Why Does Kinetic Energy of Electron in Photoelectrics Follow Classical Approach?

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SUMMARY

The kinetic energy of an electron released in photoelectric experiments follows the classical approach due to the non-relativistic speeds of the emitted electrons. The equation used is E_photon = (Planck_const)(frequency) = (Work function) + (1/2)(mass_electron)(u^2), which is valid for the low velocities typically observed. Although the relativistic equation, E_photon = (Planck_const)(frequency) = (Work function) + ((1 - ((u/c)^2)))^(-1)) - 1)(mass_electron)(c^2), is more accurate, it is unnecessary for the conditions of standard photoelectric experiments. This approximation suffices because all observed photoelectrons are highly non-relativistic.

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  • Awareness of relativistic physics principles
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Azmodan
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Hello friends.

I was just wondering why in photoelectrics, the kinetic energy of an electron released from a certain metal after being struck by a photon follows the classical approach and not the relativistic. For example...

E_photon = (Planck_const)(frequency) = (Work function) + (1/2)(mass_electron)(u^2)

instead of

E_photon = (Planck_const)(frequency) = (Work function) + ((1 - ((u/c)^2)))^(-1)) - 1)(mass_electron)(c^2)

I've read several texts, including my university's text and Serway's text but the answer isn't really there. Thanks in advance!
 
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Azmodan said:
I was just wondering why in photoelectrics, the kinetic energy of an electron released from a certain metal after being struck by a photon follows the classical approach and not the relativistic.

AFAIK all the photoelectric experiments that have ever been done have involved highly non-relativistic photoelectrons, so the classical formula works fine, but strictly speaking, it is an approximation and the relativistic formula is the strictly correct one.
 
Cool, thanks Peter.
 

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