photoelectric effect & intensity

[SaintJude]

Why does the intensity of a light beam not increase then energy that each e- possesses?

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[milski]

Increased intensity of the light beam means that it has more photons. The photoelectric effect is an interaction between a single proton and electron. If you add more photons, the interaction does not change but you have more electrons being excited.

 

[TXKnight]

Increased intensity of the light beam means that it has more photons. The photoelectric effect is an interaction between a single proton and electron. If you add more photons, the interaction does not change but you have more electrons being excited.

This. Intensity is a function of the number of photons and energy is a discrete/individual property.

 

[hellocubed]

Increased intensity of the light beam means that it has more photons. The photoelectric effect is an interaction between a single proton and electron. If you add more photons, the interaction does not change but you have more electrons being excited.

I think some of us are thinking that if there are more photons, there will be an increased chance that the SAME electron that was originally hit would get hit several more times.

Why is this incorrect?

 

[milski]

I think some of us are thinking that if there are more photons, there will be an increased chance that the SAME electron that was originally hit would get hit several more times.

Why is this incorrect?

The electron will require a different amount of energy to go from its first to its second excited state. Interactions between electrons and photons are all or nothing - you cannot use just a bit of the energy that the photon has. You will need photons with different energy (wavelength) to make that happen.

 

[chiddler]

The electron will require a different amount of energy to go from its first to its second excited state. Interactions between electrons and photons are all or nothing - you cannot use just a bit of the energy that the photon has. You will need photons with different energy (wavelength) to make that happen.

why can't an electron use all of two photons?

 

[hellocubed]

The electron will require a different amount of energy to go from its first to its second excited state. Interactions between electrons and photons are all or nothing - you cannot use just a bit of the energy that the photon has. You will need photons with different energy (wavelength) to make that happen.

So in the hypothetical situation that an electron is hit by 1000 separate photons immediately, it would be the exact same result as it being hit by just one?


I think I understand.

 

[chiddler]

So in the hypothetical situation that an electron is hit by 1000 separate photons immediately, it would be the exact same result as it being hit by just one?


I think I understand.

explain

 

[milski]

Let's say that you have three energy states: E0, E1 and E2. To move an electron from E0 to E1 you need Δ1 of energy and to move from E1 to E2 you need Δ2. Each of these transitions can be made when an electron interacts with a photon which has exactly that energy - Δ1 or Δ1. If an electron in E0 interacts with a photon which has Δ1 energy, it will move to E1. Hitting after that with more photons which have Δ1 energy will not lead to anything since the amount of energy does not much the energy needed to move to the next state. It's only when photons with Δ2 energy interact with this photon that you'll have it move from E1 to E2.

chiddler: electrons can use only the full energy from a photon. They cannot take only a part of the energy and leave the rest in the photon - that is not a valid state for the photon.

hellocubed: yes. After the first state transition happens, hitting that electron with more photons with the same energy will have no effect on it.

 

[chiddler]

but why can't they use the entirety of two photons?

the sum of thereof can be some number that fits in an energy level, can't it?

 

[theseeker4]

Let's say that you have three energy states: E0, E1 and E2. To move an electron from E0 to E1 you need Δ1 of energy and to move from E1 to E2 you need Δ2. Each of these transitions can be made when an electron interacts with a photon which has exactly that energy - Δ1 or Δ1. If an electron in E0 interacts with a photon which has Δ1 energy, it will move to E1. Hitting after that with more photons which have Δ1 energy will not lead to anything since the amount of energy does not much the energy needed to move to the next state. It's only when photons with Δ2 energy interact with this photon that you'll have it move from E1 to E2.

chiddler: electrons can use only the full energy from a photon. They cannot take only a part of the energy and leave the rest in the photon - that is not a valid state for the photon.

hellocubed: yes. After the first state transition happens, hitting that electron with more photons with the same energy will have no effect on it.

The photon hitting the electron has to have AT LEAST the energy needed to move it to the next energy level. It is true that a photon hitting an electron will not transfer part of its energy and keep the rest, but that doesn't mean that a photon must have exactly Δ1 energy to cause the excitation, it can have greater than Δ1 energy. An electron hit with a photon of (Δ1+1) will still move to the next energy level, the remaining energy will simply dissipate as heat.

 

[theseeker4]

but why can't they use the entirety of two photons?

You are talking about such small particles moving at such high speeds that multiple simultaneous collisions do not occur. When you get down to the quantum mechanics, each photon interacts with an electron separately. There is no situation where two photons can be absorbed by the same electron at the same time; one after another yes, but not at the same time.

 

[hellocubed]

but why can't they use the entirety of two photons?

If Milski is correct, because

Δ1 +Δ1 does NOT equal Δ2.
the variables represent different values.

The photon hitting the electron has to have AT LEAST the energy needed to move it to the next energy level. It is true that a photon hitting an electron will not transfer part of its energy and keep the rest, but that doesn't mean that a photon must have exactly Δ1 energy to cause the excitation, it can have greater than Δ1 energy. An electron hit with a photon of (Δ1+1) will still move to the next energy level, the remaining energy will simply dissipate as heat.

whoa whoa whoa this seems to contradict milski's statement that you can't use part of a photon's E....

