# Absorption of 2 photon by 1 electron in photoelectric effect

spaghetti3451

## Homework Statement

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 ##?

## 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.

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 