The relationship between frequency and the amount of current. (1 Viewer)

[cirs]

In our physics class our teacher explained that when you increase the frequency of light (allready above threshold freq) that the current will increase. More electrons are not emitted but rather electrons are moving with greater velocity (more energy.) and so hit the collector plate more often.

So here is my problem with that. And she did not see what i meant with this. When the electrons move with greater velocity it does not mean that the time beteen the various emmited electrons change.

So my theory is (cant find definite proof):

Thus if you only increase the frequency of the light electrons have more energy but because the current does not increase (electrons per unit time) and the amount of energy that is transferred does increase it means that the voltage across the circuit will increase due to the formula

P = I V
(where P = Energy/ time)

Chris

 

[helper]

Have a look at
PhET Photoelectric Effect - Light, Quantum Mechanics, Photons, Electrons

 

[cirs]

I dont get it. It is telling me that as i increase the frequency the current increases till a maximum, then decreases untill I reach maximum frequency?
Whats up with that?

 

[k02033]

your initial question isnt technically complete. you cant deduce what the current will be given a frequency of light, you also have to talk about its intensity)

current is charge per unit time, so this measurement depends on 2 things. How fast the electrons are travelling and how many electrons are arriving at meansurement location per unit time. (i can still get a huge current when there are many electrons travelling slowly)

the speed is controlled by frequency, and how many there are is dependent on the intensity of the light.

so i guess your question shoudl have been establish the relationship between current and frequency, given a particular intensity, or vise versa.

 

[k02033]

i guess i will answer the q "relation between current and frequency for a given intensity"

lets say we are measuringthe current of the anode, where teh electrons are arriving

intensity (I) is energy per unit area per unit of time arriving at the CATHODE metal
if A is the area of the whole CATHODE metal, then IA is the energy the CATHODE receives per unit of time
now each photon has energy of hf, total energy per unit time is IA, so there are IA/hf number of photons arriving at the CATHODE per unit of time. And since photoelectric effect is a one to one interaction, ie one photon liberates one electron, so we can conclude IA/hf number of electrons arrive at anode per unit of time.

so current =1.6*10^-19 times IA/hf
(charge magnitude) times (how many charges arrive per unit of time)

 

[cirs]

Yea, I see I should have said for given Intensity.

Each proton does have " Energy = h x f " but it needs to overcome the work function first so it arrives at the anode with less energy than the proton had.
According to " E(kinetic) = E - workfunction "

So the energy arriving at the anode is: " E(kinetic) = h * f - (workfunction)"

Also
E=Q×V
So the charge of the electron would be constant (1.60217646 × 10-19 C) and the energy arriving at the anode would increase as the frequency of light increases.

Thus voltage must increase. Current stays constant?

Please do say if you guys see somthing wrong.
Chris

 

[k02033]

wait, i am terribly sorry, i think i mixed up the terms anode and cathode haha, (cathode is the thing that the light is directed at, and electrons sprays off right? )

i guess i will answer the q "relation between current and frequency for a given intensity"

lets say we are measuringthe current of the anode, where teh electrons are arriving

intensity (I) is energy per unit area per unit of time arriving at the CATHODE metal
if A is the area of the whole CATHODE metal, then IA is the energy the CATHODE receives per unit of time
now each photon has energy of hf, total energy per unit time is IA, so there are IA/hf number of photons arriving at the CATHODE per unit of time. And since photoelectric effect is a one to one interaction, ie one photon liberates one electron, so we can conclude IA/hf number of electrons arrive at anode per unit of time.

so current =1.6*10^-19 times IA/hf
(charge magnitude) times (how many charges arrive per unit of time)

so sorry for any confusion, pretty confused myself

 

[cirs]

Yea, thats what the cathode is.

Still not sure if the current increases or not when frequency increases.