1
$\begingroup$

I know that general relationship between the frequency $f$ of the photon, the potential $U$ and thw work function $\phi$ is given by: $$eU=hf-\phi$$ I wondered if this relationship would be still true if we accounted for relativistic electrons? I'm sadly not that strong in speical relativity and searching on the web for this question doesn't give me a result which I can understand. It would be awesome if someone with a bit more knowledge on the subject could give me some thoughts.

$\endgroup$
1

1 Answer 1

0
$\begingroup$

The relationship would be very similar. The only difference is that you have to consider the conservation of the 4-momentum instead of the classical conservation of energy.

Note that in order to have a relativistic outcoming electron the energy of the incoming photon has to be in order of $\sim 30 keV$. This means that those photons are in the range of X-Rays. At those energies the cross section of the photoelectric process is much lower than the Compton cross section meaning that this process would be much more likely to happen.

If we consider the typical energies of the photoelectric effect (UB photons are around 3-4 eV) and the typical work function of metals (few eV) the kinetic energy of the electron extracted is in the order of few eV meaning that the electron is not relativistic. You could still use the relativistic formula but you would have a very small correction (less than 1‰)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.