# Vibrational states

The frequencies of a molecule’s vibrational modes, and hence the shifts in wavenumber, are determined by the reduced mass μ and the bond strength k.

As a simple example N_{2} has only one vibrational mode. The reduced mass is given by:

$$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = {{14u \times 14u} \over {14u + 14u}} = 7u = 1.162 \times {10^{ - 26}}{\rm{kg}}$$

The two nitrogen atoms are joined by a strong triple bond, with a force constant k = 22.95 N cm^{-1}. The vibrations are simple harmonic, so the frequency, and hence wavenumber, can be determined:

$$\eqalign{ & \nu = {1 \over {2\pi }}\omega = {1 \over {2\pi }}\sqrt {{k \over \mu }} \cr & \lambda = {c \over \nu } = 2\pi c\sqrt {{\mu \over k}} \cr & \tilde \nu = {1 \over \lambda } = {1 \over {2\pi c}}\sqrt {{k \over \mu }} \cr} $$

In this example the wavenumber is 235900 m^{-1} or 2359 cm^{-1}.