Morse Potential: Anharmonic Vibration, Overtones and Dissociation Energy

By Aniket Bhardwaj · 12 November 2026 · Physical Chemistry

The harmonic oscillator is a good first model for the vibration of a diatomic molecule, but it has two serious faults: it never lets the bond break, and it predicts equally spaced vibrational levels. Real molecules dissociate, and their levels crowd closer together as the energy rises. The Morse potential is the simplest realistic fix. It appears in physical chemistry and spectroscopy sections of GATE, CSIR-NET and IIT-JAM, usually as a short numerical on dissociation energy or overtone positions.

The Morse potential energy curve

V(r) = D_e [ 1 − e^{−a(r − r_e)} ]²

r_e = equilibrium bond length, where V = 0 (the bottom of the well).
D_e = depth of the well, measured from the minimum to the dissociation limit.
a = a width parameter (unit: reciprocal length). A larger a gives a narrower, stiffer well.

When r is very large the exponential goes to zero and V tends to D_e: the atoms are separated. When r is much smaller than r_e the exponential becomes large and V rises steeply, which represents repulsion. Near r_e the curve looks like a parabola, with force constant k_e = 2D_ea², so a = √(k_e / 2D_e).

Vibrational energy levels

Solving the Schrödinger equation for this potential gives an anharmonic ladder. In wavenumber units (cm⁻¹) the term values measured from the potential minimum are:

G(v) = ω_e(v + ½) − ω_ex_e(v + ½)², v = 0, 1, 2, …

ω_e = harmonic wavenumber (what the vibration frequency would be if the well were a perfect parabola).
ω_ex_e = anharmonicity constant. It is always positive for a bond, so it lowers the levels more and more as v increases.

The transition from v = 0 to v:

ν(0 → v) = G(v) − G(0) = vω_e − v(v + 1)ω_ex_e

This gives the fundamental (v = 1): ω_e − 2ω_ex_e; the first overtone (v = 2): 2ω_e − 6ω_ex_e; and the second overtone (v = 3): 3ω_e − 12ω_ex_e. In a pure harmonic oscillator, overtones would be exactly 2, 3, ... times the fundamental and would be forbidden.

Dissociation energies D_e and D₀

Two different dissociation energies are used, and mixing them is the commonest source of lost marks.

Quantity	Measured from	Formula (cm⁻¹)
D_e	Bottom of the well (no vibration)	D_e = ω_e² / (4ω_ex_e)
D₀	The v = 0 level (what chemists can measure)	D₀ = D_e − G(0) = D_e − ω_e/2 + ω_ex_e/4

D₀ is smaller than D_e because the molecule always has zero-point energy G(0). The relation D_e = ω_e²/(4ω_ex_e) comes from the Morse model, and it matches the Birge–Sponer idea: the spacing between neighbouring levels falls linearly with v, and the area under that line up to zero spacing is the dissociation energy.

Useful conversions: 1 cm⁻¹ = 0.011963 kJ mol⁻¹, and 1 eV = 8065.5 cm⁻¹.

Worked examples

The numbers below are illustrative values of the same size as those of a light hydrogen halide: ω_e = 2990 cm⁻¹ and ω_ex_e = 52 cm⁻¹. Use the values given in your own question.

Example 1: D_e and D₀.

D_e = 2990² ÷ (4 × 52) = 8 940 100 ÷ 208 = 42 981 cm⁻¹.
In kJ mol⁻¹: 42 981 × 0.011963 = 514.2 kJ mol⁻¹. In eV: 42 981 ÷ 8065.5 = 5.33 eV.
Zero-point energy G(0) = 2990/2 − 52/4 = 1495 − 13 = 1482 cm⁻¹.
D₀ = 42 981 − 1482 = 41 499 cm⁻¹ = 496.4 kJ mol⁻¹.

Example 2: where do the fundamental and overtones appear?

Fundamental: 2990 − 2(52) = 2886 cm⁻¹.
First overtone: 2(2990) − 6(52) = 5980 − 312 = 5668 cm⁻¹.
Second overtone: 3(2990) − 12(52) = 8970 − 624 = 8346 cm⁻¹.

Compare with exact multiples of the fundamental: 2 × 2886 = 5772 and 3 × 2886 = 8658. The real overtones lie lower (5668 and 8346), and the gap grows with v. This is how anharmonicity shows up in a spectrum, and the two lines can be used to solve for ω_e and ω_ex_e.

Example 3: the width parameter a. The reduced mass is 0.98 u. First find the force constant: k_e = 4π²c²ω_e²μ, with c in cm s⁻¹ and ω_e in cm⁻¹.

k_e = 4π² × (2.998 × 10¹⁰)² × (2990)² × (0.98 × 1.6605 × 10⁻²⁷ kg) ≈ 516 N m⁻¹.
D_e in joules per molecule = 514.2 kJ mol⁻¹ ÷ N_A = 8.538 × 10⁻¹⁹ J.
a = √(k_e / 2D_e) = √(516.2 ÷ 1.7076 × 10⁻¹⁸) = √(3.023 × 10²⁰) = 1.739 × 10¹⁰ m⁻¹ = 1.74 Å⁻¹.

Example 4: how many levels does the well hold? The Morse model has bound levels up to v_max ≈ ω_e/(2ω_ex_e) − ½ = 2990/104 − 0.5 = 28.75 − 0.5 = 28.25, so about 28 excited levels sit below the dissociation limit (v = 0 to about 28).

Common mistakes

Quoting D_e when the question asks for the experimentally measurable D₀ (or the reverse).
Using vω_e − v²ω_ex_e for the overtone: the correct term is v(v + 1), not v².
Reading ω_ex_e as ω_e times x_e and then dropping one of them. It is one given number (52 cm⁻¹ here), and it is the term that multiplies (v + ½)².
Mixing cm⁻¹, J and kJ mol⁻¹ without converting.
Saying the Morse potential is exact for all molecules. It is a good model of the shape of a bond, not an exact law.

Exam relevance

Exam	Typical use
GATE Chemistry	Overtone positions, zero-point energy, D₀ from spectral data
CSIR-NET	Anharmonic oscillator terms, selection rule change Δv = ±1, ±2, …
IIT-JAM	Basic idea of anharmonicity and dissociation limit

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