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5.2: The Equation for a Harmonic-Oscillator Model of a Diatomic Molecule Contains the Reduced Mass of the Molecule

For studying the energetics of molecular vibration we take the simplest example, a diatomic heteronuclear molecule AB. Homonuclear molecules are not IR active so they are not a good example to select. Let the respective masses of atoms A and B be \(m_A\) and \(m_B\). So the reduced mass \(\mu_{AB}\) is given by:

\[ \mu_{AB}=\dfrac{m_A\, m_B}{m_A+m_B} \label{5.2.1}\]

The equilibrium internuclear distance is denoted by \(r_{eq}\). However as a result of molecular vibrations, the internuclear distance is continuously changing; let this distance be called \(r(t)\). We can define the displacement \(x\) at any time away from the equilibrium length \(r_{eq}\) at any time \(t\):


When \(x\) is non-zero, a restoring force \(F\) exists which tries to bring the molecule back to \(x=0\), that is equilibrium. For small displacements this force can be taken to be proportional to \(x\).

\[ F=-kx \label{5.2.2}\]

where \(k\) is the force constant.

Reduced mass is the representation of a two-body system as a single-body one. When the motion (displacement, vibrational, rotational) of two bodies are only under mutual interactions, the inertial mass of the moving body with respect to the body at rest can be simplified to a reduced mass.

Reduced Mass

Viewing the multi-body system as a single particle allows the separation of the motion: vibration and rotation, of the particle from the displacement of the center of mass. This approach greatly simplifies many calculations and problems.

Figure \(\PageIndex{1}\): a) the individual vectors to the particles m1 and m2 in the coordinate space and the resultant vector. b) center of mass. c) reduced mass.

This concept is readily used in the general motion of diatomics, i.e. simple harmonic oscillator (vibrational displacement between two bodies, following Hooke's Law), the rigid rotor approximation (the moment of inertia about the center of mass of a two-body system), spectroscopy, and many other applications.

The reduced mass of a heteronuclear diatomic is given by:

\[\mu_{12} = \dfrac{m_{1} m_{2}}{m_{1} + m_{2}}\]

The properties of diatomic molecular vibrations are studied as they are IR active, which provides great significance in spectroscopy. 

Example \(\PageIndex{1}\):

Determine the reduced mass of the two body system of a proton and electron. (mproton = 1.6727 x 10-27 kg and melectron = 9.110 x 10-31 kg).

\[\mu_{pe} = \dfrac{(1.6727 \times 10^{-27})(9.110 \times 10^{-31})}{1.6727 \times 10^{-27} + 9.110 \times 10^{-31}} = 9.105 \times 10^{-31}\]

Reduced Mass in Frequency of Harmonic Oscillator Approximation

The Harmonic Oscillator approximation is a simple yet powerful representation of the energetics of molecular vibrations. By assuming the potential energy of a diatomic bond is analogous to that of a spring:

 \[V(x) \approx \dfrac {1}{2} kx^2 \]

The Shrodinger Equation becomes:

\[\hat{H} \psi = \dfrac {-\hbar}{2\mu} \dfrac {d^2 \psi}{d x^2} + \dfrac {1}{2}kx^2 = E\psi \]

and is solved to give quantized energy in terms of v:

\[E_v = \hbar \left(\sqrt {\dfrac {k}{\mu}} \right) \left(v + \dfrac {1}{2} \right) = h \nu \left(v+\dfrac {1}{2} \right)\]

The natural vibrational frequency of the system is given as:

\[\nu = \dfrac{1}{2 \pi}\sqrt {\dfrac {k}{\mu}}\]

and the mass, µ, is given by the reduced mass of the system. In a more practical sense, the quantized energy of the Harmonic oscillator can be used to quantify the transitions between vibrational energy levels induced by emission or absorption of radiation. These transitions only occur if there is a change in dipole moment and the energy of the radiation is equivalent to the difference in the transition between quantum states. The vibration of molecules can be "seen" through IR spectroscopy (largely absorption), which can be used to identify and study molecules. *Note: When solving simple diatomic calculations, it is important to convert the atomic mass to kg to solve in correct units.

Reduced Mass in the moment of inertia for Rigid Rotator Approximation

The Rigid Rotor model is a representation of two fixed particles as they rotate in space. To solve the Shrodinger equation for this problem, the switch from the Cartesian to Spherical Coordinate system is made. The steps to solve the problem are a bit more lengthy and can be found here. The energy levels of this system are also quantized and is represented by: \[E = J(J + 1) \dfrac {\hbar^2}{2I}\]

With the moment of inertia, I, represented by the reduced mass and bond length squared:

\[I = \mu r^2\]

Each eigenstate poseses a degeneracy of 2J+1 because of varying mJ quantum number. The transition energies of the quantized states can be detected in the microwave spectrum. These transitions can be induced by coupling the dipole moment of a molecule with an electric field. The relevant selection rules of the transitions are found by solving the moment integral and are given as: \[\Delta J = \pm 1 \] \[\Delta m_J = 0 \]

Example \(\PageIndex{2}\):

Calculate J = 0 to J = 1 rotational transition of the Omolecule. (Bond length O2 = 121pm)

\[E = \dfrac {\hbar^2}{I} = \dfrac {\hbar^2}{\mu r^2}\]

\[\mu_{O2} = \dfrac{m_{O} m_{O}}{m_{O} + m_{O}} = \dfrac{(15.9994)(15.9994)}{15.9994 + 15.9994} = 7.9997 \]

convert from atomic units to kilogram using the conversion: 1 au = 1.66 x 10-27 kg. Plug and chug.

\[E = 5.71 \times 10^{-27} \;Joules \]