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We were taught that the intensity of light is equal to the rate of flow of photons per unit area times the energy of each photon(planck’s constant times its frequency) then would that mean the saturation current is dependent on the frequency of light as well as photon flux and/or the intensity?

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  • $\begingroup$ But here’s the problem: I had this question which had three metals p,q and r having different values of work function and each of those metals was incident with different wavelengths of monochromatic light, the question further stated that the intensities of those different beams of light was the same and then it asked the I-V plots for p,q and r wherein the correct answer was the one in which the saturation currents were different. $\endgroup$ – user10738905 Dec 8 '18 at 10:42
  • $\begingroup$ Please check this link for the entire question enter link description here $\endgroup$ – user10738905 Dec 8 '18 at 10:47
  • $\begingroup$ Duplicate and links within? $\endgroup$ – Farcher Dec 8 '18 at 10:56
  • $\begingroup$ It is dependent on material, surface, and frequency. I was a photoemission experimentalist. I wrote an answer before: physics.stackexchange.com/questions/401526/… $\endgroup$ – Pieter Dec 8 '18 at 21:31
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The following makes gross simplifications which turn out to be wrong, for details see Pieter's answer on a similar question

Under the assumption that all electrons leave the metal with $KE = hf - W$ we would have:

  1. No photo current if $hf \le W$ regardless of intensity.
  2. For fixed $f$ so that $hf > W$, the saturation photo current will increase if intensity is increased because increased intensity -> more photons hitting the cathode in unit time -> more electrons are ejected in unit time.
  3. If you somehow manage to maintain constant intensity while varying frequency: intensity of monochromatic light can be written as $I_{\text{ph}} = \frac{Nhf}{t}$. If you increase $f$, the quantity $\frac{N}{t}$ must decrease to have the same intensity. So less photons strike the cathode in unit time -> less photo current.
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  • $\begingroup$ That is the basic information which I think most books would provide on photoelectric effect but then I wanted to know about photon flux (dN/dt) and its relation to intensity and/or frequency with intensity? $\endgroup$ – user10738905 Dec 8 '18 at 10:49
  • $\begingroup$ My second paragraph addresses exactly this: if dN/dt goes up, so does the current I. $\endgroup$ – Jasper Dec 8 '18 at 11:23
  • $\begingroup$ Would it be correct to say that Intensity=[(dN/dt)hf]/A ? Please explain. $\endgroup$ – user10738905 Dec 8 '18 at 12:15
  • $\begingroup$ yes that would be correct. $\endgroup$ – Jasper Dec 8 '18 at 13:34
  • $\begingroup$ I don’t get it sir. Does the equation not imply intensity is directly proportional to the frequency? Please elaborate sir. $\endgroup$ – user10738905 Dec 8 '18 at 13:45

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