diatomic molecules.
2
• The rotational constant of
I-Cl bond length.
The rotational constant is
I Cl is 0.1142 cm−1 . Calculate the
127 35
B=
h̄
,
4πI
I=
h̄
.
4πB
thus the moment of inertia
On the other hand, for a diatomic molecule:
I = µR2 ,
where
µ=
126.094 · 34.9688
mI mCl
= 27.4146 a.m.u.
=
mI + mCl
126.094 + 34.9688
is the effective mass. Comparing the two expressions for the moment
of inertia, we find
h̄
µR2 =
,
4πB
and we can express the interatomic distance (bond length) as
s
h̄
R=
.
4πBµ
To convert the rotational constant from cm−1 to Hz, multiply it by
c (in cm/s):
B = 0.1142 cm−1 · 2.998 · 1010 cm/s = 3.424 · 109 Hz.
Plugging this value and the value of µ that we calculated (in kg) into
the expression for R, we find:
s
1.0546 · 10−34 J · s
R=
= 2.32 Å.
4 · 3.1416 · 3.424 · 109 Hz · 27.4146 · 1.661 · 10−27 kg
• The microwave spectrum of 1 H127 I consists of a series of lines separated by 384 GHz. Compute its bond length. What would the
separation of the lines be in the spectrum of 2 H127 I?
The selection rule for rotational spectra is ∆J = ±1. The transition
energy for levels J and J + 1 is
∆EJ→J+1 = hB (J + 1) (J + 2) − hBJ (J + 1) =
= hB J 2 + 2J + J + 2 − J 2 − J = 2hB (J + 1) .
3
Lines in the spectrum will appear at energies 2hB, 4hB, etc., with
energy spacing 2hB between each pair of lines or frequency spacing
2B between each pair of lines. Thus,
2B = 384 · 109 Hz,
and
B=
384 · 109 Hz
= 192 · 109 Hz.
2
Because
B=
h̄
h̄
,
=
4πI
4πµR2
we can express R as
s
R=
h̄
,
4πµB
where
µ=
mI mH
126.094 · 1.008
·1.66054·10−27 kg = 1.660·10−27 kg.
=
mI + mH
126.094 + 1.008
Plugging in the values of B and µ, we obtain
R=
1.0546 · 10−34 J · s
= 1.62 Å.
4 · 3.1416 · 1.660 · 10−27 kg · 192 · 109 Hz
Replacing 1 H by 2 H changes the effective mass,
µ=
mI mH
126.094 · 2.014
=
·1.66054·10−27 kg = 3.292·10−27 kg.
mI + mH
126.094 + 2.014
However, it almost does not effect the bond length, which we have
calculated to be 1.62 Å. Thus, the rotational constant for 2 H127 I can
be calculated as
B=
h̄
1.05457 · 10−34 J · s
= 96.9 GHz
=
4πµR2
4 · 3.1416 · 3.292 · 10−27 kg · 1.62 · 10−10 Å
and the separation between spectral lines of 2 H127 I is
2B = 194 GHz.
Assume m
a.m.u., m
H = 1.008 a.m.u., m 2 H = 2.014 a.m.u., m
127
I = 126.905 a.m.u. (20%)
1
35
Cl = 34.969
5. Suppose the C=O group in a peptide bond can be regarded as isolated
from the rest of the molecule. Given the force constant of the bond in
a carbonyl group is 908 N/m, calculate the vibrational frequency of (a)
12
C=16 O and (b) 13 C=16 O. (15%)
4
The vibrational frequency is
1
ν=
2π
where for
µ=
12
s
k
,
µ
C=16 O
mC mO
12.0000 · 15.9949
=
· 1.661 · 10−27 kg = 1.1385 · 10−26 kg.
mC + mO
12.0000 + 15.9949
Plugging in the values of k and µ into the expression for ν, we find ν =
4.49 · 1013 Hz.
Similarly, for 13 C=16 O the effective mass is calculated to be µ = ·10−26 kg
and ν = 4.39·1013 Hz. Thus, vibrational spectroscopy is able to distinguish
between different isotopes in a molecular group.
6. The wavenumber of the fundamental vibrational transition of Cl2 is 565
cm−1 . Calculate the force constant of the bond. (15%)
The wavenumber is defined in spectroscopy as
s
1
k
ν
.
νe = =
c
2πc µ
For a homonuclear molecule like Cl2 , the effective mass is simply one half
of the atomic mass:
1
µ = m 3 5Cl .
2
Expressing the force constant, we obtain
2
2
k = (2πce
ν ) µ = 12 (2πce
ν ) m 3 5Cl =
= 12 2 · 3.1416 · 2.998 · 108 m/s · 5.65 · 104 m−1 · 34.9688 · 1.661 · 10−27 kg = 329 N/m.
5