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Suppose I have a cathode with a work function of 3eV and an anode at a potential of 2V above the cathode. If a photon having 2eV of energy hits the cathode, what happens?

A. An electron is emitted from the cathode because it only takes 1eV to go from the cathode to the anode.

B. An electron is not emitted because 3eV are required to escape the cathode.

Sources seem unclear on this and I need to know whether it is possible to knock electrons off of a surface using light of a lower energy than the work function given a sufficient electrical potential.

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  • $\begingroup$ What do you mean by "anode at a potential of 2V above the cathode"? Do you mean that anode is at potential 2V greater than cathode, in other words potential difference is 2V? $\endgroup$
    – Sensebe
    Commented Feb 16, 2014 at 9:19
  • $\begingroup$ I think potential difference also contributes for the emission of electron along with the photon's energy. $\endgroup$
    – Sensebe
    Commented Feb 16, 2014 at 10:01

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The work function, $\phi$ is the amount of energy required to free the electron from the pull of the nuclei of the atoms of the photosurface. Here $\phi=3\text{ eV}$.

Since the kinetic energy of the electron is given by, $$E_k=hf-\phi$$ it becomes evident that the condition of electron emission is when $hf>\phi$. Clearly this is not the case, which is why the electron is not emitted.

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The answer is $B$.(as explained by Ruben)
A voltage of 2 Volts is applied at anode. Some electrons can be generated at cathod due to thermoionic emission. These electrons will acquire kinetic energy $eV_{applied}$ only when they reach the anode i.e they are accelerated due to electric field and there K.E increses untill they reach anode.

If an electron(coming from cathode) having energy 2eV and a photon of energy 2ev hits the anode's surface simultaneously then an electron having kinetic energy 1ev will be ejected from anode.

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  • $\begingroup$ This answer is misleading, as it gives the impression that electrons are indeed emitted because of the presence of the light. While it is OK to say "No, photoelectrons are not emitted, but there's this other mechanism that [...]", it's important to make it clear that it is a separate mechanism. In particular, this thermionic emission is independent of whether there is light shining on the metal, so when doing photoelectric effect measurements it should be taken as dark counts and as a background to be subtracted from the signal. $\endgroup$
    – Emilio Pisanty
    Commented Mar 14, 2014 at 12:29
  • $\begingroup$ Further, you seem to indicate that the answer to "is it possible to knock electrons off of a surface using light of a lower energy than the work function given a sufficient electrical potential?" is yes, where the light has some role in the process. This is why the answer is incorrect. $\endgroup$
    – Emilio Pisanty
    Commented Mar 14, 2014 at 12:30
  • $\begingroup$ @EmilioPisanty: I think It is possible to knock electrons off of a surface using light of a lower energy than the work function given a sufficient electrical potential. My answer may be technically wrong. I read this on somewhere on SE like here:physics.stackexchange.com/questions/95411/… $\endgroup$
    – user31782
    Commented Mar 14, 2014 at 13:30
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    $\begingroup$ OK, so that explains the downvotes. Your answer is wrong. (Re: John Rennie's post, read in particular the last paragraph.) $\endgroup$
    – Emilio Pisanty
    Commented Mar 14, 2014 at 13:36

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