Quantum Chemistry/Example 11

Example 11
Use the 1D particle in a box model to estimate the wavelength of light required to excite an electron from a pi to pi* MO in ethene.

Solution:

The energy levels of a particle in a 1D box with a specific quantum number $$ n $$, are as follows.

$$ E_n = \frac{h^2n^2}{8mL^2} $$

In this equation $$ h $$ represents Planck's constant, $$ m $$ is the mass of the particle, and $$ L $$ is the length of the box.

The pi electron in the double bond between the carbon atoms in ethene can be approximated to the particle in a 1D box model. This means that the mass of the particle in this question will be the mass of an electron, and the length of the box corresponds to the bond length between the carbon atoms in the molecule ethene.

Additionally, the energy equation above needs to be transformed into a equation for $$ \Delta E $$ since the electron is moving from one energy level to another.

$$ \Delta E = E_f - E_i = \frac{h^2n_i^2}{8mL^2} - \frac{h^2n_f^2}{8mL^2} = \frac{h^2}{8mL^2}(n_f^2-n_i^2) $$

The change in energy between the pi and pi* MO in ethene can now be calculated knowing that the bond length between doubly bonded carbon atoms is 133pm and the mass of an electron is 9.1093856x10-31 kg. Moving from the ground state n=1 to an excited state of n=2 :

$$ \Delta E = \frac{h^2}{8mL^2}(n_f^2-n_i^2) $$

$$ \Delta E = \frac{(6.626\text{x}10^{-34}\frac{\text{m}^2\text{kg}}{\text{s}})^2}{8(9.109\times 10^{-31}\text{kg})(133\times 10^{-12}\text{m})^2}(2^2-1^2) $$

$$ \Delta E = 1.0217 \times 10^{-17}\text{J} $$

Now that the energy required to excite the electron to the pi* orbital is known, the wavelength of light can be calculated through the following equation, where c is the speed of light in a vacuum and $$ \lambda $$ is the wavelength of light.

$$ \Delta E = \frac{hc}{\lambda} $$

The equation can then be re-arranged to solve for the wavelength of light.

$$ \lambda = \frac{hc}{\Delta E} $$

By plugging in the known constants and the value for $$ \Delta E $$ that has been calculated above, the wavelength can be found.

$$ \lambda = \frac{hc}{\Delta E} $$

$$ \lambda = \frac{(6.626 \times 10^{-34}\frac{\text{m}^2\text{kg}}{\text{s}})(2.998 \times 10^8\frac{\text{m}}{\text{s}})}{1.0217 \times 10^{-17}\text{J}} $$

$$ \lambda = 1.9 \times 10^{-7} \text{m} $$

$$ \lambda = 19 \text{nm} $$

Therefore the wavelength of light required to excite an electron from a pi to pi* molecular orbital in ethene is 19nm.