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DISCLAIMER: I have not yet fully entered into the vast field of quantum mechanics and and was reading about the photoelectric effect as a part of chemistry, however I feel, the topic of structure of atom formally comes under physics so I will put forward my question over here. Please answer the question in the most rudimentary way as possible. Although I have completed sufficient amount of classical mechanics and electromagnetism to be able to understand certain terminologies so feel free to include that.

DISCLAIMER 2.0 (edit): the question which is linked is, whether or not the number of photoelectrons increase on increasing frequency.. My question is how does the change in frequency affect the photocurrent. (this was to differentiate the questions) In photoelectric effect experiment, why will the number of photoelectrons not change on increasing frequency?

MAIN QUESTION: Why doesn't the frequency affect the photocurrent? I understand that photointensity effects the photocurrent as number of photons striking in a unit area in unit time is increased, hence number of charged particles increased and hence, current flow increased. I also understand the fact that changing the frequency will only change the energy of the emitted electron and not the number of electrons. But my argument is, as the energy has been increased, so has the velocity, which suggests in unit time, more charged particles will flow. Shouldn't that increase the current flow? As per what I understand kinetic energy of particle increases then stopping potential too increases; current should also increase. Any help would be appreciated.

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    $\begingroup$ Does this answer your question? In photoelectric effect experiment, why will the number of photoelectrons not change on increasing frequency? $\endgroup$
    – Jon Custer
    May 25 at 15:32
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    $\begingroup$ @JonCuster Not actually, the question which you sent is asking, whether or not the number of photoelectrons increase on increasing frequency.. My question is how does the change in frequency affect the photocurrent $\endgroup$
    – Ambro234
    May 25 at 16:24
  • $\begingroup$ @Ambro234 I think my answer or d_b's answer address your particular concern about the velocity of the ejected photoelectrons. $\endgroup$
    – hft
    May 25 at 19:43

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MAIN QUESTION: Why doesn't the frequency affect the photocurrent?

It does affect it, in general. It just does not affect the number of ejected photoelectrons (to zeroth order).

If you google for the emission spectrum of your favorite metal, you will see that the emitted intensity does indeed change with incident frequency. This is because the "optical constants" of a material are not really "constant." They change with frequency. But we ignore this in the elementary discussion of the photoelectric effect for various reasons, some explained below.

But my argument is, as the energy has been increased, so has the velocity, which suggests in unit time, more charged particles will flow. Shouldn't that increase the current flow?

If you were able to measure the current in such a way that the ejected electron velocity matters, then yes.

But in a typical elementary photoelectric effect experiment, you collect the electrons into a collector apparatus that is very blunt. It's slurping up whatever electrons get kicked out of the material and is not really sensitive to their individual (or average) speeds. It is basically counting the number of electrons that are ejected in a given time, not how long it takes for those electrons to get from the material to the collector.


In addition, I think the point of the elementary discussion of the photoelectric effect is that if a single photon has enough energy to kick out a single electron, then increasing the frequency doesn't change that--i.e., a single electron still gets kicked out of the material by a single photon. (Caveat: There are higher order processes that could provide small corrections to this effect.)

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    $\begingroup$ A complication is that the photon is generally absorbed at some depth within the material, and the photoelectron loses energy on its way out. When it reaches the surface, it may not have enough energy to escape. $\endgroup$
    – John Doty
    May 25 at 20:17
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None of the above answers have addressed your point because your point, in fact, does not exist.

I'll give you an analogy.

Consider a highway where you have, in one part of it, a lower speed limit, not bad enough to cause a jam, and assume that cars enter it at the same rate all day long.

The "current", i.e., the number of cars crossing any point of the highway per hour, is the same whether in the regular part, 55 mph limit, as in the slower part, because there is no jam : the number of cars per hour is everywhere the number of cars that enter.

In the slower part, cars move slower, but this causes them to be closer to each other (though not so close as to cause jams). You have the same current but with slower cars, closer to each other. See ?

So this is the same with electrons. The current depend only on the rate the electrons are kicked out of the metal and thus is the same, whatever the frequency. If the frequency is higher, the electrons move faster, but they will just be farther from each other, because the last one to be ejected will have moved farther away from its source if it is faster, before the next one is ejected.

Same current, with faster electrons farther away from each other.

Just like cars on a highway.

