Photoelectric effect photocells

[alciefrederic]

Homework Statement

Given two photocells, using light sources with identical frequency and intensity, but with different metal anodes with different work functions. Which of the followings are correct:

A. The stopping voltage will be identical in both cells
B. The current will be identical in both cells
C. The stopping voltage will be higher in the cell with lower work function
D. The metal with the larger work function would result in larger kinetic energy of the photoelectrons.

Homework Equations

E(e) = hf - W = 0.5mv^2
I = q/t

The Attempt at a Solution

The answers given in the book are B and C

However B seems dubious as:
E(e) = hf - W , since the photocells have different work functions, the kinetic energy of electrons from the two cells should be different, and so, electrons should have different speed and so, results in different current. However B claims the current to be the same. So I need help in verifying this question (as my textbook has been proven to be untrustable many times before).

 

[Doc Al]

What determines how many electrons will be released per second?

 

[alciefrederic]

I know intensity does. But, as I said, since these two metals have different work functions, their electrons emitted should have different kinetic energy, considered all other things equal, thus different current, or not? (E = VIt, with V and t constant (assumed), the larger E is the larger I, or as I said above, higher kinetic energy, faster, and so higher current)

 

[Doc Al]

I know intensity does. But, as I said, since these two metals have different work functions, their electrons emitted should have different kinetic energy, considered all other things equal, thus different current, or not?

Not. Who cares how much energy the electrons have? Current is the number of electrons per second. If only 100 photons hit the metal each second, then you'll only have 100 electrons emitted per second (or less), regardless of their kinetic energy.

(E = VIt, with V and t constant (assumed), the larger E is the larger I, or as I said above, higher kinetic energy, faster, and so higher current)

This E is the total energy not the energy per electron, which is what we're talking about here. All this says is that if you double the current you'll double the total energy.

 

[alciefrederic]

Wait a minute, isn't the electrical energy produced is the energy that photons that gave (E(e) = hf - W), so if let say, 100 photons were used, 100 electrons emitted, total electrical energy produced should be 100(hf-W) = VIt, no? And since hf - W are different, VIt should be different and so I different?

 

[Doc Al]

Wait a minute, isn't the electrical energy produced is the energy that photons that gave (E(e) = hf - W), so if let say, 100 photons were used, 100 electrons emitted, total electrical energy produced should be 100(hf-W) = VIt, no? And since hf - W are different, VIt should be different and so I different?

The maximum KE per released electron will be hf - W. If you have 100 photons, then the maximum total KE of all the electrons would be 100(hf-W). (Not sure what you are getting at by setting that equal to VIt. I suppose that It would equal 100e, thus VIt = 100Ve. Ve would correspond to the energy per electron, which would be hf-W.)

In any case, if 100 photons hit the cell each second, the current can only be 100 electrons per second regardless of their energy.

 

[alciefrederic]

I see it like this: When photoelectric effect occurs, energy from light sources are used to:
- Free electrons
- Produce electricity
Since the energy from the light sources here are equal, if the amount of energy used to free electrons of one photocell is higher than the other one, then the electrical energy produced of that cell will be less.
Since electrical energy depends directly on either voltage, current or time (E = VIt), assuming these photocells have identical voltage and function time, then we can deduced that the current must be different.

And you said: "I suppose that It would equal 100e, thus VIt = 100Ve. Ve would correspond to the energy per electron, which would be hf-W." (i don't know how to quote, sorry)
Which is precisely my point VIt = 100Ve = 100(hf-W). And since hf - W are different, VIt are different, assuming photocells have the same voltage and all, they will have different current.

 

[Doc Al]

Since electrical energy depends directly on either voltage, current or time (E = VIt), assuming these photocells have identical voltage and function time, then we can deduced that the current must be different.

Don't assume they have the same voltage.

Which is precisely my point VIt = 100Ve = 100(hf-W). And since hf - W are different, VIt are different, assuming photocells have the same voltage and all, they will have different current.

Again, you assume the cells have the same voltage. But no:
100Ve = 100(hf-W)
Ve = hf-W

The voltage depends on hf-W.

 

[alciefrederic]

What if we let E = IR^2t? Since these cells are indentical, they have the same resistance, so, the current should be different.
(And since voltage is the work required for an electron to go through a certain passage, if the two cells are identical, it should require the same amount of energy to go through them, since photons do not increase a photocell's resistance to make it harder to go through, and therefore safe to assume that voltage is identical)