Photoelectric work function Watch

'bob'
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Report Thread starter 7 years ago
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Hey guys. This past exam question has stumped me a bit. Its based on the equation off the photoelectric effect ( hv = work function of metal + kinetic energy of emitted electrons).

We are given the frequency of the em radiation, its intensity (in Watts per square metre) and kinetic energy of the emmited electrons.

The first part asks for the energy of each photon. So i multiply v by h and that gives me an energy value in J but is that the energy for a single photon?

Next up it asks me the number of photons reaching the surface if the area is (so and so) metres squared. (No idea on this one)

The last one seems simple enough. It just asks the work function of the metal ( =hv - Ek)

Thanks. my main problem is the relationship between intensity, area and number of photons
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Sheldon
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1. Yes it is
2.The intensity is how many photons you have, so basically times it by Surface area
3.Yep
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'bob'
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(Original post by Sheldon)
1. Yes it is
2.The intensity is how many photons you have, so basically times it by Surface area
3.Yep
Thanks. As for part 2 my intensity is 4.5 micro watts per metre squared and my surface area was 2.5 micrometres squared. Needless to say this gave me a very small number.

So what i did is divided the intensity by the energy of each photon (1.34*10 -18) and multiplied that by the surface area. Got a value of 8.4million which seems about right but not sure if it actually is.
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