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I don't think the other answers really get at the point you are making. You are right that the rate at which electrons reach the collector electrode determines the current. However, changing the frequency of the incident light doesn't change this rate to any measurable degree because regardless of the frequency, the electrons are reaching the collector more or less instantaneously.

Let's look at the time it takes for electrons to travel from an emitting electrode made of sodium to the collector electrode. I will take the distance between the electrodes to be 5 mm as in the original photoelectric experiments of Lenard (see here).

Sodium has a work function of about 2.3 eV, so the cutoff wavelength for a photocurrent to flow is about 539 nm. Let's so we use a wavelength close to but below the cutoff wavelength, say 535 nm. The maximum kinetic energy of the emitted electrons will be about 0.018 eV, so the maximum speed will be about 79 km/s, and the fastest electrons will reach the collector in about 60 ns; the slower electrons will not be far behind.

This is a small enough time that effectively the electrons are getting to the collector as soon as they are emitted. So the only thing that affects the rate of electrons reaching the collector is the rate at which they are emitted, and that only depends on the light intensity. (I am sure someone else who knows more about electrical engineering than me could tell you exactly how slowly the electrons would have to be moving for the travel time to matter.)

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  • $\begingroup$ I think this is still misleading. if you make the simplifying assumption that every electron has the same velocity, the time delay between photon absorption and current registering is just that: a time delay. Even if it's an hour, it's not going to affect the measured current, other than having to wait an hour between turning the light on and recording the current you measure. A distribution of velocities will change that sharp turn-on into a curve, but once you wait long enough for the slow tail to get there the current's still independent of frequency, no? $\endgroup$
    – llama
    May 26 at 17:50
  • $\begingroup$ @llama I did realize after I posted this that even for a simple back of the envelope calculation, it is a little naive. However, I do still think that within this simplified picture, the speed of the electrons affects the current. The current density between the electrodes is just $J = env$, so based on that it seems like the speed matters. But I understand your point and am not sure I’ve convinced myself that this works. $\endgroup$
    – d_b
    May 26 at 18:10
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    $\begingroup$ yeah, that equation only applies if you're not limited by the number of electrons at the input so to speak (or rather $n$ will effectively go down in proportion to $v$ going up with higher photon energy). If you only have one photon leaving the surface per second and you quadruple the photon energy, it's definitely not going to double the current. $\endgroup$
    – llama
    May 26 at 18:15
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The key here is that changing the frequency of the incoming photons changes their energy, and to photoeject an electron from an atom requires a certain precise amount of energy to knock the electron away from the atom. If the photon has too low a frequency, it will not have enough energy and the photoemission current will be zero- no matter how many photons you throw at the atom.

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  • $\begingroup$ True but OP seems to want to know why raising the frequency beyond the threshold for surface electron release will not result in significantly more electrons being released per unit time . . . $\endgroup$
    – Trunk
    May 26 at 10:56
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Most answers here do not address the question of the OP, which was relating to the dependence of the photocurrent on the photon energy/frequency for frequencies already above the photoionization threshold. The OP was not asking for quaentum mechanical explanations of the threshold nor of the connection between photoelectron energy and photon frequency.

The intuitive expectation of the OP that the photocurrent increases with increasing photon frequency is simply incorrect because the electron flux/current depends solely on the number of electrons released per second, which in turn depends on the number of photons absorbed but not on their frequency (assuming the latter is above the ionization threshold). The point is that the flux is proportional not only to the speed $v$ but also to the density $n$ of the particles and both must be inversely proportional to each other because of the particle conservation law (you must get as many particles out of the system than are created in the first place) .

As an analogy, imagine cars from a car park setting off at intervals of 1 minute. It does not matter how fast they are going, if they are all going at the same speed, they will still pass you on the road in 1 minute intervals 10 miles later. It is only that their distance on the road would be larger if they are going faster. The 'current' of the cars does not depend on the speed though. It only depends on the rate they set off from the car park.

However, a further aspect to consider is that in case of high particles densities (like in metals) the photoelectrons won't move in a straight line, but their energy becomes thermalized over very short distances due to collisions with each other as well as other particles. So the energy of the photoelectrons is effectively disordered energy i.e. thermal energy. It is not related to the photocurrent (the net bulk motion of the electrons), which is given by the number of photoelectrons and the externally applied voltage only. As for a a normal current in a wire, one should actually expect the current to get smaller with increasing energy/temperature of the conducting electrons as the resistivity increases.

