we all have been studying that when photon is incident on alkali metal surface, photo currents are created due to ejection of electron after exceeding the work function.but the electron do have certain k.e revolving in the nucleus.why dont we consider that energy in einstein photoelectric equation?
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We do consider that energy. It reduces the amount of energy needed for the electron to be freed from the surface.
An analogy:
If a satellite is already in orbit, you need less energy to make it escape earth's gravity than if you started with the satellite on a launch pad on earth. The energy of the satellite in orbit is like the energy of the electron before the photoelectric interaction. Note that the average electron energy at room temperature is about 0.025 eV, which is very small compared to the energy needed for the electron to escape (this is why electrons prefer to stay inside the metal; and also why you can get spontaneous electron emission in hot cathodes in vacuum tubes - there, the energy of a handful of electrons is much higher than the mean at room temperature).
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$\begingroup$ ok..lets talk about velocity... the velocity of electron is tangential to the revolution..so now does the kinetic energy of electron depend on the direction of incident photon....?...coz if photon are incident in exactly opposte direction of electrons....the electron should be even forced inside the metal! $\endgroup$ Commented Feb 12, 2015 at 4:03
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2$\begingroup$ @AlokSubedi At the level of individual electrons bound states one cannot talk of velocity in a classical model way. We can deduce a velocity from the energy assuming a classical model, but it is a wrong model . The photon does not meet the electron but the whole atom/lattice. The only constraints are conservation of total energy , momentum and angular momentum for the total interaction, and it is a quantum mechanical interaction. $\endgroup$– anna vCommented Feb 12, 2015 at 4:44