Morse Potential: Dissociation Energy and Anharmonicity
Vibrational spectroscopy is high-scoring in CSIR-NET, GATE, and IIT-JAM. While harmonic oscillator models treat chemical bonds as ideal springs, real bonds dissociate when stretched. The Morse potential accurately models dissociation and vibrational anharmonicity. Mastering Morse potential relations, De versus D0, and Birge–Sponer extrapolation allows students to solve exam numericals quickly and reliably.
Why the Harmonic Model Fails for Real Bonds
The harmonic oscillator approximates potential energy as a parabola: V(r) = (1/2)k(r - re)2. Near equilibrium bond distance re it works, but fails for real bonds. Nuclear repulsion at short distances (r < re) rises steeper than a parabola, while at large distances (r > re), bonds weaken and dissociate into neutral atoms. Real potential wells widen as energy rises, causing vibrational levels to converge. This deviation is called anharmonicity.
The Morse Potential Function
Philip M. Morse proposed an empirical potential reproducing real diatomic curves:
Morse Potential Function:
V(r) = De [1 - e-a(r - re)]2
V(r) = Potential energy; re = Equilibrium bond length
De = Equilibrium well depth; a = √(ke / (2De))
At r = re, V(re) = 0. When r → ∞, V(r) → De. If zero energy is set at dissociation, V(r) = De [1 - e-a(r - re)]2 - De.
Vibrational Term Values and Dissociation Energy
Solving the Schrödinger equation for a Morse oscillator gives quantized vibrational term values G(v) in cm-1:
Energy Level Formulas:
G(v) = ωe (v + 1/2) - ωexe (v + 1/2)2
De = ωe2 / (4ωexe)
D0 = De - G(0) = De - [(1/2)ωe - (1/4)ωexe]
v = Vibrational level; ωe = Harmonic wavenumber (cm-1)
xe = Anharmonicity constant; ωexe = Parameter (cm-1)
De = Well depth; D0 = Ground dissociation energy
To convert cm-1 to kJ/mol, multiply by NAhc = 0.01196266 kJ/(mol·cm-1).
Transitions and Birge–Sponer Extrapolation
Anharmonicity relaxes harmonic selection rules, allowing overtones:
- Fundamental (0 → 1): ν̃1 = ωe - 2ωexe
- First overtone (0 → 2): ν̃2 = 2ωe - 6ωexe
Level spacing is ΔGv+1/2 = ωe - 2ωexe(v + 1). Plotting ΔG against (v + 1) gives the linear Birge–Sponer plot. At dissociation (ΔG = 0), vmax + 1 = ωe / (2ωexe). The plot area gives De = ωe2 / (4ωexe).
Worked Numerical Examples
Question: A molecule has ωe = 3000.0 cm-1 and ωexe = 50.0 cm-1. Calculate: (a) De in cm-1 and kJ/mol, (b) zero-point energy G(0) in cm-1, and (c) D0 in cm-1 and kJ/mol. (1 cm-1 = 0.01196266 kJ/mol).
Step-by-step solution:
Step 1: Calculate De.
De = ωe2 / (4ωexe) = (3000.0)2 / [4 × 50.0] = 9,000,000 / 200.0 = 45000.0 cm-1
Step 2: Convert De to kJ/mol.
De = 45000.0 cm-1 × 0.01196266 kJ/(mol·cm-1) = 538.32 kJ/mol. Second route: (45000.0 / 8065.54) × 96.485 = 538.32 kJ/mol.
Step 3: Calculate zero-point energy G(0).
G(0) = (1/2)ωe - (1/4)ωexe = (1/2)(3000.0) - (1/4)(50.0) = 1500.0 - 12.5 = 1487.5 cm-1
Step 4: Calculate D0.
D0 = De - G(0) = 45000.0 - 1487.5 = 43512.5 cm-1 = 520.53 kJ/mol
Answer: De = 45000.0 cm-1 (538.32 kJ/mol), G(0) = 1487.5 cm-1, and D0 = 43512.5 cm-1 (520.53 kJ/mol).
Question: An infrared spectrum shows fundamental band ν̃1 = 2886.0 cm-1 and first overtone ν̃2 = 5668.0 cm-1. Find: (a) ωe and ωexe, (b) De in cm-1, and (c) maximum bound level vmax.
Step-by-step solution:
Step 1: Set up transition equations.
ν̃1 = ωe - 2ωexe = 2886.0 cm-1 (1)
ν̃2 = 2ωe - 6ωexe = 5668.0 cm-1 (2)
Step 2: Solve for ωexe and ωe.
2 × (1) - (2): 2ωexe = 2(2886.0) - 5668.0 = 5772.0 - 5668.0 = 104.0 cm-1
ωexe = 104.0 / 2 = 52.0 cm-1
From (1): ωe = 2886.0 + 2(52.0) = 2990.0 cm-1
Double-check via (2): 2(2990.0) - 6(52.0) = 5980.0 - 312.0 = 5668.0 cm-1. Confirmed.
Step 3: Calculate De.
De = ωe2 / (4ωexe) = (2990.0)2 / (4 × 52.0) = 8,940,100 / 208.0 = 42981.25 cm-1
Step 4: Find vmax.
At dissociation, dG/dv = 0 &implies; vmax + 1/2 = ωe / (2ωexe) = 2990.0 / 104.0 = 28.75
vmax = 28.75 - 0.50 = 28.25 &implies; integer vmax = 28.
Answer: ωe = 2990.0 cm-1, ωexe = 52.0 cm-1, De = 42981.25 cm-1, and vmax = 28.
Common Mistakes to Avoid:
- Equating De with D0: De is from the well bottom; D0 is from v = 0. Always use D0 = De - G(0).
- Denominator error: Use De = ωe2 / (4ωexe), dividing by 4, not 2.
- Confusing xe with ωexe: xe is dimensionless; ωexe has units of cm-1.
- Overtone formula error: For 0 → 2, ν̃2 = 2ωe - 6ωexe, not 2ωe - 4ωexe.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CSIR-NET, GATE, IIT-JAM, CUET-PG
- Topic: Morse Potential, Anharmonicity and Bond Dissociation Energy
- Difficulty: Advanced post-graduate entrance level
| Exam | Question Format | Key Focus Areas |
|---|---|---|
| CSIR-NET | MCQs (Parts B and C) | Birge–Sponer plots, zero-point energy, and De calculations |
| GATE | NAT and MCQs | Exact numerical calculation of De, D0, and force constant k |
| IIT-JAM | MCQs and NAT | Anharmonic curves, overtone positions, and De = ωe2 / (4ωexe) |
| CUET-PG | MCQs | Selection rules, definition of xe, and De versus D0 |
Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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