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The energy of light is calculated by the formula $$E=hf$$ where $h$ is the Planck constant and $f$ is frequency. The intensity of light, on the other hand, is proportional to the number of photons (emitted from the source or reaching to the surface of the material). For example, one x-ray photon (frequencies around $10^{17}\ \text{Hz}$) is more energetic than ten UV photons (frequencies around $10^{15}\ \text{Hz}$) but ten UV photons has a higher intensity than one x-ray photon.

Let's assume that I have a material from which photoelectrons can be emitted by UV light (which means that the energy of the UV light is a threshold for the photoelectric effect for this material and that any other light with higher energy will also work). If I shine UV light onto this material I will have photoelectrons emitted. If I increase the intensity of the UV light the photocurrent increases. But if I shine IR radiation (which is less energetic compared with UV light) to the same material then I will not have photoelectrons emitted from the material. Because the IR light will not have enough energy. In this case increasing the intensity of light is not going to change anything in terms of photocurrent, it will we always zero.

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In the simplest of terms: It is the kinetic energy of an individual photon that determines whether a single photoelectron is released. So it is the number of photons (Intensity) with appropriate kinetic energy that affects photocurrent and the kinetic energy of a photon is proportional to its frequency.

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The key thing is that current is charge per unit time, so the observed current is only a function of the number of electrons reaching the collector, not their energy (or speed).

Imagine a situation where a person is tossing ball bearings at a target, and throws one ball bearing every second. Now imagine a situation where a person has a gun, and shoots the target once every second. In both cases the target is getting hit one time every second.

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Some history first.

The photoelectric effect was the first observed energy exchange that illustrated the discrete (or 'quantized') nature of energy.

This was unlike macroscopic energy exchanges, e.g. collision of billiard balls, where any increase in incident kinetic energy however small (e.g. from the moving ball) always resulted in an increase the energy of the receiving matter (e.g. the post-impact kinetic energy of the stationary ball).

With photonic energy from a light source to a metal surface, it was observed that the light frequency had to be above a minimum 'threshold' value before any resulting current flowed from the metal surface towards a nearby anode.

The explanation proffered for this was that in atomic level interactions, light energy was only transferable to matter via a single charged particle and in discrete amounts - and the size of each such amount (or 'quantum') was in proportion to the light's frequency.

Removing a free electron from a metal surface requires a minimum amount of energy, the latter being related to the electron's potential energy within its associated metal lattice / electron-gas situation plus/minus any kinetic energy it may have in the direction towards/away from the bulk metal. If the incident photon has adequate frequency then an electron is freed from the parent metal and towards the nearby anode. This results in a current being observed in the photo-cell circuit.

You ask: why doesn't a higher frequency (= more energy) in the incident light photons result in more photo-cell current ? Simply because the new (more energetic) photon will still be transferred in whole to a single outer electron - not in part to one electron and in part to another. The 'excess' energy will, as you already understand, be used to increase the kinetic energy of the freed electrons as they advance towards the anode.

With higher than threshold frequencies, we have the same number of electrons escaping from the metal surface per unit time, although each of them has more energy. Current is charge flow per unit time, i.e. the number of photo-electrons passing through a circuit point per unit time - not the speed with which individual charges cross that point. Under steady-state conditions electrons cannot accumulate within the circuit. So the number of photo-electrons passing any point in the photo-cell circuit has to be equal to the number of photo-electrons generated at the metal surface. Therefore photo-cell current is unaffected by increase in the light frequency, although electron velocity increases - and electron density within a volume of the circuit correspondingly decreases - as the 'activating' light's frequency is increased.

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The basic idea is that intensity affects photocurrent(i) not frequency, and for photocurrent to start a minimum “threshold” frequency (t.f.) of the radiation is needed. If the frequency is below the t.f. Then i = 0 irrespective of the intensity. The particle nature of light holds true here, a packet of energy of the photon is entirely supplied to one electron only which is ejected, it is not spread out as a wave. You may want to refer to NCERT class 12 phy Textbook and HC Verna concepts of Physics.

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  • $\begingroup$ All true but if all your answer only states facts and references to explanations - which the OP has probably already read - then it is of little help. Might be better rewording it as a conceptual explanation without equations. $\endgroup$
    – Trunk
    May 26 at 12:47
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    $\begingroup$ Don't recommend anyone NCERT textbooks. They are riddled with mistakes and conceptual errors. $\endgroup$ May 28 at 11:13

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