(i) In the explanations of photoelectric effect, we assume one photon of frequency `nu` collides with an electron and transfer its energy. This leads to the equation for the maximum energy `E_(max)` of the emitted electron as `E_(max)=hnu-phi_(0)` Where `phi_(0)` is the work function of the metal. if an electron absorbs 2 photons (each of frequency v) what will be the maximum energy for the emitted electron?
(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Options

A. `2hv-phi_(0)`
B. `2(hv-phi_(0))`
C. `2hv`
D. None

Answer

A

Explanation

Given ,`E_(max)=hv-phi_(0)`
As energy of `2` photons =`2hv,E'_(max)=2hv-phi_(0)`
As there is one to one interaction, probability of absorbind `2` photons by same electron is very low.Thus, emission of two photon absorption is negligible.