- #1

Malamala

- 299

- 27

$$A_{E_2} = \frac{32\pi^5\alpha c a_0^4}{15 \lambda g'}S_{E_2} = \frac{1.11995 \times 10^{18}}{\lambda g'}S_{E_2}$$

where ##\lambda## is the transition wavelength in angstroms, ##g'## is the degeneracy of the excited state and ##S_{E_2}## is the E2 line strength in atomic units. ##S_{E_2}## is known theoretically to be 1.43 and in the end I get ##A_{E_2} \sim 2\times 10^{-4} s^{-1}##. I also have the (pulsed) laser power which is 100 mW and beam area which is 1mm##^2##. Using the Plank's formula, I get that the number of photons per unit second per mm##^2## is ##\sim 10^{18}##. If I have a given atom, interacting with such a pulse, I would like to know the probability of being excited. However, it seems like I would need to plug in some sort of area of the atom. Basically the formula I would use is:

$$2 \times 10^{18} \times 10^{-4} \times A_{atom} \times mm^{-2}$$

But I am not sure what to use for that. I guess I can use ~##10^{-10}##m for the atomic radius, but that seems like a classical approximation. I was wondering if I am doing something wrong. Why do I get this extra area in the formula and how do I deal with it? Thank you!