 

[chiddler]

You are talking about such small particles moving at such high speeds that multiple simultaneous collisions do not occur. When you get down to the quantum mechanics, each photon interacts with an electron separately. There is no situation where two photons can be absorbed by the same electron at the same time; one after another yes, but not at the same time.

oh this makes sense. thank you.

 

[theseeker4]

If Milski is correct, because

Δ1 +Δ1 does NOT equal Δ2.
the variables represent different values.




whoa whoa whoa this seems to contradict milski's statement that you can't use part of a photon's E....

That is the point, Milski is mistaken that the photon must be exactly Δ1. Δ1 is the threshold frequency, meaning MINIMUM energy to produce an excitation. More energy in a photon can still cause the excitation, the remaining energy doesn't remain in the form of a photon of different E, but the excess energy dissipates as heat.

 

[milski]

but why can't they use the entirety of two photons?

the sum of thereof can be some number that fits in an energy level, can't it?

Because the electron can interact only with one photon at a time, you cannot sum the energies of two photons together.

 

[theseeker4]

This is from http://en.wikipedia.org/wiki/Photoelectric_effect

Mathematical description The maximum kinetic energy

of an ejected electron is given by

where

is the Planck constant and

is the frequency of the incident photon. The term φ =

is the work function (sometimes denoted

), which gives the minimum energy required to remove a delocalised electron from the surface of the metal. The work function satisfies

where

is the threshold frequency for the metal. The maximum kinetic energy of an ejected electron is then

Kinetic energy is positive, so we must have

for the photoelectric effect to occur.[11]

 

[milski]

The photon hitting the electron has to have AT LEAST the energy needed to move it to the next energy level. It is true that a photon hitting an electron will not transfer part of its energy and keep the rest, but that doesn't mean that a photon must have exactly Δ1 energy to cause the excitation, it can have greater than Δ1 energy. An electron hit with a photon of (Δ1+1) will still move to the next energy level, the remaining energy will simply dissipate as heat.

That's only partially true too. If the photon has more energy than the binding energy of the electron, it will make it escape the atom and will make an ion. If its energy is less than that and does not much the energy to move to another energy state nothing will happen. You cannot use only part of the energy of a photon - there is no mechanism to "dissipate that as heat."

 

[milski]

That is the point, Milski is mistaken that the photon must be exactly Δ1. Δ1 is the threshold frequency, meaning MINIMUM energy to produce an excitation. More energy in a photon can still cause the excitation, the remaining energy doesn't remain in the form of a photon of different E, but the excess energy dissipates as heat.

If you are moving between excitation states the energy has to be exact, there would not be photoelectric effect otherwise. Once you have more energy that the ionization energy (the energy that you need to break the electron free from the atom), it can absorb any amount of energy - some of it goes to move up the quantum states and the rests ends up as additional kinetic energy of the electron.

 

[theseeker4]

That's only partially true too. If the photon has more energy than the binding energy of the electron, it will make it escape the atom and will make an ion. If its energy is less than that and does not much the energy to move to another energy state nothing will happen. You cannot use only part of the energy of a photon - there is no mechanism to "dissipate that as heat."

Not necessarily. If you are moving from energy level 1 --> 2 and absorb a little more energy than is needed to move from 1 to 2, but not enough to move it to 3, the electron will still move from 1 to 2.

http://en.wikipedia.org/wiki/Absorption_%28electromagnetic_radiation%29 seems to indicate that there is, in fact, a mechanism for excess energy to dissipate as heat. See bolded below:

In physics, absorption of electromagnetic radiation is the way by which the energy of a photon is taken up by matter, typically the electrons of an atom. Thus, the electromagnetic energy is transformed to other forms of energy for example, to heat. The absorption of light during wave propagation is often called attenuation. Usually, the absorption of waves does not depend on their intensity (linear absorption), although in certain conditions (usually, in optics), the medium changes its transparency dependently on the intensity of waves going through, and the saturable absorption (or nonlinear absorption) occurs.

Light absorbed by matter does generate heat, as observed by any kid with a magnifying glass and a piece of paper in the sun.

If the energy is less than the energy required to move the electron to the next energy level, it is correct that it will not move.

 

[theseeker4]

Milski, I could be wrong about the above, so if you have a source that contradicts me I would gladly accept the correction. I am confident that the energies required were always talked about as being minimum numbers, not absolute (how could you get an EXACT energy like that, after all?) but am not so sure of myself that I am unwilling to concede being wrong

 

[milski]

Not necessarily. If you are moving from energy level 1 --> 2 and absorb a little more energy than is needed to move from 1 to 2, but not enough to move it to 3, the electron will still move from 1 to 2.

http://en.wikipedia.org/wiki/Absorption_%28electromagnetic_radiation%29 seems to indicate that there is, in fact, a mechanism for excess energy to dissipate as heat. See bolded below:
Light absorbed by matter does generate heat, as observed by any kid with a magnifying glass and a piece of paper in the sun.

