Intensity of Light understanding

  • Thread starter Dba18
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Homework Statement


When light of wavelength 630 nm is directed onto a photocell, electrons are emitted at a rate of 2.6 *10 raised -12 C/s. Assume that each photon that impinges on the photocell emits one electron. How many photons per second are striking the photocell? How much energy per second is the photocell absorbing?



Homework Equations



c = nu * lambda
E = h * nu

The Attempt at a Solution

(1) I determine the frequency. (2) I already have the energy given off. How do I relate this to number of electrons given off and the amount of energy the photocell is absorbing? I don't see the connection here. And I didn't see anything like this in the text.
 

Answers and Replies

  • #2
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Are the units in the problems statement Coulombs per second? I suspect that to be the case and therefore refer you to a constant that you should become familiar with: Faraday Constant. That, plus some dimensional analysis, should get you the answer.
 
  • #3
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Yes, the problem is given in Coulombs/second. However, it is another two chapters before we have any introduction to Faraday's Constant, so I don't understand how that, charge per mole of electrons, would fit in here. This is my thinking so far and I have no clue if it's correct. 1: I find the Energy for one photon. There is a ratio of one photon per electron. If I divide the rate of electrons per second by the charge on one electron, I will know how many electrons are emitted, then I know how many photons are hitting the photocell (1:1). But if this were the case, wouldn't the amount of energy absorbed equal the amount of energy given off?
 
  • #4
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Yes, you can use the charge per electron or the charge per mol of electrons. These two constants differ by a factor of avogadro's number. Sometimes it's good to write out your terms with the units to see where you need to get. You simply need a way to get from coulombs per second to number of electrons per second, this is where the dimensional analysis comes in, the conversion factor can be chosen to be the most convenient but I never remember the charge per electron whereas I have seem to always remember Faraday's constant and Avogadro's number.

Once you have worked out the number of electrons emitted you can work out the energy absorbed from the equations you quoted in the OP. Simple energy conservation should dictate that energy absorbed = energy emitted. In the real world you have losses and such (I'll take a guess and say there is resistance somewhere in the system) so you won't get current which corresponds exactly to the energy absorbed but I suspect that may be too involved of an analysis to take up for this question.
 
  • #5
Borek
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If you know electron charge and Avogadro's number, you already know Faraday's constant - you just don't know its name. Faraday's constant is just a mole of electrons, so it easily links C/s with number of electrons per second. But you don't need even that - think, how many electrons in one Coulomb?
 

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