Current produced by photolectrons

[Oliver321]

Homework Statement

In perspective to the photoelectric effect I found following question:
Electric current is charge flowing per unit of time. If we increase the kinetic energy of the photoelecrons (by increasing the energy of the incident photons), shouldn’t the current increase, because the charge flows more rapidly? Why doesn’t it?

Homework Equations

E_kin = h*f - φ
φ: work function

The Attempt at a Solution

Maybe if I increase the energy of the photons, the same number of electrons are emitted in the same unit of time but with greater kinetic energy and therefore there is a greater difference between the separate electrons getting emitted? But I have no idea why this should occur ( there should be the same number of photons in a lighthbeam which could knock out electrons).

 

[kuruman]

You are on the right track but you have to say it better. Think of the current as electrons being ejected per unit time. Can you increase the number of ejected electrons per unit time if you increase the energy of the incident photons without changing the number of incident photons per unit time?

 

[Oliver321]

You are on the right track but you have to say it better. Think of the current as electrons being ejected per unit time. Can you increase the number of ejected electrons per unit time if you increase the energy of the incident photons without changing the number of incident photons per unit time?

Thanks for the awnser!

Yes I could. The number of incident photons should be constant, no matter how large the energy of the photon is. But this would mean a increased current? Hmm May you give me another hint :‘D

 

[BvU]

How many electrons can one incident photon knock out of the metal ?

 

[Oliver321]

How many electrons can one incident photon knock out of the metal ?

Only one electron.
But if there are as many high energetic photons as there are low energetic, wouldn’t that mean that the lightbeam consists out of less high energetic electrons?

 

[BvU]

Yes it does. The low-energy ones don't contribute.

If we increase the kinetic energy of the photoelecrons (by increasing the energy of the incident photons), shouldn’t the current increase, because the charge flows more rapidly

Well, it does. What causes the electrons, once liberated from the cathode, to reach the anode ? Their kinetic energy (partially obtained from an incident electron with sufficient energy), the fact that the anode is positive, or both ?

 

[Oliver321]

Yes it does. The low-energy ones don't contribute.

Well, it does. What causes the electrons, once liberated from the cathode, to reach the anode ? Their kinetic energy (partially obtained from an incident electron with sufficient energy), the fact that the anode is positive, or both ?

I would think both. Nevertheless there is also a stopping potential applied to measure the kinetic energy of the electrons. But I think the whole thought experiment should neglect this stopping potential. Because we cannot measure the kinetic energy and the current at the same time (or can we)? When we applied a stopping potential that is equal to the kinetic energy, no electrons would reach the other end and so the current is 0.

 

[BvU]

(or can we)?

Looks as if we can: potential is voltage and current is amount ...

 

[BvU]

fact that the anode is positive

note that if we apply a stopping potential, this means the anode isn't positive wrt the cathode anymore ...

There's lots of threads on https://www.physicsforums.com/search/95675532/?q=photoelectric&o=relevance on PF. Did you inspect a few of them ?

 

[Oliver321]

note that if we apply a stopping potential, this means the anode isn't positive wrt the cathode anymore ...

There's lots of threads on https://www.physicsforums.com/search/95675532/?q=photoelectric&o=relevance on PF. Did you inspect a few of them ?

No I didn’t. But I will!

But what is now the awnser?
Following is true (?):
1. the kinetic energy of the electrons right after the emission is higher
2. there are as many electrons emitted with less energetic photons as there are with high energetic photons
3. Slow electrons get stopped by the stopping potential

I conclude that fact 2. is the one who solves all this? Slow electrons don’t account for the current because the stopping potential stops them?

 

[BvU]

But what is now the answer?

I forgot the question
1. higher than what ?
2. the book says no
3. true

 

[ZapperZ]

No I didn’t. But I will!

But what is now the awnser?
Following is true (?):
1. the kinetic energy of the electrons right after the emission is higher
2. there are as many electrons emitted with less energetic photons as there are with high energetic photons
3. Slow electrons get stopped by the stopping potential

I conclude that fact 2. is the one who solves all this? Slow electrons don’t account for the current because the stopping potential stops them?

You need to figure out what is being kept constant in that question.

If the source is monochromatic (i.e. the light source provides only ONE frequency), then all the photons hitting the cathode will have the same energy. So now, if you increase the energy of every single photon, BUT, keeping the number of photons hitting the cathode per second constant, then will you get a higher current?

Note that you have already said that each photon causes the emission of one electron. So will increasing the energy of the photons make a difference in this standard photoelectric effect scenario?

Zz.