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Explaining the photoelectric effect, our teacher told us that back in that time, they calculated the time that an electron would take to be emitted from an atom if the energy got to the atom in a continuum way. We calculated it for potassium and a source of 1 Watt at 1 meter, giving approximately a couple of minutes. He told us then that the real time could not be calculated because of how small it was, being it $t<10^{-9}$ seconds.

I don't know if he meant they couldn't do it back then, or it's still impossible nowadays becuase of how small that time is? Because that number $10^{-9}$ looks big to me having in mind the times they were able to (incorrectly) measure at the CERN when all that neutrino stuff happened. Is there a smaller bound?

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I'm unsure of how you'd calculate that, but note that 1 ns is quite a long time by today's experimental standards. Laser pulses lasting ~1 ps were available in the late 80s, and since the 2000s we have few-fs pulses. This can stimulate high-harmonic generation to produce pulses as short as tens of attoseconds. That's a factor of $10^{-8}$ on top of a nanosecond. –  Emilio Pisanty Feb 24 '13 at 16:31
    
@EmilioPisanty That's what I was thinking. So today, do we know that value? Is it still smaller than our smallest precission? BTW the value was estimated by supposing all the energy that went through the atom would be absorbed, so we know that the actual time would have to be larger or equal to that. –  MyUserIsThis Feb 24 '13 at 17:38
    
Are you wondering whether it's experimentally possible to measure the time that an electron would take to be emitted from an atom if the energy got to the atom in a continuum way due to technological advances? If it isn't actually continuum in nature, we can't do it experimentally regardless of the technological advances. (though we might use computers to calculate...) –  raindrop Feb 24 '13 at 21:01
    
That's a part of my question. So theoritically, Einstein said that it would be completely instantanous, like $t=0$? that looks like an infinite aceleration. –  MyUserIsThis Feb 24 '13 at 22:22
    
An estimate of the time an electromagnetic interaction takes (photons interact electromagnetically) can be seen in the decay of the pi_0 which goes into two photons, and is of order of 10^-16 seconds en.wikipedia.org/wiki/… . –  anna v Feb 26 '13 at 19:14

1 Answer 1

up vote 3 down vote accepted

It's common to think of the light as photons that behave like billiard balls colliding with and knocking out electrons. However this isn't a good model for what actually happens. The photon is really the quantised energy transfer from the photon field to the metal. Localisation of the light into a photon only happens at the moment the electron is ejected.

The mechanism of the photoelectric effect is that the free electrons in the metal respond to the light by oscillating at the same frequency of the light. At some unpredicatable moment a single photon's worth of energy will be transferred from the light into the oscillations of the free electrons, and this energy ends up concentrated in a single electron that gets ejected. This obviously isn't just an interaction between a single photon and a single electron because the electron emerges in a different direction and therefore with a different momentum to the incident photon. There is some interaction of the emitted electron with the metal lattice.

I couldn't find any info on exactly how long this process takes, but I'd guess it would be of the order of a few periods of the light, or a few femtoseconds. I did find this paper that describes emission of photoelectons by a femtosecond laser. See in particular the section headed "Measurement of Transverse Velocity Spreads" on page 2. The experiment did not report a delay in the emission of the electrons and describes the emission as "nearly instantaneous".

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+1 on a very clear explanation. –  Emilio Pisanty Feb 26 '13 at 11:09
    
Thank you for the information –  MyUserIsThis Feb 26 '13 at 14:12

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