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:

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

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

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

$$ v = $$ (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:

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

where $$ \lambda$$ is the wavelength.

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

This equation can be rearranged to solve for wavelength:

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

= $$ {1 \over2143 cm^{-1} } $$

= 4.666356*10-4 cm

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

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

= 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:

$$ E = hv_0 (v+\left ( \frac{1}{2} \right ))$$

where E is the energy, h is Planck's constant, $$ v_0 $$ is the frequency of the ground state, and $$ v$$ is the quantum number of the energy level.

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

$$ E = hv_0 (v+\left ( \frac{1}{2} \right ))$$

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

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

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.

$$ E = hv_0 (v+\left ( \frac{1}{2} \right ))$$

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

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

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

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

Rearrange the equation to solve for λ:

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

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

$$ \lambda = 0.0001866542 cm $$

Convert the wavelength to nm.

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

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

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.