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  • (i) In the explanation of photoeletric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads . to the equation for the maximum energy Emax of the emitted electron as E max equal to hv minus phi 0 where

(i) In the explanation of photoelectric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads. to the equation for the maximum energy Emax 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?

 

Answers (1)

(i) In the question, 2 photons transfer its energy to one electron as E=hv

  Ee=Ep 

  hve=2hv 

 ve=2v

v is the frequency

 The maximum energy of the emitted electron is  Emax=hve - \phi_{0}

=h(2v) - \phi_{0}

=2hv - \phi_{0}

 (ii) Due to the mass difference, the probability that the electron absorbs 2 photons is extremely low.  Therefore the chances of such emission of electrons is close to negligible.
 

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