The adsorption of infrared radiation by diatomic molecules increases the vibrational energy fo molecules and gives information about the force constant for the "spring" of the molecule.;
The molecular motion that has the next larger energy level spacing after the rotation fo molecules is the vibration of the atoms of the molecules with respect to each other.
The allowed energies for a single particle of mass m vibrating against a spring with force constant k, that is, experiencing a potential energy U = ½ kx^{2}, where x is the displacement from equilibrium.
ε_{vib} = (v + ½ ) h/2∏ √k/m = (v + ½ )hv_{vib} v = 0, 1, 2 ...
Where v vib, the frequency fo the classical oscillator, represents the term [1/ (2∏)]√k/m. this quantum mechanical result indicates a pattern of energy levels with a constant spacing [h/ (2∏)]√k/m. it is this result that was used for the calculation of the average vibrational energy per degree of freedom.
Classical analysis: now let us investigate the details of the vibrational motion of the atoms of a molecule. The simplest case of a diatomic molecule is our initial concern.
The harmonic oscillator treatment results when we assume that the potential energy of the bond can be described by the function
U = ½ k (r - re)^{2}, where r is the distance between the nuclei of the bonded atoms and re is the value of r at the equilibrium internuclear distance. The constant k enters as a proportionality constant, the force constant. It is a measure of the bond.
The classical solution for a vibrating two particle diatomic molecule system can be obtained from Newton's f = ma relation. If the bond is distorted from its equilibrium length re to a new length r, the restoring forces on each atom are - k (r - r_{e}). These forces can be equated to the ma terms for each atom where r_{1} and r_{2} are the postions of atoms 1 and 2, respectively, relative to the center of mass of the molecule. These forces can be equated to the ma terms for each atom as:
m_{1} × d^{2}r_{1}/dt^{2} = - k (r - r_{e}) and m^{2} × d^{2}r_{2}/dt^{2} = - k (r -r_{e})
Where, r_{1} and r_{2} are the positions of atoms 1 and 2 respectively, relative to the center of mass of the molecule. The relation that keeps the center of mass fixed is r_{1}m_{1} = r_{2}m_{2}, and with r = r_{1}+ r_{2} this gives:
r_{1} = m_{2}/(m_{1} + m_{2}) × r and r_{2} = m_{1}/(m_{1} + m_{2}) × r
Substitution in either of the ƒ = ma equation gives:
m_{1}m_{2}/(m_{1} + m_{2}) × d^{2}r/dt^{2} = - k (r - r_{e})
Since r, is a constant, this can also be written:
m_{1}m_{2}/(m_{1} + m_{2}) × d^{2} (r- r_{e})/dt2 = - k (r- r_{e})
The term r - r_{e} is the displacement of the bond length from its equilibrium position. If the symbol xis introduced as x = r - r_{e} and the reduced mass of μ is inserted for the mass term becomes:
μ × d^{2}x/dt^{2} = - kx
This expression is identical to the corresponding equation for a single particle, except for the replacement of the mass m by the reduced mass. A derivation like the classical vibrational frequency for a two particle system would give the result,
V_{vib} = 1/2∏ √k/μ