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.

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

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

Solutions

a) The example provides the wavenumber as an IR spectrum in a plot of wavenumber vs intensity. In this example the wavenumber is 2143 cm^-1. This can be converted to frequency using the following equation:

${\nu ={\tilde {\nu }}\times c}$

where $\nu$ is the frequency, ${\tilde {\nu }}$ is the wavenumber and $c$ is the speed of light. The speed of light is a constant of 2.9979248x10¹⁰ cm/s.

$\nu =(2143{\text{ cm}}^{-1})\times (2.99792458\times 10^{10}{\text{ cm/s)}}$

$\nu =6.424552375\times 10^{13}{\text{ s}}^{-1}$

This value then must be converted to the SI units for frequency. One Hz is equal to one s^-1, so the conversion is simple by adding the corresponding prefix for $1\times 10^{12}={\text{T}}$

$\nu =64.25{\text{ THz}}$

Next in order to determine the wavelength of the vibration we can use the equation:

${\tilde {\nu }}={\frac {1}{\lambda }}$

where $\lambda$ is the wavelength.

This equation can be rearranged to solve for wavelength:

$\lambda ={\frac {1}{\tilde {\nu }}}$

$\lambda ={\frac {1}{2143{\text{ cm}}^{-1}}}$

$\lambda =4.6663555\times 10^{-4}{\text{ cm}}$

This should be converted to the SI units of wavelength which is nm.

$\lambda =4.666356\times 10^{-4}{\text{ cm}}\times {1{\text{ nm}} \over (1\times 10^{-7}{\text{ cm}})}$

$\lambda =4.666\times 10^{3}{\text{ nm}}$

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

$E=h\nu _{0}{\biggl (}v+\left({\frac {1}{2}}\right){\biggr )}$

where $E$ is the energy, $h$ is Planck's constant $(6.425\times 10^{13}{\text{ s}}^{-1})$, $\nu _{0}$ 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. As this is the ground state the quantum number $v$ is equal to 0.

$E=(6.62607015\times 10^{-34}{\text{ m}}^{2}{\text{kg/s}})\times (6.425\times 10^{13}{\text{ s}}^{-1})$${\biggl (}0+\left({\frac {1}{2}}\right){\Biggr )}$

$E=2.128625036\times 10^{-20}{\text{ J}}$

$E=2.129\times 10^{-20}{\text{ J}}$

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

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

$E=(6.62607015\times 10^{-34}{\text{ m}}^{2}{\text{kg/s}})\times (6.425\times 10^{13}{\text{ s}}^{-1})$${\biggl (}2+\left({\frac {1}{2}}\right){\Biggr )}$

$E=1.064312518\times 10^{-19}{\text{ J}}$

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:

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

$\lambda =\left({\frac {(6.62607015\times 10^{-34}{\text{ m}}^{2}{\text{kg/s}})\times (2.99792458\times 10^{10}{\text{ cm/s}})}{1.064312518\times 10^{-19}{\text{ J}}}}\right)$

$\lambda =1.86641219\times 10^{-4}{\text{ cm}}$

Convert the wavelength to nm for proper SI units

$\lambda =1.86641219\times 10^{-4}{\text{ cm}}\times \left({\frac {1{\text{ nm}}}{1\times 10^{-7}{\text{ cm}}}}\right)$

$\lambda =1866.41219{\text{ nm}}$

Conclusion

After working thought part a-c you should be able to see how the conversion between wavelength, frequency, energy and wavenumber of electromagnetic radiation occurs.