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4.31: Calculating the Pi-electron HOMO-LUMO Electronic Transition for Benzene

  • Page ID
    151348
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    Calculate the wavelength of the photon required for the first allowed (HOMO-LUMO) electronic transition involving the π−electrons of benzene.

    Screen Shot 2019-05-22 at 12.23.28 PM.png

    Energy conservation requirements:

    \[ \frac{n_i^2 h^2}{2 m_e C^2} + \frac{hc}{ \lambda} = \frac{n_f^2 h^2}{2 m_e C^2} \nonumber \]

    Fundamental constants and conversion factors:

    \[ \begin{matrix} pm = 10^{-12} m & aJ = 10^{-18} J \end{matrix} \nonumber \]

    \[ \begin{matrix} h = 6.6260755 (10^{-34}) \text{joule sec} & c = 2.99792458 (10^8) \frac{m}{sec} & m_e = 9.1093897 (10^{-31}) kg \end{matrix} \nonumber \]

    Calculate the photon wavelength for the HOMO-LUMO electronic transition.

    \[ \begin{matrix} \text{HOMO:} & n_i = 1 & \text{LUMO:} & n_f = 2 & \text{Benzene circumference:} & C = 6(140) pm \end{matrix} \nonumber \]

    \[ \begin{array}{c|c} \lambda = \frac{n_i^2 h^2}{2 m_e C^2} + \frac{hc}{ \lambda} = \frac{n_f^2 h^2}{2 m_e C^2} & _{ \text{solve, } \lambda} ^{ \text{float, 3}} \rightarrow .194e-6m^3 \frac{kg}{ joule~ sec^2} ~ \lambda = 194 nm \end{array} \nonumber \]

    Calculate the photon energy and frequency.

    \[ \begin{matrix} \text{energy} & \frac{c}{ \lambda} = 1.024 aJ & \text{frequency} & \frac{c}{ \lambda} = 1.545 \times 10^{15} Hz \end{matrix} \nonumber \]

    Plot Wave Functions

    See Figure 7.6 (page 111) in Quantum Chemistry and Spectroscopy, by Engel.

    The real part of the wave function is plotted below.

    Quantum number: n = 5

    \[ \begin{matrix} \text{numpts} = 100 & i = 0 .. \text{numpts} & j = 0 .. \text{numpts} & \phi_i = \frac{2 \pi i}{ \text{numpts}} \\ x_{i,~j} = \cos \left( \phi_i \right) & y_{i,~j} = \sin \left( \phi_i \right) & z_{i,~j} = \frac{1}{ \sqrt{2 \pi}} \text{exp} \left(i n \phi_i \right) & zz_{i,~j} = 0 \end{matrix} \nonumber \]

    Screen Shot 2019-05-22 at 12.36.27 PM.png

    The square of the absolute magnitude for all the wave functions (for all values of the quantum number n) is 1/2π, as shown below.

    \[ \begin{matrix} \left( \left| \frac{1}{ \sqrt{2 \pi}} \text{exp} (i n \phi ) \right| \right)^2 & \text{simplifies to} & \frac{1}{2 \pi} \end{matrix} \nonumber \]

    The wave functions for the electron on a ring are eigenstates of the momentum operator. In other words the momentum is precisely known: p = nh/C, where n is the quantum number and C is the ring circumference. According to the uncertainty principle, the elctron position must be uncertain. The result above confirms this; the electron density is distributed uniformly over the entire ring.


    This page titled 4.31: Calculating the Pi-electron HOMO-LUMO Electronic Transition for Benzene is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.