Concept about photoelectric effect

[songoku]

Homework Statement

Please see below

Relevant Equations

##hf=\phi+KE_{max}##

I have trouble understanding these two separate questions:

I know the answer for this one, it is (D). The KE of electron depends on the frequency of the light so since the frequency of the light does not change (I assume so), the KE wil stay the same. Increasing the intensity will increase the no. of photons so the no. of eletrons emitted will also increase.

But for the next question:
"The frequency of the incident light is increased but the intensity remains constant. Explain why this increase in frequency results in a change to the maximum photoelectric current (saturation current)"

I don't understand why there is change in photocurrent while the intensity is constant. My answer would be the current does not change but the electrons will move with higher speed.

Why can't I use the same logic as the 1st question? Or maybe there is something wrong with my assumption in 1st question (where the frequency is constant)?

Thanks

 

[haruspex]

What is the source of these questions and answers?

 

[apostolosdt]

We cannot expect photoemission to occur on identical orbitals (shells) only. Although you mention nothing about metal substrate, increasing the frequency may allow emissions from deeper subshells. That’s too generic an assumption, as more info is needed.

 

[apostolosdt]

On a second thought, though, since you give no clue about the level of your source, one may propose that with higher energy, more electrons can overcome the surface barrier, collectively described by the ##\phi## potential.

 

[nasu]

If you have the same intensity and higher frequency there are fewer photons reaching the surface in any time interval. So, fewer electrons are ejected. Saturation current means that all the ejected electrons are collected. Their KE does not matter. But their number does.

 

[apostolosdt]

If you have the same intensity and higher frequency there are fewer photons reaching the surface in any time interval. [...]

That higher frequency presumably refers to the energy of the photons as acquired by them on the source; not to the rate of the photons emitted by the source. That rate of emission is the intensity of the source and depends on the latter's physical/chemical characteristics, don't you think? Or am I misreading your post? Could you clarify?

 

[Steve4Physics]

That higher frequency presumably refers to the energy of the photons as acquired by them on the source; not to the rate of the photons emitted by the source. That rate of emission is the intensity of the source and depends on the latter's physical/chemical characteristics, don't you think? Or am I misreading your post? Could you clarify?

Hi @apostolosdt. You may be misinterpreting the context/wording of the question.

We have a (nearly) monochromatic light (photon) source, e.g. a laser.

We have a (let’s say) metal with a work function ##\phi##. The work function is the minimum amount of energy needed to remove a conduction electron (not an electron bound to an atom) from the metal’s surface (or from just below the surface). Not a rigorous definition though!

The laser beam is incident on the metal’s surface. The ‘intensity’ refers to the intensity of the light incident on the metal’s surface; it is rate of arrival of energy per unit area at the metal’s surface.

We now increase the photon frequency by some factor k (e.g. use different laser). Consequently the photon energy increases by a factor k.

To keep the intensity unchanged, we need to reduce the number of photons hitting the surface each second (e.g. with a filter) by the factor k.

 

[apostolosdt]

What is the source of these questions and answers?

Could the OP provide just that? Otherwise, the discussion may prove to be fruitless.

 

[hutchphd]

more electrons can overcome the surface barrier,

Why would this occur? The work function is relative to infinity.... not a localized potential barrier. Please explain.

 

[nasu]

That higher frequency presumably refers to the energy of the photons as acquired by them on the source; not to the rate of the photons emitted by the source. That rate of emission is the intensity of the source and depends on the latter's physical/chemical characteristics, don't you think? Or am I misreading your post? Could you clarify?

This is what I understood by frequency in my post. Not the "the rate of emission of photons" which is related to intensity. I have never seen the 'frequency of light" to refer to rate of emission of photons.

 

[apostolosdt]

Why would this occur? The work function is relative to infinity.... not a localized potential barrier. Please explain.

I still believe we should hear from the OP about the level of explanations he/she is expecting. If, as I'm guessing, that is at best freshman-level, then a discussion around the original expression of a photon's energy ##h\nu## should suffice, and so all the previous posts as well.

