Photoelectric Effect: Max Speed Dependency & Graphical Representation

[AlchemistK]

In what manner is max speed of photo electron dependent on the incident ray?

Could i see a graphical representation of it?

I thought it would be directly proportional but it isn't so.

Thank you

 

[sophiecentaur]

Google "Photoelectric effect" and you'll find hundreds of KE vs Photon Energy graphs. KE is related to speed in a very easy way.

 

[AlchemistK]

According to the equation, h( v - v' ) = max. K.E.

where v is the frequency of incident photon and v' is the threshold frequency.

Since h and v' are constant, the frequency of the incident photon must be directly proportional to the kinetic energy, forming a straight line graphical representation. Am i correct?

 

[invic]

Not directly proportional - as long as v' > 0, the graph will intercept the y-axis below 0. But it will be a straight line, at least this is the way I understand it.

 

[AlchemistK]

So if the max kinetic energy is proportional, so is the maximum speed of the ejected photo electron.

Thats what i thought too, but the answer of this problem is a curved graph. (see attachment)


There isn't an upper limit to the maximum velocity a photo electron can have right?

 

[sophiecentaur]

Not directly proportional - as long as v' > 0, the graph will intercept the y-axis below 0. But it will be a straight line, at least this is the way I understand it.

Proportional but not directly proportional.
But speed is proportional to root KE, remember.

 

[jtbell]

Right, KE is proportional to v^2, so v is proportional to the square root of KE.

There is, in principle, an upper limit to the electron speed, namely the speed of light (from relativity theory), but we're only dealing with a few electron volts of KE, which leads to speeds much too small for relativistic effects to be observed.

 

[AlchemistK]

Oh alright. Thanks.

 

[sophiecentaur]

So if the max kinetic energy is proportional, so is the maximum speed of the ejected photo electron.

Thats what i thought too, but the answer of this problem is a curved graph. (see attachment)


There isn't an upper limit to the maximum velocity a photo electron can have right?

I wonder why you are looking for a "maximum velocity". Photo electrons are emitted from the region, a few atoms thick, near the surface of the metal. You can get electrons off any surface if you want to use a high enough energy photon (you can even blast them out of a nucleus with an appropriate gamma photon and you'd call them Beta particles) but is this what you want?

Conservation of energy tells us that the surplus energy of an emitted electron will be equal to the photon energy minus the binding energy (called the Work Function in the context of photo electricity) but taking things to an extreme would not be called "Photoelectric" any more, I think.
For very high energy photons, you might not even get an interaction with a significant number of the conduction electrons (??) so a different graph could result.

 

[AlchemistK]

Imagine this situation, a metal plate, supposedly containing more than two shells of electrons, and a beam of green light incident on it is able to release the electrons.
If instead of the green light an intense indigo light is shone on the plate, can it release more electrons than the green light?

 

[Bloodthunder]

Imagine this situation, a metal plate, supposedly containing more than two shells of electrons, and a beam of green light incident on it is able to release the electrons.
If instead of the green light an intense indigo light is shone on the plate, can it release more electrons than the green light?

Same intensity? No.
But the electrons released can move faster.

 

[ZapperZ]

Imagine this situation, a metal plate, supposedly containing more than two shells of electrons, and a beam of green light incident on it is able to release the electrons.
If instead of the green light an intense indigo light is shone on the plate, can it release more electrons than the green light?

1. A typical photoelectric effect does not involved individual atoms. It is done on metals, which means that there are conduction bands. Electrons in this conduction band are the ones involved in the photoelectric effect (i.e. this is not a core-level photoemission).

2. You need to change ONE parameter at a time, or else you'll confuse yourself. Going from a "green light" to "an intense indigo light" means that you're changing not only the photon energy, but also the intensity! So what is it that you're trying to find the relationship of?

3. I believe the original question has been answered. The max. "speed", corresponding to the most energetic photoelectrons, is definitely dependent on the photon energy. The "number of electrons" emitted is NOT that easy because now, it depends on the nature of the material, resulting on its quantum efficiency. While there is a general trend that the higher the photon energy, the larger the number of photoelectrons emitted, this is not true all the time and for all material. The only comparison that can be done is using the same material and same photon energy, but with different intensity.

Zz.

