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.

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

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

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

$E=h{\nu }_0(v+\left(\frac{1}{2}\right))$

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})$$(0+\left(\frac{1}{2}\right))$

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

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

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.

$E=h{v}_0(v+\left(\frac{1}{2}\right))$

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

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

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. Template:BookCat