# Quantum Chemistry/Example 31

Write an example problem related to the conversion between wavelength, frequency, energy and wavenumber of electromagnetic radiation. Show each step in detail and explain the conversion.

Example: Carbon monoxide shows a sharp peak in its IR spectrum at 2143 cm-1.

a) Calculate the frequency and wavelength of the molecule's vibration.

The question gives us the wavenumber which can be converted to frequency through the equation:

${\displaystyle v={\widehat {v}}*c}$

where v is the frequency, ṽ is the wavenumber and c is the speed of light.

${\displaystyle v={\widehat {v}}*c}$

${\displaystyle v=}$ (2143cm-1) * (2.99792458*1010cm s-1)

= 6.42455237494*1013 s-1

= 64.24 THz

To find the wavelength of the vibration we can use the equation:

${\displaystyle {\hat {v}}}$= ${\displaystyle {1 \over \lambda }}$

where ${\displaystyle \lambda }$ is the wavelength.

${\displaystyle {\hat {v}}}$= ${\displaystyle {1 \over \lambda }}$

This equation can be rearranged to solve for wavelength:

${\displaystyle \lambda }$= ${\displaystyle {1 \over {\hat {v}}}}$

= ${\displaystyle {1 \over 2143cm^{-1}}}$

= 4.666356*10-4 cm

This should be converted to nm since wavelength is usually written in nm.

λ = 4.666356*10-4 cm * ${\displaystyle {1nm \over (1*10^{-7}cm)}}$

= 4.6*103nm

b) Calculate the energy of carbon monoxide at its ground state.

To find the energy of a molecule at its ground state we can use the equation:

${\displaystyle E=hv_{0}(v+\left({\frac {1}{2}}\right))}$

where E is the energy, h is Planck's constant, ${\displaystyle v_{0}}$ is the frequency of the ground state, and ${\displaystyle v}$ is the quantum number of the energy level.

We can use this equation since carbon monoxide's vibrations act as a harmonic oscillator.

${\displaystyle E=hv_{0}(v+\left({\frac {1}{2}}\right))}$

${\displaystyle E=h(6.424*10^{13}s^{-1})}$${\displaystyle (0+\left({\frac {1}{2}}\right))}$

${\displaystyle E=2.12847674*10^{-20}J}$

c) Calculate the wavelength of a photon that excites an electron from the ground state up 2 levels.

In this case, the quantum number v is equal to 2. We can use the same equation used in b) to first solve the change in energy caused by the photon.

${\displaystyle E=hv_{0}(v+\left({\frac {1}{2}}\right))}$

${\displaystyle E=h(6.424*10^{13}s^{-1})}$${\displaystyle (2+\left({\frac {1}{2}}\right))}$

${\displaystyle E=1.06423837*10^{-19}J}$

Now that we have found the change in energy we can calculate the wavelength of the photon using the equation:

${\displaystyle E=\left({\frac {hc}{\lambda }}\right)}$

Rearrange the equation to solve for λ:

${\displaystyle \lambda =\left({\frac {hc}{E}}\right)}$

${\displaystyle \lambda =\left({\frac {hc}{1.06423837*10^{-19}J}}\right)}$

${\displaystyle \lambda =0.0001866542cm}$

Convert the wavelength to nm.

${\displaystyle \lambda =0.0001866542cm*\left({\frac {1nm}{1*10^{-7}cm}}\right)}$

${\displaystyle \lambda =1.86*10^{-6}nm}$

Therefore, a photon that is able to cause a transition from ground state to v = 2 must have a wavelength of 1.86 * 10-6 nm.
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