# Absorption of 2 photon by 1 electron in photoelectric effect

1. Dec 31, 2014

### spaghetti3451

1. The problem statement, all variables and given/known data

In the photoelectric effect, it is assumed that a single electron absorbs a single photon. But, there is a certain probability that a single electron may simultaneously absorb two identical photons from a high-intensity laser. How would such an occurrence affect the threshold frequency, the stopping potential and the equation $eV_{0} = hf-\phi$?

2. Relevant equations

3. The attempt at a solution

The threshold frequency is halved because it is possible to have the most loosely bound electron (with binding energy = work function) absorb the two photons.

The equation $eV_{0} = hf-\phi$ is modified to become $eV_{0} = 2hf-\phi$ because hf is the energy of one photon.

Therefore, the stopping potential for a given frequency increases from its corresponding earlier value.

Are my answers correct?

2. Dec 31, 2014

### PWiz

The threshold frequency of any metal on which monochromatic electromagnetic radiation which causes two photon absorption(or interaction) is incident will be half the value of what it would normally be for single photon interactions, which you have correctly evaluated. So the equation $h(f-f_0) = \frac{1}{2}m_e v^2 = e V_0$ becomes equal to $h(2f-f_0)$ where $f_0 = 2 f_x$ (fx is the threshold frequency for the TPA effect), which is essentially the same as the last equation you came down to in the end. So yes, for E.M. waves with the same frequency, with the TPA effect, you will have a higher stopping potential value.(The ratio of the stopping potentials will be $\frac{2f-f_0}{f-f_0}$ to be precise).
P.S. Let me know if you think there's somewhere I went wrong.

3. Dec 31, 2014

### PWiz

Might I add that if there are $n$ photons with distinct energies ($n$ frequencies) in a mixed frequency E.M. wave and, hypothetically speaking, if all of them could simultaneously interact with one electron, then the stopping potential would be given by $eV_0 = h(f_1+f_2+f_3....+f_n -f_0)$ , where photoelectric emission just starts happening when $f_0 = ∑f_i$. The ratio of stopping potentials in this case would be $\frac{∑f_i - f_0}{f_x-f_0}$ where $f_x$ is the frequency of the monochromatic E.M. wave with which we are comparing the mixed frequency wave with. ($f_i$ is the set of frequencies in the mixed frequency E.M. wave). I really hope I'm not over-complicating this