(i) In the explanation of photoelectric effect we assume one photon of frequency with v collides with an electron and transfers its energy.This lead to the equation for the maximum energy `E_(max)` of the emitted electron as
`E_(max)=hv-phi_(0)`
Where `phi_(0)` is the work function of the metal If an electron absorbs 2 photon (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 consueration in our discussion of the stopping potential?

Explanation

(i)Suppose one electron absorbs two phtons each having frequency f,out of which it spends W amount of energy for its emission and remaining amount (2hf-W) is possessed by it in the form of kinetic energy after getting emitted .Hence,
K=2hf-W
`therefore K_(max)=2hf-W_(min)`
`therefore K_(max)=2hf-phi_(0)(becauseW_(min)=phi_(0))`
(ii)If above assumption is true then according to work energy theorem,
Taking `K_(max)=v_(0)e`,
(where `V_(0)`=stopping potential)
`therefore 2hf-phi_(0)=V_(0)e`
`therefore V_(0)=((2h)/(e))f+(-phi_(0))`
Above equation is like equation of a straight line y=mx+c.Hence if we plot graph of `V_(0)toF` experimentally then we should get its
Slope equal to `((2h)/(e))`.But experimentally we
get this slope only `((h)/(e))`.Hence above
assumption is proved to be wrong.That is why only such assumption is not considered in the discussion of stopping potential.