Otherwise, one should be careful not to take the relation ##h\nu = \phi + {\rm KE}_{\rm max}## too literally. It is only a book-keeping formula for some energies involved. Discussions about photoemission have appeared in the past. The phenomenon is a many-body effect on relaxed metallic surfaces; the latter can also be polycrystalline and never without any gaseous contamination, however of low-concentration. That work function was just some parameter representing the 1920s' ignorance on crystal physics; people needed at least a decade before they started recognizing the crystalline state, and another twenty years before they introduced quantum field theory concepts in condensed matter.

These are well known topics, of course, but also useless for a freshman's question about photoemission.

 

[songoku]

What is the source of these questions and answers?

From practice questions given by teacher, answer also given. The level of the question is high school.

For 1st question, answer key is (D). For 2nd question, the answer is something like the photon per second will be lower so the photocurrent will decrease.

This is what in my notes:

So I am confused why the answer for 2nd question is not be the current does not change but the electrons will move with higher speed.

Is there something wrong with my notes?

Thanks

 

[Steve4Physics]

So I am confused why the answer for 2nd question is not be the current does not change but the electrons will move with higher speed.

You have to remember that the current you are considering is the saturation current.

A few made-up example figures (no claims to be realistic) may help.

Suppose ##10^8## photons/second hit a metal surface. If the photoemission efficiency is 1% then ##10^6## electrons/second are knocked-out. If you collect all of them - i.e. measure the saturation current - you get a current of ##10^6## electrons/second.

The speed of the electrons as they leave the surface – slow or fast – is irrelevant. The important thing is that you collect all of the emitted electrons (by applying a sufficiently large collection-voltage which, in fact, accelerates them).

You now double the frequency so the photon energy is also doubled (E=hf). To keep the intensity the same, you must now have only ##5 \times 10^7## photons/second hitting the metal surface.

Assuming the efficiency is unchanged, that means ##5 \times 10^5## electrons/second are knocked-out. If all of them are collected, the current is now ##5 \times 10^7## electrons/second. So the current has been halved.

 

[songoku]

You have to remember that the current you are considering is the saturation current.

A few made-up example figures (no claims to be realistic) may help.

Suppose ##10^8## photons/second hit a metal surface. If the photoemission efficiency is 1% then ##10^6## electrons/second are knocked-out. If you collect all of them - i.e. measure the saturation current - you get a current of ##10^6## electrons/second.

The speed of the electrons as they leave the surface – slow or fast – is irrelevant. The important thing is that you collect all of the emitted electrons (by applying a sufficiently large collection-voltage which, in fact, accelerates them).

You now double the frequency so the photon energy is also doubled (E=hf). To keep the intensity the same, you must now have only ##5 \times 10^7## photons/second hitting the metal surface.

Assuming the efficiency is unchanged, that means ##5 \times 10^5## electrons/second are knocked-out. If all of them are collected, the current is now ##5 \times 10^7## electrons/second. So the current has been halved.

What I understand so far is:

For my 1st question ##\to## frequency is constant (presumably) but intensity increases so rate of photon increases

For my 2nd question ##\to## frequency increases but intensity stays the same so rate of photon needs to decrease

I have some other questions to clear my doubts:
1)
for 2nd question, are the emitted electrons moving faster? My answer would be yes (based on photoelectric formula) because energy of photon is higher but work function is constant so KE of electrons should increase

2)
I want to ask about my notes in post #12: "The photoelectric current depends on intensity of the light because as the number of photons that hit the metal increase, there will also be increasing in number of electrons emitted so that the photoelectric current will increase".

Is this concept correct only if the intensity increases but frequency stays constant?

3)
If the frequency increases, does it have any effect to intensity or rate of incoming photon? I think increasing the frequency will also increase the intensity of the light but does not change the rate of incoming photon so the photocurrent also does not change.

Thanks

 

[Steve4Physics]

For my 1st question frequency is constant (presumably) but intensity increases so rate of photon increases

Agreed.

For my 2nd question frequency increases but intensity stays the same so rate of photon needs to decrease

Agreed.

I have some other questions to clear my doubts:
1)
for 2nd question, are the emitted electrons moving faster? My answer would be yes (based on photoelectric formula) because energy of photon is higher but work function is constant so KE of electrons should increase

Yes. The electrons leaving the metal surface are moving faster. (And this is true whenever frequency increases - whatever happens to the intensity (as long as it's non-zero).)

