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Photoelectric effect?
A couple short questions:
If I increase the frequency of light (decrease the wavelength), and keep the intensity constant, will the rate of photons onto the metal increase? I thought it might have stayed the same, but I'm not sure. It could also decrease. I'm really confused on this.
Also, same situation. If I increase the frequency of light (decrease the wavelength), and keep the intensity constant, does the rate of photoelectrons stay the same or decrease. I'm pretty sure it doesn't increase.
Also, the intensity can be defined as the number of photons in the light. For example, increasing the light intensity increases the number of photon in the light.
Thanks for your time and help today!
1 Answer
- Anonymous1 decade agoFavorite Answer
If you subscribe to the theory that the photoelectric effect is due to a photon being absorbed by an electron then what you need to consider is how the photon energy is related to its frequency, what energy is needed to free an electron and how intensity and photon flux are related. You should then be able to work out the answer.
If you are interested, the photoelectric effect has evolved since Einstein's formulation. It has been found that the effect can be described without introducing photons by calculating transition probabilities associated with a Hamiltonian describing the interaction of the electron with a classical EM wave.
Source(s): W. E. Lamb, Yale, Feb '68: The Photoelectric Effect Without Photons.