Photocurrent's dependence on frequency [duplicate]

Question

Sounds like a rookie question, this, but could someone please explain to me why doesn't photocurrent increase when we increase the frequency of the incident radiation? I mean, an increase in frequency would mean that the photons would have higher energy (E=hf) and this increased energy should correspond to the emitted photoelectron. Now since photoelectrons have higher kinetic energy, they would obviously have higher velocity and since they'd have a higher velocity, then acc. to this formula, I=nea(vd), the current should increase but it doesn't (at least, that's what's written in my school textbook). It'd be great if someone could please explain this to me!

Edit: I didn't assume that a higher frequency would knock off multiple electrons. What I am asking can be explained like this. Let's say, for simplicity, that the photodetector is a kilometer away from the photoelectrons, and it shows the current as the number of electrons which reach it each second. Now, let's take a case in which, say, the radiated light emits a total of 10 electrons from a given photosensitive material at a given frequency and intensity, 5 with a velocity of 1 km/s and the other 5 with a velocity of 500 m/s. Obviously, after a second, only 5 electrons would have reached the photodetector and it'd show the current as 5 electrons per second. Now, in another case with the same apparatus, let's increase the frequency of the light without changing its intensity such that the velocities of the emitted electrons roughly double. Even though the emitted electrons are still the same, he velocities of the electrons would now be 2 km/s for 5 electrons and 1 km/s for the other 5 electrons. Now, obviously after a second, all the 10 electrons would have reached the photodetector as compared to only 5 when the frequency was low, and the photodetector would show the current as 10 electrons per second. This certainly contradicts the fact that photocurrent is independent of frequency (given the frequency is above the work function of the photosensitive material) so what I'm asking is simply how to explain this contradiction.

Thank you!

Answer 1

The photocurrent is the steady state current.

Let's take your example and suppose your light source is emitting 10 photoelectrons per second. You can use two frequencies - at the lower frequency electrons travel at 1km/s and at the higher frequency the electrons travel at 2km/s.

When you first turn on the light there is a delay before the first photoelectrons reach the detector. For the lower frequency this delay is one second while for the higher frequency it is 0.5 seconds. So there is a difference between the frequencies when the light is first turned on.

However, once the current has reached a steady state 10 photoelectrons per second are emitted from the metal and 10 electrons per second are received at the detector. It doesn't matter how fast the electrons travel because that only affects the initial delay and not the steady state. That's why there is no difference between the frequencies.

The only way the frequency would make a difference is if the number of electrons emitted per second changed i.e. the quantum efficiency changed. This is what Whit3rd is referring to in his answer. In general the quantum efficiency does change with frequency, but for metals at frequencies close to (and greater than) the work function the changes in the quantum efficiency are negligable.

Answer 2

In some cases, it DOES increase: about one ion per 36 eV, is the rule of thumb for X-rays generating free charges, as for instance in an ion chamber detector. That only happens when the amount of energy in the photon is so high that it generates a cascade of ions when absorbed, though, and the simple photoelectric effect one-photon-produces-one-electron generates higher kinetic energy electrons, but not MULTIPLE electrons (later, a fast enough electron can knock into atoms and free more electrons, but those are secondary to the initial photon-absorption event).

That rule-of-thumb implies a visible light photon (circa 2 eV) doesn't produce multiple free electric charges in a photoelectric interaction, because the secondary collisions won't be energetic enough.

Answer 3

Well, it depends. The photoelectron can have a kinetic energy between zero to
Max. Kinetic energy = hv - (the work function of metal)
It depends on the electrons. You never know how many collisions did it underwent before leaving the metal. So if you strictly say that all the electrons will have maximum kinetic energy then theoretically the photocurrent will definitely increase. But you can't comment strictly in a practical situation.