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In photoelectric emission, "The number of photoelectrons emitted per unit area per unit time is directly proportional to the intensity of light used." Is this true if the frequency is not held constant? If yes, please explain how. If a light of frequency f and intensity I emits n photoelectrons from a metal surface(of threshold frequency f/4) per unit area per unit time, how many photoelectrons will be emitted from the same metal surface per unit area per unit time when a lights of frequency and intensity 1)f/2, I 2)2f, 2I 3)f/2, 2I are separately used.

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  • $\begingroup$ See physics.stackexchange.com/questions/340592/…. If the energy of a single photon is above the material work function then the number of photoelectrons is equal to (or directly proportional to) the number of photons. For fixed intensity the number of photons is inversely proportional to frequency. For fixed frequency the number of photons is directly proportional to intensity. $N \propto \frac{I}{f}$ $\endgroup$ – Jagerber48 Jan 9 '18 at 15:29
  • $\begingroup$ I would say the statement old true for a source with a given spectrum. Once above threshold. $\endgroup$ – Alchimista Jan 9 '18 at 19:13
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Yes, the number of photoelectrons is proportional to the light intensity. At a fixed frequency.

The rest of this question is impossible to answer. The workfunction threshold is given by the OP as $f/4$, so $f$ should be in the far UV. It is then likely that the photoelectric yield is higher at $f/2$. And that it would be lower at $2f$ (likely above the plasmon cutoff).

But photoemission intensities depend on details of the electronic structure of the material: its density of occupied states and also on the final states reached by vertical transitions in $k$-space. And at $2f$ (eight times the work function) there might be emission from a shallow core level.

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The number of photoelectrons emitted per unit area per unit time is directly proportional to the intensity of light used.

is a correct statement. The frequency of the light is proportional to the energy of the light (through $E=hf$) and therefore only affects the kinetic energy of the photoelectrons coming out from the surface not their numbers.

For example, if you use light with low energy (that is low frequency) than the work function of the metal then you can increase the intensity as much as you want; no photoelectrons would be emitted. On the other hand, if you have light with energy higher than the work function of the metal even a very dim light source would emit photoelectrons.

So, the number of electrons emitted is only proportional to the intensity of the light not the energy (or frequency) as long as the energy of the photons is higher then the work function of the metal.

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  • $\begingroup$ I correct. Frequency does affect the number of photoelectrons coming out from the surface. $\endgroup$ – user137289 Jan 9 '18 at 18:51
  • $\begingroup$ @Pieter will you also tell how? $\endgroup$ – physicopath Jan 9 '18 at 20:49
  • $\begingroup$ It depends... things go up and down with frequency, depends on material, that is what spectroscopy measures. $\endgroup$ – user137289 Jan 9 '18 at 20:53
  • $\begingroup$ well, I guess what we are talking here is not an XPS, ARPES or an UPS measurement. Of course number of electrons emitted from a core level will depend on photon energy through photoemission cross section etc. In the context of photoelectric effect number of electrons emitted from the surface is independent of the frequency of the light. $\endgroup$ – physicopath Jan 9 '18 at 20:58
  • $\begingroup$ Strange belief. Just google "photocathode spectral response" and you will see the curves. $\endgroup$ – user137289 Jan 9 '18 at 21:06

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