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(i) In the explanations of photoelectric...

(i) In the explanations of photoelectric effect, we assume one photon of frequency v 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)=hv-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?

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(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?

(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?

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