Ive understood that the saturation current is the current obtained in a photoelectric experiment when all the photo electrons emitted at the cathode are able to pass through the repulsive space charge and eventually make it at the anode. Although i don't understand the exacts factors on which the saturation current depends on: a) Frequency of incident light? b) Intensity of light?

For example, take the below graph of a photoelectric experiment on the same metal. What can we comment about the frequency and intensity of A B C and why? Thank you!

enter image description here

  • $\begingroup$ What do you think? What role do the frequency and the intensity play in the PE? Sorry, nobody is gonna do your homework for you ;) $\endgroup$
    – Samuel
    Commented Mar 15, 2023 at 16:46
  • $\begingroup$ Guess I'm wrong. $\endgroup$
    – Samuel
    Commented Mar 15, 2023 at 16:57

1 Answer 1


Here is a simple rule(not exactly rule but it helps me remember this):

1. In photoelectric effect the intensity decides the number of electron only and more electron emitted more will be the photocurrent

2. The frequency decide only the energy(Kinetic Energy) with which the electrons will be emitted.(means it would require more -ve potential $V_s$ to stop the electrons)

So, more is the intensity more will be the photocurrent

Now coming to your question:

a) From graph you can see that stopping potential is same for A and B so their frequency will be same which will be less than frequency of C as for C it require more $V_s$ to release the electron as compare to B and C

b) Coming to Intensity from graph you can see that for B and C photocurrent is same and more than photocurrent of A so intensity of B and C should be more(as I said in rule 2) as compared to intensity of A

You can comment if you didn't understand or have confusion in any statement


  • $\begingroup$ Intensity is defined as energy per unit area per unit time. Current of B and C is same so number of photoelectrons is same, so number of incident photons should be same for both cases (100% efficiency of photons) , but since C has more energetic photons, shouldn't the intensity of C be more than B? (same number of photons and energy of C photons > energy of B photons) $\endgroup$
    – Shridp
    Commented Mar 16, 2023 at 3:46
  • $\begingroup$ intensity only decides number of electron to be emitted and the energy with which they will be emitted is decided by frequency.. Photocurrent will depend upon number of electron striking at collectors plate (cause electron have same charge regardless of speed or energy) so if current is more then more electrons should strike the collector plate and in order to increase number of electron more intensity is required so that's why B and C have more intensity.. Photoelectric effect was a experiment you can see here en.wikipedia.org/wiki/Photoelectric_effect $\endgroup$
    – Shardul
    Commented Mar 16, 2023 at 3:53
  • $\begingroup$ But intensity is defined as energy per unit area per unit time right, so why is it being compared to only number of photons, it also depends upon frequency. Intensity = nhv/At (Also, the unit of intensity is watt/m^2, thus it must also depend upon frequency of light, and not just the number of photons) $\endgroup$
    – Shridp
    Commented Mar 16, 2023 at 4:53
  • $\begingroup$ if frequency is high that means the energy of incident photon is high (one photon will transfer energy to only 1electron means for every single photon only 1 electron will be emitted) if frequency is higher it means that the energy of each photon is also higher so it will transfer more energy to electron and by definition intensity is the energy transferred to a surface per unit area per unit time by the photons striking on the surface more intensity means more photon are striking the metal surface per unit area $\endgroup$
    – Shardul
    Commented Mar 16, 2023 at 6:53

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