Question

1. 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 E_max of the emitted electron as E_max = 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?
2. Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Answer 1

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 + K_max

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.