Photoelectric Effect Question (1 Viewer)

[Alchemist K]

Can someone please explain why increasing the frequency of the incident light (assuming it is above the threshold frequency) does not increase the photocurrent?

 

[jake2.0]

because the frequency of the light determines the energy of each photon, while the intensity of the light determines the number of photons emitted. the photocurrent is basically the number of electrons being liberated, and this is dependent on the intensity not the frequency.

 

[Alchemist K]

So photocurrent does not refer to the amount of charge passing a certain point in the circuit per unit of time? I thought photocurrent was simply the current resulting form the photoelectric effect, and wasn't anything THAT special.

So my problem arises when you give electrons more kinetic energy and thus they have a higher speed.. doesn't this then mean more will pass a given point per unit of time.. and hence you have a higher photocurrent.

 

[menty]

OK my *theory* is this

Frequency does not affect the photocurrent.
Itll emit a photocurrent if KE > work function (so its like a SWITCH)
Because its quantum theory, say the metal can only absorb 1 quanta. If you give it 3 quanta, or 4 quanta, or 100 quanta (thats freq...) , it will only absorb a max of one and reject the others. If u give it 0.5, nothing wil happen.

Now. Im not sure myself about my logic, becasue according to jacaranda pg 215, dotpoint 5. it says "If the energy of the photon is geater than the work fucntion, the additional energy above th work fucntion WILL PROVIDE KE OF THE PHOTOELECTRON" My logic says that energy above the work function is REJECTED.

 

[jake2.0]

energy above the work function is NOT rejected. that energy becomes thr kinetic energy of the electron, use eintein's photoelectric equation

 

[menty]

Then explain why the current is 0

 

[jake2.0]

^sorry i dont quite understand what you are asking

 

[batigol]

So photocurrent does not refer to the amount of charge passing a certain point in the circuit per unit of time? I thought photocurrent was simply the current resulting form the photoelectric effect, and wasn't anything THAT special.

So my problem arises when you give electrons more kinetic energy and thus they have a higher speed.. doesn't this then mean more will pass a given point per unit of time.. and hence you have a higher photocurrent.

although u may giv it more KE, and so they will travel faster, but theres not many electrons 'passin a point per unit time'. the number of electron then, is dependent upon the intensity as explained by einstein. so similarly, if u hav a bigger intensity but lower frequency (providede its >f0) the current magnitude is somewhat similar to what i said in the 1st place.

 

[Captain Gh3y]

Can someone please explain why increasing the frequency of the incident light (assuming it is above the threshold frequency) does not increase the photocurrent?

Increasing the frequency gives photons greater energy, according to E = hƒ. Now, if the frequency is such that the energy of the photon is greater than the work function of the material, an electron will be ejected by each photon absorbed.

If the frequency is increased beyond this cutoff frequency, E_k = hƒ - ø, where ø is the work function. So each electron ejected just has more kinetic energy as it comes out.

Photocurrent, as the name implies, is current, I, measured in amps, which is a measure of how much charge flowing past a fixed point. You are actually RIGHT in saying that photocurrent is just current due to the photoelectric effect.

It's just that the amount of kinetic energy an electron has doesn't affect the amount of 'current' it's worth (i'm not phrasing this very well).

Current is increased by having more electrons flowing past a certain point, and the way to get more electrons, is to have more photons, ie. greater intensity. It doesn't matter whether these photons are just at the cuttoff frequency, or if they're super high energy gamma rays, if they have enough energy to eject electrons, the photocurrent resulting will be proportional to the INTENSITY of the photons.

SImilarly, if the photons don't have enough energy (ie. frequency is too low), all the intensity in the world won't make electrons come out.

(This is off the top of my head and it's midnight, so...)

 

[darkwarrior2]

Intensity of light

if E=hf and c=f x wavelength , then as the intensity of light increases the energy per photon decreases ?

Here is a question from 2003 phys paper

25
A physics student was conducting an investigation on the photoelectric effect. THe student used an infrared laser with a wavelength of 1.55 x 10xy 6.

(b) (3 Marks) When the laser light was shone onto a phot-cell no current was detected. The student increased the intensity of the light but still detected no current.
Explain this observation

I just said that the intesity of light must be greater than the work function (Energy required to release an electron from the surface of a metal to produce the photoelectric effect), is this enough?

 

[richz]

if E=hf and c=f x wavelength , then as the intensity of light increases the energy per photon decreases ?

no, no, no the energy per photon will not decrease, the intensity does not effect the energy of photons. It just affects the amount of photocurrent.

Here is a question from 2003 phys paper

25
A physics student was conducting an investigation on the photoelectric effect. THe student used an infrared laser with a wavelength of 1.55 x 10xy 6.

(b) (3 Marks) When the laser light was shone onto a phot-cell no current was detected. The student increased the intensity of the light but still detected no current.
Explain this observation

I just said that the intesity of light must be greater than the work function (Energy required to release an electron from the surface of a metal to produce the photoelectric effect), is this enough?

well the intensity affects the photocurrent, it doesnt not affect the frequency. Since we know that the work function is the min amount of ENERGY needed to release an electron, if we increase the intensity the photocurrent increases so energy of the photons will not increase so it wont exceed the threshold frequency and so no electrons can be emitted.

 

[Abtari]

Photocurrent Determined By Frequency As Well As Intensity?

light, has been shown to be consisting of particles, i.e. photons. Each photon has an energy that is dependent upon Planck's constant and the frequency of the photon. Now, a photon will liberate a photoelectron from a metal only above a certain frequency, called the threshold freqency. This is because below the threshold frequency, the energy of the photon will be less than the binding energy of the electron. i.e. it won't exceed or equal to the work function and hence cannot liberate the photoelectron.

consider the case when it exceeds the binding energy, (or the work function). IN this case, any extra energy carried by the photon, because of its higher frequency, say because it is UV light or something along those lines, then the additional energy is contributed towards the kinetic energy of the released photoelectron. Thus, photon energy = kinetic energy of photoelectron + work function (said by einstein)

increasing the frequency, does not increase the no. of photoelectrons released because one photon, no matter how much energy it carries can only liberate ONE photoelectron. Photocurrent is basically the rate of flow of charges (recall the basic definition of current). Thus, photocurrent is determined by the no. of charges flowing and the time period in which they 'flow' across a certain point in the circuit. INcreasing the frequency thus does not increase photocurrent cos u haven't increased the no. of electrons emitted. Intensity increases the no. of photoelectrons emitted, hence it would increase the photocurrent.

The point of conjecture for me lies in this:

increasing the frequency increases the Kinetic energy of the photoelectrons...

wouldn't that mean they go faster? i.e. cover the same instant in lesser time...

so would frequency not increase photocurrent then?

 

[helper]

It is because of the rate determining step. The phocurrent is the current leaving the surface of the metal. The number of photons determine how many electrons can leave the surface at a time, so determine the size of the photocurrent.

If only 4 photons hit a surface in a second, only four electrons can leave in a second, no matter what their KE is