# Photoemission in diatomic molecules

1. Jul 8, 2008

### Diracn

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 $$\alpha$$ is the fine-structure constant (dimentionless), $$a_0$$ is the Bohr's radius ( units meters), $$\omega$$ is the Energy of radiation (units Jules) and $$D_{lmm_\gamma}$$ 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 ( $$k = \sqrt{2E}$$ )

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

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 !!!