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:

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

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

= 4.666356*10^-4 cm

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

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

= 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:

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

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

$\lambda =0.0001866542cm$

Convert the wavelength to nm.

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

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

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.