Quantum Chemistry/Example 27

Question

Find the wavenumber (in cm^-1) of the photon required to excite a CO molecule from its ground vibrational state to its first excited vibrational state, given that the force constant for the CO bond is k=1860 N.m^-1.

Solution

The frequency of the photon required for this transition corresponds to the fundamental vibrational frequency of the molecule, $v_{\text{0}}$. To determine the wavenumber of this transition, the first step is to calculate the reduced mass of the molecule. The reduced mass ($\mathrm{\mu }$) for a diatomic molecule is given by the following equation:

$\mu =\frac{m_1\cdot m_2}{m_1+m_2}$

The obtained atomic masses of Carbon and Oxygen are as below:

C= 12.011u, O =15.999u

Plugging these values into the reduced mass equation gives:

$\mu =\frac{12.011\text{u}\cdot 15.999\text{u}}{12.011\text{u}+15.999\text{u}}$

$\mu =6.861\text{u}$

The unit conversion from atomic mass units (u) to kilograms (kg) is as follows:

$1u=1.66054\times 10^{-27}\text{kg}$

$\mu =1.139\times 10^{-26}\text{kg}$

Now, we can calculate the frequency of the photon using the following equation:

$v_0=\frac{1}{2\pi }\cdot (\frac{k}{\mu }{)}^{\frac{1}{2}}\text{where}\mu =1.139\times 10^{-26}\text{kg}$

Plugging in the values for the reduced mass and force constant gives:

$v_0=\frac{1}{2\pi }\cdot (\frac{1860\text{N/m}}{1.139\times 10^{-26}\text{kg}}{)}^{\frac{1}{2}}$

$v_0=6.43\times 10^{13}{\text{s}}^{-1}$

Finally, we need to covert the frequency of the photon ($v_{\text{0}}$) in s^-1 to the wavenumber (ṽ) in cm^-1:

$v=\frac{v_0}{c}\text{where}c=2.998\times 10^{10}\frac{\text{cm}}{\text{s}}$

By substituting the known values, the wavenumber of the photon can be determined as follows::

$v=\frac{6.43\times 10^{13}{\text{s}}^{-1}}{2.998\times 10^{10}\frac{\text{cm}}{\text{s}}}\therefore v=2.14\times 10^3{\text{cm}}^{-1}$