Photoelectric Problems
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Photoelectric Problems
Is the photoelectric effect just conceptual or is there actual computation for these equations? Do we just need to understand the graphs that depict the photoelectric effect and how this was revolutionary as a photon behaved as both a particle and a wave? Thank you-Roger.
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Re: Photoelectric Problems
it is conceptual but also need to do problems involving equation
energy of photon = work function + KE of ekectron
1/2mv^2
E=hv
these are equations used for the calculations
energy of photon = work function + KE of ekectron
1/2mv^2
E=hv
these are equations used for the calculations
Re: Photoelectric Problems
In addition to the photoelectric effect equation, you should be able to understand the parameters that can be manipulated. For example, if electrons are being expelled from the metal surface, what will happen if I increased intensity? How can I make the electrons be expelled at a higher velocity? Keep these questions in mind.
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Re: Photoelectric Problems
To answer the questions listed above...
1)What will happen if I increased intensity?
Intensity (the number of photons) has no effect on the velocity of the electron because the photoelectric effect illustrates a 1-to-1 interaction between photons and electrons of the metal. This is shown in the example where a high-intensity light of low frequency does not eject any electrons. The intensity of a light has no effect on the ejection of electrons.
2)How can I make the electrons be expelled at a higher velocity? Keep these questions in mind.
So, if intensity has no effect on the velocity of the electron, then what does? It would be to increase the energy of the incoming photon. Since the metal will always have a set work function (the energy required to eject an electron), increasing the electron increases the kinetic energy of the ejected photon since Energy of photon = work function + kinetic energy of the ejected electron.
1)What will happen if I increased intensity?
Intensity (the number of photons) has no effect on the velocity of the electron because the photoelectric effect illustrates a 1-to-1 interaction between photons and electrons of the metal. This is shown in the example where a high-intensity light of low frequency does not eject any electrons. The intensity of a light has no effect on the ejection of electrons.
2)How can I make the electrons be expelled at a higher velocity? Keep these questions in mind.
So, if intensity has no effect on the velocity of the electron, then what does? It would be to increase the energy of the incoming photon. Since the metal will always have a set work function (the energy required to eject an electron), increasing the electron increases the kinetic energy of the ejected photon since Energy of photon = work function + kinetic energy of the ejected electron.
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