Just checking to make sure I understood this concept;
If the photons energy is the same as the work energy needed to remove an electron, how would the electron move away from the metal? Would its kinetic energy just be 0 and not leave the metal?
Photoelectric Effect
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Re: Photoelectric Effect
I believe so. The electron would not have any velocity (through the kinetic energy equation KE = 1/2 mv^2) to fully remove itself from the metal, but it is considered "ejected".
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Re: Photoelectric Effect
If the photon energy matches that of the work function (phi) then the electron only has sufficient energy to be ejected not to be fully removed from metal.
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Re: Photoelectric Effect
I believe that it is only momentarily "removed" from the metal, but the only possible way for it to "leave" the metal is if there is excess energy (if the photon's energy is greater than the energy to remove the electron) and in that situation, the electron is ejected from the metal, but Professor Lavelle said that the electron detector is attached to an electric circuit that brings the electron back to the metal.
Re: Photoelectric Effect
The electron would move to the surface of the metal, but because in the experiment the detector that records the KE of electrons has a positive charge the electron would be pulled away by charge of the detector.
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Re: Photoelectric Effect
The detector also has a slight positive charge which helps electrons in this situation drift towards it because opposite charges attract. As a result, the electron is counted as "ejected".
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Re: Photoelectric Effect
to clarify, when the energy level drops thats when it emits a light as a photon and when it gains energy, it is absorbed?
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