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Question
- In the explanation of photo electric effect, we assume one photon of frequency ν collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as Emax = hν – φ0 where φ0 is the work function of the metal. If an electron absorbs 2 photons (each of frequency ν) what will be the maximum energy for the emitted electron?
- Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?
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Solution
According to Einstein, photoelectric effect is the result of one to one inelastic collision between photon and electron in which the photon is completely absorbed.
Einstein's photoelectric equation is E = W0 + Kmax
Where `K_("max") = 1/2 mv_("max")^2` = maximum kinetic energy of emitted electrons
And W0 = Work function (or threshold energy)
`W_0 = hv_0 = (hc)/λ_0` Joules; v0 = Threshold frequency and λ0 = Threshold wavelength
i. According to the question, an electron absorbs the energy of two photons each of frequency v then v' = 2v where v' is the frequency of emitted electron.
Here, `E_("max") = hv - phi_0`
Thus, maximum energy for emitted electrons is `E_("max") = h(2) - phi_0 = 2hv - phi_0`
ii. The probability of absorbing two photons by the same electron is very low. Hence such emissions will be negligible.
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