2)
I want to ask about my notes in post #12: "The photoelectric current depends on intensity of the light because as the number of photons that hit the metal increase, there will also be increasing in number of electrons emitted so that the photoelectric current will increase".

Note that we must (to avoid ambiguity) refer to the saturation current. (A photoelectric current's value can be smaller than the saturation current if the collecting-voltage is too small.)

Is this concept correct only if the intensity increases but frequency stays constant?

No - these are not necessary conditions.

The photoelectric saturation current increases if 2 conditions are both met:

a) the photon frequency does not fall below the threshold frequency (if this happened, there would be no photoemission);

b) the number of incident photons/second increases.

So, for example, the intensity could be reduced by 10% and the frequency could be reduced by 20%. Providing condition a) is still met, the number of photons/second has increased - so the saturation current increases (surprisingly!).

3)
If the frequency increases, does it have any effect to intensity or rate of incoming photon? I think increasing the frequency will also increase the intensity of the light but does not change the rate of incoming photon so the photocurrent also does not change.

It depends on the equipment. E.g. if you are using coloured filters, changing the filter changes the frequency and could increase or decrease the rate of photons.

To tell if the saturation current will increase, you need to know what happens to both the intensity and to the frequency.

If the frequency doesn't change, then clearly the number of photons/second is proportional to intensity. Increasing intensity then increases saturation current.

If the intensity doesn't change, then the number of photons/second is inversely proportional to frequency. Increasing frequency then decreases saturation current.

But if both the frequency and intensity change, then you need to find if the number of photons/second increases or decreases (or stays the same) for your particular situation.

Note, the term ‘intensity’ is being used rather loosely. The correct meaning is the power per unit area normal to a surface. We can pretend that the normal incident area of the metal is unit area. Then

Number of photons/second = ##\frac {\text {intensity}}{\text {photon energy}}##.​

Edit - typo's and minor improvements.

 

[songoku]

No - these are not necessary conditions.

The photoelectric saturation current increases if 2 conditions are both met:

a) the photon frequency does not fall below the threshold frequency (if this happened, there would be no photoemission);

b) the number of incident photons/second increases.

So, for example, the intensity could be reduced by 10% and the frequency could be reduced by 20%. Providing condition a) is still met, the number of photons/second has increased - so the saturation current increases (surprisingly!).


It depends on the equipment. E.g. if you are using coloured filters, changing the filter changes the frequency and could increase or decrease the rate of photons.

To tell if the saturation current will increase, you need to know what happens to both the intensity and to the frequency.

If the frequency doesn't change, then clearly the number of photons/second is proportional to intensity. Increasing intensity then increases saturation current.

If the intensity doesn't change, then the number of photons/second is inversely proportional to frequency. Increasing frequency then decreases saturation current.

But if both the frequency and intensity change, then you need to find if the number of photons/second increases or decreases (or stays the same) for your particular situation.

Note, the term ‘intensity’ is being used rather loosely. The correct meaning is the power per unit area normal to a surface. We can pretend that the normal incident area of the metal is unit area. Then

Number of photons/second = ##\frac {\text {intensity}}{\text {photon energy}}##.​

Edit - typo's and minor improvements.

Is what in my notes incomplete? Because if the intensity increases, I also need to know about the frequency to be able to draw any conclusions about the saturation current.

Thanks

 

[Steve4Physics]

Is what in my notes incomplete? Because if the intensity increases, I also need to know about the frequency to be able to draw any conclusions about the saturation current.

Your Post #12 notes aren’t quite complete/accurate IMO. For example, compare this (improved IMO) version to the Post #12 version.

For a given material (work function ##\phi##) and light of a given frequency (##f##), the maximum kinetic energy of a photoelectron (##KE_{max}##) does not depend on the intensity of the light. That’s because each photon carries the same energy (##hf##) and ##KE_{max}## depends only on photon energy and the work function: ##KE_{max} = hf - \phi##.​

​

The photoelectric saturation current is proportional to the number of incident photons per second (##N##). ##N## depends on both the intesity and the photon energy of the light. For a fixed photon energy (frequency), ##N## is proportional to the intensity of the light; in this case the saturation current is proportional to the intensity.​

 

[songoku]

I understand.

Thank you very much for all the help and explanation haruspex, apostolosdt, nasu, Steve4Physics, hutchphd