 

[AlchemistK]

If the electrons are ejected by green light they must also be ejected by indigo light, that is confirmed and the fact that the electrons ejected by indigo light will have more K.E.
It is known that a single photon is sufficient to release an electron if it has the required energy, so a beam with lower intensity will remove as much electrons as a beam with higher intensity, right?

 

[Bloodthunder]

If the electrons are ejected by green light they must also be ejected by indigo light, that is confirmed and the fact that the electrons ejected by indigo light will have more K.E.
It is known that a single photon is sufficient to release an electron if it has the required energy, so a beam with lower intensity will remove as much electrons as a beam with higher intensity, right?

Assuming the light can release the electrons in the first place...
Higher intensity = more photons = more electrons liberated.
Higher energy of EACH photon = higher velocity of EACH electron liberated.

Intensity = number of photons per area
Color = wavelength = energy of photon

 

[AlchemistK]

Higher intensity = more photons = more electrons liberated.

How does a higher intensity lead to more electrons liberated?
Suppose on a metal plate the number of electrons are fixed,for example, n electrons, so if n number of photons with sufficient energy strike those electrons, they will all get ejected, so if light of intensity higher than that is used, it won't matter.

 

[sophiecentaur]

Same intensity? No.
But the electrons released can move faster.

Of course there will be more electrons released. Electrons that couldn't be shifted by the green light will be released by the indigo. I think there is some confusion with the "maximum" idea. There are electrons with many different binding energies (an energy distribution) near the surface. The Work Function corresponds to electrons with the minimum binding energy. Many photons with the threshold energy will interact with the metal but not necessarily release a photoelectron. Many more of the 'indigo' frequency photons will interact with electrons that can be released.

 

[Bloodthunder]

Of course there will be more electrons released. Electrons that couldn't be shifted by the green light will be released by the indigo. I think there is some confusion with the "maximum" idea. There are electrons with many different binding energies (an energy distribution) near the surface. The Work Function corresponds to electrons with the minimum binding energy. Many photons with the threshold energy will interact with the metal but not necessarily release a photoelectron. Many more of the 'indigo' frequency photons will interact with electrons that can be released.

Haha, you got me there. I was assuming that the light could only liberate the outermost electron(s), otherwise, yes, higher energy light would be able to release more electrons as well.

and @AlchemistK: highly unlikely we'll run out of electrons in the material. number of incident photons are too few compared to metal.

 

[sophiecentaur]

I remember in all the notes I took about this sort of thing (a Monk taught us in Latin, just before we did the illuminated manuscripts and plainsong ) involved graphs of energy distribution with a vertical line, to show some critical energy level. Everything to the right would 'do it' (whatever it was) and everything to the left wouldn't. The level, in this case, would correspond to the work function.

Re the number of available electrons - you'd need a photo current of many amps before availability would be a problem, I think.

 

[AlchemistK]

Wait, let me get this straight, the thing is if green light can remove electrons on the surface, indigo light with the same intensity or higher can remove more electrons, from deeper in the atoms of the metal?

(see attachment)

 

[sophiecentaur]

"deeper" is a bit too classical / intuitive a term. Electrons have all sorts of energies - some may be 'deeper' but happen to be traveling towards the surface. Others may be nearer the surface but traveling away. Some may be right next to a metal ion whilst others may be mid-way between a group of them. I don't even like those analogies but perhaps you'll see where I'm coming from on this; they would all have different requirements for energy to remove them. It's statistics at work and nothing like a hydrogen atom model (no 'simple orbitals', for a start).

 

[Drakkith]

Wait, let me get this straight, the thing is if green light can remove electrons on the surface, indigo light with the same intensity or higher can remove more electrons, from deeper in the atoms of the metal?

(see attachment)

Yes. The green light might remove electrons from deeper in the metal, but they will not have enough energy to "escape" the metal before it runs out of energy and is captured by the metal again. Higher frequency light COULD make that same electron posess enough energy to escape to the surface and beyond.

 

[AlchemistK]

Alright, thanks. I didn't think that a shift in frequency of just a few angstroms would make such a difference.

 

[Bloodthunder]

Frequency is in hertz (reciprocal of second). It's wavelength that's in angstroms (or meters, or nanometers).

 

[AlchemistK]

oh..err..yeah wavelength, got that.