SME
Physical Chemistry II: Quantum & Spectroscopy · Problem type
A bond’s vibration and its zero-point energy
Use this for “The force constant of H–³⁵Cl is 480 N/m. Find its vibrational frequency, its wavenumber and its zero-point energy.”
Diagram
Eᵥ = (v + ½)ħω: the levels are evenly spaced by ħω, starting from ½ħω, never 0. ψ₁ has 1 node; each ψ leaks past its turning points.
Absorbed:
λ
=
hc
ΔE
.
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
k
Force constant
example · N/m
ν
Frequency
calculated · Hz
ν̃
Wavenumber
calculated · cm⁻¹
E₀
Zero-point energy
calculated · J
E₀,m
Zero-point energy per mole
worked out (not typed) · kJ/mol
Formulas
μ
=
m₁
×
m₂
m₁
+
m₂
×
1.661 ×
10
-27
1.63 ×
10
-27
=
1.008
×
34.97
1.008 + 34.97
×
1.661 ×
10
-27
ν
=
√
k
÷
μ
2 × π
8.64 ×
10
13
=
√
480 ÷ (1.63 ×
10
-27
)
2 × π
ν̃
=
ν
2.998 ×
10
10
2,883
=
8.64 ×
10
13
2.998 ×
10
10
E₀
=
0.5
×
6.626 ×
10
-34
×
ν
2.86 ×
10
-20
=
0.5
×
6.626 ×
10
-34
×
(8.64
×
10
13
)
E₀
,m
=
E₀
×
6.022 ×
10
23
1,000
17.2
=
2.86 ×
10
-20
×
6.022 ×
10
23
1,000
Assumptions
•
The bond is a harmonic spring of force constant k between the two atoms, which move as one reduced mass μ.
•
The levels are (v + ½)hν, so the lowest, v = 0, keeps E₀ = ½hν; infrared light lifts it from v = 0 to 1.
•
Stronger bonds and lighter atoms vibrate faster and absorb at a higher wavenumber.
•
u = 1.661 × 10⁻²⁷ kg, h = 6.626 × 10⁻³⁴ J·s, c = 2.998 × 10¹⁰ cm/s, NA = 6.022 × 10²³ mol⁻¹.
Step-by-step
WE KNOW
m₁ = 1.008 u, m₂ = 34.97 u, k = 480 N/m
FIND
reduced mass (μ), frequency (ν), wavenumber (ν̃), zero-point energy (E₀), zero-point energy per mole (E₀,m)
Step 1 · Find reduced mass (μ)
Use μ = m₁ × m₂ ÷ (m₁ + m₂) × 1.661 × 10⁻²⁷
Product
over
sum
gives
μ
in
u;
×
1.661 ×
10
-27
turns
it
into
kilograms.
μ
=
1.008
×
34.97
1.008 + 34.97
×
1.661 ×
10
-27
μ
=
1.008
×
34.97
35.978
×
(1.661
×
10
-27
)
μ
=
35.25
35.978
×
(1.661
×
10
-27
)
μ
=
0.9798
×
(1.661
×
10
-27
)
μ
=
1.6274 ×
10
-27
kg
Step 2 · Find frequency (ν)
Use ν = √(k ÷ μ) ÷ (2 × π)
A
spring
swings
at
√(
k
μ
)
radians
per
second;
÷
2π
gives
cycles
per
second.
ν
=
√
480 ÷ (1.6274 ×
10
-27
)
2 × π
ν
=
√
2.949 ×
10
29
2π
ν
=
5.43 ×
10
14
2π
ν
=
8.644 ×
10
13
Hz
Step 3 · Find wavenumber (ν̃)
Use ν̃ = ν ÷ (2.998 × 10¹⁰)
Divide by c in cm/s to count the waves in each centimeter.
ν̃
=
8.644 ×
10
13
2.998 ×
10
10
ν̃
=
2,883
cm⁻¹
Step 4 · Find zero-point energy (E₀)
Use E₀ = 0.5 × 6.626 × 10⁻³⁴ × ν
The
lowest
level
sits
half
a
step
up:
E₀
=
1
2
hν,
never
zero.
E₀
=
0.5
×
6.626 ×
10
-34
×
(8.644
×
10
13
)
E₀
=
(3.313
×
10
-34
)
×
(8.644
×
10
13
)
E₀
=
2.86 ×
10
-20
J
Step 5 · Find zero-point energy per mole (E₀,m)
Use E₀,m = E₀ × 6.022 × 10²³ ÷ 1,000
A mole of bonds is NA of them; ÷ 1000 turns J into kJ.
E₀
,m
=
2.86 ×
10
-20
×
6.022 ×
10
23
1,000
E₀
,m
=
17,222.92
1,000
E₀
,m
=
17.2
kJ/mol
Check
Put the numbers back in. Both sides should match.
1.6274 ×
10
-27
=
1.008
×
34.97
1.008 + 34.97
×
1.661 ×
10
-27
✓
8.644 ×
10
13
=
√
480 ÷ (1.6274 ×
10
-27
)
2 × π
✓
2,883
=
8.644 ×
10
13
2.998 ×
10
10
✓
2.86 ×
10
-20
=
0.5
×
6.626 ×
10
-34
×
(8.644
×
10
13
)
✓
17.2
=
2.86 ×
10
-20
×
6.022 ×
10
23
1,000
✓
Topic check
See what you know: 5 questions on this topic
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