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Photoemission in diatomic molecules

  1. Jul 8, 2008 #1
    Hello !!!

    I have a question that breaks the head to me, jejeje

    In the process of photoemission, the total cross section is defined by:

    \sigma(\omega) = \frac{4\pi}{3}\alpha a_0^2 \omega \sum_{lmm_\gamma} \left| D_{lmm_\gamma}(\omega)\right|^2

    where [tex]\alpha[/tex] is the fine-structure constant (dimentionless), [tex]a_0[/tex] is the Bohr's radius ( units meters), [tex]\omega[/tex] is the Energy of radiation (units Jules) and [tex]D_{lmm_\gamma}[/tex] are the transition coefficients expressed of the following way

    D_{lmm_\gamma}(\omega) = \sqrt{\frac{4\pi}{3}}\bigl< \psi_{lm}(k, r)\bigl| r Y_{1}^{m_\gamma}(\theta,\phi) \bigl | \psi_0(r) \Bigr>

    where the wave function for the emitted electron is ( [tex]k = \sqrt{2E}[/tex] )

    \psi_{lm}(k, \mathbf{r}) = \frac{1}{kr}f_{lm}(kr)Y_{lm}(\theta,\phi)

    and the wave function for the ionized orbital is

    \psi_0( r) = \frac{1}{r}\sum_{l m}\psi_{l m}(r)Y_{l}^{m}(\theta,\phi)

    my simple question is:

    What units must have each one of the involved terms so that the total cross section is expressed in barns?

    Thanks !!!
  2. jcsd
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