John's Puzzling Question: Why Frequency is Responsible for Photoelectric Effect?

[jaumzaum]

Hi, I'm studying quantum theory now and one thing I couldn't understand is why the frequency (and not the intensity) is the only respon sible for starting the photo eletric effect?

I didn't understand this expression

E total = E0 + E kinetic

E0 = work functio n

[]'s
John

 

[Ken G]

The idea is, a process that can knock an electron out is a quantum process (indeed, this was Einstein's Nobel prize-winning discovery that it is a quantum process), so requires a quantum of light to make it happen. It can't happen a little at a time, like sliding a boulder, it has to be all or nothing, like throwing a ball through a window. So if the photon does not have enough energy (> E0), then no electron comes off at all, regardless of intensity (how many photons).

Also note the logic originally went the other way-- not why it had to work that way because it was a quantum process, but that because it did work that way, it must be a quantum process.

 

[ibysaiyan]

The equation is : Ekmax = hf - w

The respective letters represent the following energies :
Ekmax = Max kinetic energy of an electron
Hf = Energy of a photon
W - work function of the emitting surface/metal /atom

Now the conditions required to eject an electron are the following:
Incoming photon must have energy greater than the work function.
Threshold frequency is the when the incoming photon ejects electron with minimum momentum /Ekmax. This occurs when : Ek max = hf -w , since ekmax = 0 , we end up with hf = w.
Now if the incoming photon has energy less than the work function then no electron is ejected.
^Interestingly the above phenomenon tells us that light is in packets/quanta...Now to answer your question regarding intensity/frequency... that's because since light's energy in packets , photons act individually,each photon hits a single electron.

Also I would like to clear the definition of intensity to you , in case you're thinking it different to how PEF suggests.Here (PFE) when we talk about varying 'intensity' what we are essentially doing is, increasing or decreasing the number of photons /s being hit to the target metal.The only effect this has is change in the value of current between the circuit.

-ibysaiyan

 

[jaumzaum]

But let's say a light has a minimum frequency f0 to make the electron k n ock, and a intensity I. And another light has 1/3 of this frequency (1/3 of the energy per phothon), but an intensity 3I

So if the fisrt light has 1 photon per electron, and a energy h.f0 that makes the electron knock, the second has 3 photons per electron, and consequently a energy 3h.f0/3 = h.f0.

So why the second light can't start photoeletric effect?

 

[Ken G]

So if the fisrt light has 1 photon per electron, and a energy h.f0 that makes the electron knock, the second has 3 photons per electron, and consequently a energy 3h.f0/3 = h.f0.

So why the second light can't start photoeletric effect?

Because what I mean by a "quantum effect" is that the photons cannot "gang up" on the electron-- it's just one photon and one electron at a time, and if the electron isn't ejected from the metal, the metal grabs it again, and the next photon that comes along is back to square one.

 

[ibysaiyan]

But let's say a light f0 to make the electron k n ock, and a intensity I. And another light has 1/3 of this frequency (1/3 of the energy per phothon), but an intensity 3I

So if the fisrt light has 1 photon per electron, and a energy h.f0 that makes the electron knock, the second has 3 photons per electron, and consequently a energy 3h.f0/3 = h.f0.

So why the second light can't start photoeletric effect?

Sorry but I am finding it hard to understand your use of notation.

By f0 do you mean light with no frequency or first photon in a set-up ?
Also if we're talking of one photon then intensity is irrelevant in this case,I maybe wrong...
'Another light' ,there can be other photon, which could be from different sources , UV,IR,etc.. but I don't think it's possible to have 1/3 of frequency here and 2/3 on the other.
Are you suggesting that if two photons which I have just mentioned above hit the same electron overtime, then it should be ejected,provided that 1/3 +2/3 = minimum energy required ? :s

No

 

[jaumzaum]

kkkkkkk I'm sorry , this keyboard sucks (I've just bought it yesterday and it doensnt type some letters correct ly)


Let me try writing it again


In 2 experiments, with the same material. In first we have a frequency f0 (let's say, 9.10^10 Hz) that is the minimum to start photoelectron effect). And a intensity I.

In the second we have another light, with frequency f0/3 (3.10^10), and a intensity 3I.

So in the second case we have 3 photons per electron if in the first case we had 1. So the energy per electron would be the same. So why the second case does not begin photoeletric effect as the 1st?

But I think Ken G has already answered that. 3 photons can't reach AT THE EXACTLY same time a electron, so if one only has not energy enough, it is captured by the metal (is this correct?).

But let's say the energy is not dissipated and as the time pass the metal gets more energy. So after a time all his electrons would be with the same energy right? If the energy reach h.f0 wouldn't the electrons knock? So if it's correct, even with a f0/3 light , after a time we would get each electron charged with h.f0 energy, so why the photoeletric effect does not start ?

