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I did a question in which, the intensity of the incident radiation on a metal surface was kept constant but the wavelength of the photons has been reduced. The question inquired what will be the effect on the maximum photoelectric current? The initial wavelength was smaller than threshold wavelength of the metal surface.

My thinking was since the intensity remains constant, thus the number of photons emitted from the source remains constant and thus the number of electrons emitted from the metal surface. And since number of electrons per unit time isn't changed, the current will remain the same.

However, the answer key stated "Fewer photons (per unit time) so (maximum) current is smaller"

How does decreasing wavelength (equivalent to increasing the energy of photons) result in a fewer photon emission?

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It is the interface between classical electrodynamics and quantum mechanics.

Intensity is a classical electromagnetic wave measure of energy, measured by the average electric field in the wave :

the average intensity for a plane wave can be written

intensity

So for a given value of intensity/energy for a classical wave of frequency $\nu$, there are $N$ number of photons with energy $E=h\nu$. If the intensity is constant and the frequency gets larger , fewer photons are needed to add up to the classical intensity .

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I assume in reality everything is much more complicated, but in the simple model it is quite clear: A (bound) electron interacts with a photon, and can gain (at most) the photon's energy E. If E is large enough, it can overcome the bound and become a free electron. If E gets even larger, then the electron could even overcome an additional potential difference (i.e., we can produce higher voltage), but a single photon will still interact with a single electron only.

So if you wavelength gets shorter (i.e., energy for each photon increases), then given constant intensity you will have fewer photons, i.e., fewer electrons are removed from the material, i.e., smaller current (but those electrons will have higher energy).

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  • $\begingroup$ Can you please clarify, how does a shortened wavelength yet a constant intensity provide fewer number of photons? $\endgroup$
    – mathnoob123
    Commented Mar 22, 2017 at 11:46
  • $\begingroup$ Well, a (probably terribly naive, but hopefully suitable) description could be: Think of intensity as the energy that flows through a specific area per second. In the photon picture, this means: Numer of photons (that pass per second) times energy of each photon. If you increase the energy of each photon while keeping the intensity constant, this means fewer photons. $\endgroup$
    – Jakob
    Commented Mar 22, 2017 at 11:59
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Intensity of light means total energy per unit time per unit area.As in later case wavelenght of light was decreased so the energy of individual photon will increases but the intensity was kept constant so there should Be less number of photon falling per unit time per unit area.Just remember intensity of light is not determined by either the energy of each photon or the number of photon but as combination of both because total energy depend upon total number of photon and energy of each photon.

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  • $\begingroup$ I do not think that is correct. Current is the charge that flows through a surface in some time (1s, say). Whether you shoot 1 lelectron/s through the surface with near light speed, or gently and slowly push 1 electron/s through the surface; the current in both cases will be 1 elementary charge per second. As explained in the other answers, decreased wavelength (at constant intensity) will actually decrease the current. $\endgroup$
    – Jakob
    Commented Mar 22, 2017 at 20:33
  • $\begingroup$ If the wavelength decrease the energy of photon, how will a photon with decreased energy give rise to an increased kinetic energy of electron. $\endgroup$
    – mathnoob123
    Commented Mar 23, 2017 at 10:38

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