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 (${\ce {\mu}}$) 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:

$v_{0}={\frac {1}{2\pi }}\cdot {\biggl (}{\frac {k}{\mu }}{\biggr )}^{\frac {1}{2}}\qquad {\text{where}}\ \mu =1.139\times 10^{-26}{\text{kg}}$

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

$v_{0}={\frac {1}{2\pi }}\cdot {\biggl (}{\frac {1860{\text{N/m}}}{1.139\times 10^{-26}{\text{kg}}}}{\biggr )}^{\frac {1}{2}}\qquad$

$v_{0}=6.43\times 10^{13}{\text{s}}^{-1}$

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

$v={\frac {v_{0}}{c}}\qquad {\text{where}}\ c=2.998\times 10^{10}{\frac {\text{cm}}{\text{s}}}$

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

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