 

[Ken G]

If the energy reach h.f0 wouldn't the electrons knock? So if it's correct, even with a f0/3 light , after a time we would get each electron charged with h.f0 energy, so why the photoeletric effect does not start ?

There would always be a chance of what you are suggesting happening, but the chance is very small. The problem is that the electron does not keep the energy the photon gives it, there is a very short timescale for the electron to give up that energy to the metal. So you need multiple photon hits on the same electron during that short time, and it just doesn't happen. It takes one really big hit by a single photon to knock it right out of the metal.

 

[jaumzaum]

But the metal keeps charging as the time passes right?

So even if this short-time photon-hit doesn't happen, after some time the metal gets more charge and ALL its electrons gets the same energy. So after a long time, couldn't each electron reach h.f0 (3 photon per electron, not in the same time). Let's say the photon of that ligh has 10^-14 J of energy. So if the light has a power of 20W, 20.10^14 photons reach the metal each second. Let's say the metal has 10^17 electrons, so after 50 seconds each electron has the qeuivalent of 1 photon of energy. so after 150 second, wouldn't all the electrons has 3 photons?

 

[Ken G]

I'm not sure what you mean, the metal does not increase its charge. The only reason a potential is involved is so they can collect the electrons that leave the metal-- you'd get the same effect with no potential at all, just shine light on the metal. In other words, the photons that knock the electrons out of the metal are something completely different from the voltage that helps detect those electrons once they are knocked off. The light might heat the metal up a little, but you can assume that heat is being conducted away-- no energy is building up over time in the electrons in the metal. If it were, then yes, it would be easier to knock them out, and the energy barrier would drop with time. It just wouldn't be the right experiment to detect the photoelectric effect.

 

[jaumzaum]

Thanks Ken

It was what I mean, build up energy over time. But now I thought about it and the metal as the metal temperature raises, it power of losing energy raises too, and when power of light = power of losing energy, there's no energy charge

And by charge in that answer I mean energy charge (not coulomb charge)

One thing I can't understand yet. By energy conservation, if only a parcel of energy is trasnsformed in kinetic, where does the other parcel go (the work function)? Does it stay in the eletron? If so, in what type of energy?

 

[Ken G]

One thing I can't understand yet. By energy conservation, if only a parcel of energy is trasnsformed in kinetic, where does the other parcel go (the work function)? Does it stay in the eletron? If so, in what type of energy?

That's an excellent question. It's essentially a form of potential energy-- the electrons in metal have a relationship to all those positive ions in the metal, and they don't really like to get separated from them, even though there are lots of other electrons in there too. I don't know exactly but it might be a lot similar to the "latent heat" of evaporating water from a liquid to a gas-- there are interparticle forces that make the molecule want to stay with the rest of the water, and evaporation causes some of the heat to be lost to getting the particle out of the "potential well" of those forces.

 

[jaumzaum]

That's what I was thinking

Couldn't it be a kinetic energy too, but rotational. As you know kinetic energy are rotational and transational, the mv²/2 is the transational and the Iw²/2 is the rotational, that's responsible for the oscillation of the atom (and it's a type of "potential" energy), the heat.

Is it correct?

 

[Ken G]

Potential energy is quite a bit different from either translational or rotational kinetic energy. The latter are both energies of motion, but potential energy is just the ability to do work, so it's pretty much the place where kinetic energy goes when nothing is moving. It has to do with the locations of things, not how they are moving, and those locations (like being in the metal) correspond to work that was done on the particle that first has to be undone to move it somewhere else.

 

[jaumzaum]

But could it be in the form of heat too ? As far as I know, heat is the state of vibration/oscilation of the atoms, So, in the atom we have energy, so it can vibrate (rotate the atom in each self, the electrons in the atom, make a disorderly movement). All this waste energy. But we don't see this energy in pratical. The only explanation I can see to that is that it's heat . So couldn't the energy be heat?

 

[ibysaiyan]

But could it be in the form of heat too ? As far as I know, heat is the state of vibration/oscilation of the atoms, So, in the atom we have energy, so it can vibrate (rotate the atom in each self, the electrons in the atom, make a disorderly movement). All this waste energy. But we don't see this energy in pratical. The only explanation I can see to that is that it's heat . So couldn't the energy be heat?

Which energy, do you mean the left over energy after an electron escapes at velocity 'v' or the work function ?

 

[jaumzaum]

If all energy is transformed in kinetic, all the "left over" will be work function right?

 

[Ken G]

It sounds like the answer is "yes", if I understand your question. As for whether or not some of the work function could be heat, I suppose some of it always would be, as no storage of potential energy is completely efficient.