SME
Physical Chemistry II: Quantum & Spectroscopy
Spectroscopy
Use this for “The microwave lines of ¹²C¹⁶O (12.000 and 15.995 u) are 3.845 cm⁻¹ apart. Find B, the moment of inertia and the bond length.”
Diagram
Lines at ν̃ = 2B(J + 1), 2B = 2 × 10.58 = 21.2 cm⁻¹ apart.
Heights
follow
the
lower
level’s
share,
(2J
+
1)
e
−1.4388BJ(J + 1) ÷ T
at
298.15
K:
the
strongest
line
starts
at
J
=
3.
Lit: J = 0 → 1 at 2 × 10.58 × 1 = 21.2 cm⁻¹.
This is an example. Type your own values to start a new problem.
m₁
Mass of atom 1
example · u
m₂
Mass of atom 2
example · u
μ
Reduced mass
calculated · kg
r
Bond length
example · pm
I
Moment of inertia
calculated · kg·m²
B
Rotational constant
calculated · cm⁻¹
2B
Line spacing
calculated · cm⁻¹
J
Lower level
example
ν̃
Line
calculated · cm⁻¹
Formulas
μ
=
m₁
×
m₂
m₁
+
m₂
×
1.661 ×
10
-27
1.627 ×
10
-27
=
1.008
×
34.969
1.008 + 34.969
×
1.661 ×
10
-27
I
=
μ
×
(
r
×
10
-12
)
2
2.646 ×
10
-47
=
1.627 ×
10
-27
×
(127.5 ×
10
-12
)
2
B
=
2.799 ×
10
-46
I
10.58
=
2.799 ×
10
-46
2.646 ×
10
-47
2
B
=
2
×
B
21.16 = 2 × 10.58
ν̃
=
2
×
B
×
(
J
+
1)
21.16 = 2 × 10.58 × (0 + 1)
Assumptions
•
A rigid rotor: the bond length stays fixed as the molecule turns, and its levels are BJ(J + 1).
•
Absorption takes J to J + 1 only, so the line from J sits at 2B(J + 1) and neighboring lines are 2B apart.
•
Only polar molecules show these lines: H₂ and N₂ have none.
•
h ÷ (8π²c) = 2.799 × 10⁻⁴⁶ kg·m²·cm⁻¹ with c in cm/s; 1 u = 1.661 × 10⁻²⁷ kg.
Step-by-step
WE KNOW
m₁ = 1.008 u, m₂ = 34.969 u, r = 127.5 pm, J = 0
FIND
reduced mass (μ), moment of inertia (I), rotational constant (B), line spacing (2B), line (ν̃)
Step 1 · Find reduced mass (μ)
Use μ = m₁ × m₂ ÷ (m₁ + m₂) × 1.661 × 10⁻²⁷
Product
over
sum
of
the
two
masses
gives
μ
in
u;
×
1.661 ×
10
-27
turns
it
into
kilograms.
μ
=
1.008
×
34.969
1.008 + 34.969
×
1.661 ×
10
-27
μ
=
1.008
×
34.969
35.977
×
(1.661
×
10
-27
)
μ
=
35.25
35.977
×
(1.661
×
10
-27
)
μ
=
0.9798
×
(1.661
×
10
-27
)
μ
=
1.6274 ×
10
-27
kg
Step 2 · Find moment of inertia (I)
Use I = μ × (r × 10⁻¹²)²
The reduced mass times the bond length squared, with picometers turned into meters.
I
=
1.6274 ×
10
-27
×
(127.5 ×
10
-12
)
2
I
=
(1.6274
×
10
-27
)
×
(1.6256
×
10
-20
)
I
=
2.646 ×
10
-47
kg·
m
2
Step 3 · Find rotational constant (B)
Use B = 2.799 × 10⁻⁴⁶ ÷ I
h
8π²c
,
with
c
in
cm/s,
divided
by
the
moment
of
inertia
gives
B
in
cm⁻¹.
B
=
2.799 ×
10
-46
2.646 ×
10
-47
B
=
10.58
cm⁻¹
Step 4 · Find line spacing (2B)
Use 2B = 2 × B
Neighboring
lines
differ
by
2
B
.
2
B
=
2
×
10.58
2
B
=
21.16
cm⁻¹
Step 5 · Find line (ν̃)
Use ν̃ = 2 × B × (J + 1)
The
J
→
J
+
1
line
is
the
gap
B
(
J
+
1)(
J
+
2)
−
BJ(
J
+
1)
=
2
B
(
J
+
1).
ν̃
=
2
×
10.58
×
(0
+
1)
ν̃
=
21.16
cm⁻¹
→
The
J
=
0
→
1
line;
its
neighbors
sit
21.16
cm⁻¹
(
2
B
)
either
side.
Check
Put the numbers back in. Both sides should match.
1.6274 ×
10
-27
=
1.008
×
34.969
1.008 + 34.969
×
1.661 ×
10
-27
✓
2.646 ×
10
-47
=
1.6274 ×
10
-27
×
(127.5 ×
10
-12
)
2
✓
10.58
=
2.799 ×
10
-46
2.646 ×
10
-47
✓
21.16 = 2 × 10.58 ✓
21.16 = 2 × 10.58 × (0 + 1) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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