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For a diatomic molecule, let $V(r)$ be the electronic potential energy at bond length $r$, with the atomic nuclei treated as fixed while finding the electronic energy. Choose zero at infinitely separated chemical atoms in the specified electronic quantum states. If the minimum occurs at $r_e$, the bond potential-well depth isThe bottom of this curve describes an equilibrium bond length, not an attainable motionless vibrational quantum state. Changing the zero of potential energy shifts both endpoints equally and leaves $D_e$ unchanged. A quantum harmonic oscillator approximates the curve near its minimum but rises without limit, so that local approximation alone cannot supply a dissociation asymptote. A Morse potential supplies both a minimum and a finite separated-fragment limit.^[ref]

A diatomic molecule contains two chemical atoms; hydrogen chloride and carbon monoxide are examples. Their fixed-separation moment of inertia is $I=\mu r^2$ in the linear rigid rotor model.

The ground-state dissociation energy starts at the lowest vibrational quantum state, with the lowest allowed rotational quantum state and the fragment states explicitly specified. For a diatomic molecule whose fragments are ground-state chemical atoms, omitting any additional rotational correction,where $E_{\mathrm{ZP}}$ is the vibrational zero-point energy measured upward from the well minimum. The initial molecule already has this energy, so less additional energy is needed to reach the same final limit.^[ref]

For fragments that are themselves molecules, include their zero-point energies too. Define $D_e$ as the electronic energy difference between separately relaxed products and reactant; then, with consistent rotational conventions,Thus “subtract the parent's zero-point energy” is sufficient for atomic products but generally incomplete for molecular products. An excited final quantum state raises its channel threshold by its excitation energy; that threshold does not become the ground-product $D_0$ just because a detector selects it.

A linear molecule has collinear atomic nuclei in its idealized molecular geometry. Both a diatomic molecule and carbon dioxide can supply linear rigid rotor examples, although their electric-dipole selection rules need not give an observable permanent-electric dipole spectrum.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

Take an invented diatomic molecule with a Morse potential, setting the separated-atom energy to zero:Here $a$ is an inverse length, $r_e$ the equilibrium bond length, and ${\omega }_e=a\sqrt{2D_e/\mu }$, with reduced mass $\mu$. These stipulated parameters describe a simple model, not a measured chemical substance. For the one-dimensional Morse oscillator, energy eigenvalues above the minimum areConsequently,The harmonic estimate $A/2=0.1000\mathrm{eV}$ is slightly larger than this model's exact zero-point energy. The physical bound-level restriction matters: the quadratic expression is not an infinite ladder of vibrational quantum states.

For a separate thermal illustration using this same stipulated $D_0$, assume an ideal gas, fully active molecular rotation of the parent, negligible thermal molecular vibration, and negligible electronic excitation for all species. Each monatomic product contributes $\delta h=5RT/2$; the diatomic molecule contributes $7RT/2$. Thus the net increment is $3RT/2$, with molar gas constant $R$. At $T=300\mathrm{K}$,This last result follows from the stated thermal model, not a universal correction for every chemical bond. Frozen rotations, accessible excited states, different fragment composition, or a condensed phase change the thermal increments.