If the energy is less than the energy required to move the electron to the next energy level, it is correct that it will not move.

That's the whole point - you cannot absorb just a little more energy when you move from E1 to E2. The absorption of energy results in the disappearance of the photon. There is no way to make heat (movement of particles) from the 'left over energy' - you can either convert the whole photon to energy and move the electron one level up or you cannot.

Once the photons carry enough energy to remove the electrons from the atom all this changes - now the electron can absorb enough to leave the atom and use the rest as kinetic energy. The example with the magnified glass and the rest of your quote discuss this situation.

If you could absorb any amount of energy you would not have a photoelectric effect. The energies to move from 2-3, 3-4 and so on are smaller than the energy to move from E1 to E2. That would mean that the electrons can absorb any wavelength shorter than the one that moves an electron from E1->E2 and that increased intensity (more photons) would indeed lead to more absorption (moving more electrons to higher energy states).

 

[milski]

Milski, I could be wrong about the above, so if you have a source that contradicts me I would gladly accept the correction. I am confident that the energies required were always talked about as being minimum numbers, not absolute (how could you get an EXACT energy like that, after all?) but am not so sure of myself that I am unwilling to concede being wrong

I am very sure about what I am saying - it was a big deal when physicist eventually managed to explain the photoelectric effect and came up with the energy quantas.

What they expected was what you are describing - that all energies above a certain number will be absorbed. What was observed was that only specific energies/wavelengths were absorbed up to a point after which everything was absorbed.

I know that just repeating what I am saying is not a very convincing argument. I did try to quickly look up what I have as sources but the information is too scattered and verbose to directly post. I will try to put something a bit more concise and convincing with some quotes but it will take a bit of time and right now is a really bad time for that (finals/end of quarter for me). Do call me out if I don't do something to support my claims in the next few day though.

Getting exact energy - you just have an exact wavelength - that translates to an exact energy for a photon.

 

[theseeker4]

I am very sure about what I am saying - it was a big deal when physicist eventually managed to explain the photoelectric effect and came up with the energy quantas.

What they expected was what you are describing - that all energies above a certain number will be absorbed. What was observed was that only specific energies/wavelengths were absorbed up to a point after which everything was absorbed.

I know that just repeating what I am saying is not a very convincing argument. I did try to quickly look up what I have as sources but the information is too scattered and verbose to directly post. I will try to put something a bit more concise and convincing with some quotes but it will take a bit of time and right now is a really bad time for that (finals/end of quarter for me). Do call me out if I don't do something to support my claims in the next few day though.

Getting exact energy - you just have an exact wavelength - that translates to an exact energy for a photon.

I found what you were looking for: http://en.wikipedia.org/wiki/Absorption_spectra

You are correct in saying transition between energy levels requires a specific, NOT minimum, frequency. It is only in ionization that the photon energy (frequency) has a minimal value.

I was mistaken in thinking the decay of energy states produced specific spectra of light (which it does, corresponding to the energy drop of the electron) but that absorption of energy would require only a minimal energy, not an absolute energy.

 

[milski]

I found what you were looking for: http://en.wikipedia.org/wiki/Absorption_spectra

You are correct in saying transition between energy levels requires a specific, NOT minimum, frequency. It is only in ionization that the photon energy (frequency) has a minimal value.

I was mistaken in thinking the decay of energy states produced specific spectra of light (which it does, corresponding to the energy drop of the electron) but that absorption of energy would require only a minimal energy, not an absolute energy.

Perfect, that will save me some typing. I suspected that Wikipedia will have something appropriate but I did not find it on the first try. Thanks again for looking it up!

It's a really weird concept since we normall work with energy in fairly large quantities and are used to treating it as a non-discrete quantity. It helps if you think about the photon as the energy itself and consider that there should not be parts of photons at the end of the process.

 

[BerkReviewTeach]

I can't recall whether this is from the BR book or the lecture handout, but he gave a great analogy that completely cleared this up for me.

The photoelectric effect is like a vending machine. Let's say it gives sodas.

If you add one dollar, hit the button, you get nothing. So the soda must cost more than a dollar. You add two dollars and hit the button, you get a soda and fifty cents change. From this we conclude that the soda costs $1.50.

So just like the photoelectric effect, we get an equation of: Input = Cost + Change (hfin = work function + KEelectron). The metal is just like an electron vending machine.

So what if we put three dollars into the soda machine? Do we get two sodas or one soda and more change? We get only one soda per purchase, with more change back this time. This is again like the photoelectric effect where we get only one electron per incident photon (if it has enough energy), but with more kinetic energy if the incident photon has greater energy.

The way to get more sodas is not to increase the amount of money we add in, but instead we must increase the number of times we make a purchase. Each purchase is unique, and change cannot be applied to a future purchase. To get two sodas, I need to make two separate $1.50 purchases. This is again like the photoelectric effect, where more photons (purchases of electrons from the surface of the metal) results in more ejected electrons (but with the same kinetic energy for each).

I hope this analogy helps, because at the MCAT level of thinking, it worked well for me and I was glad I knew it on test day.