[calvert11]

Question:
The charge of an electron is -1.602e-19 C. How many microamperes of electrical current are produced by a photoelectric cell that ejects 2.50e-13 electrons each second? How many photons must be absorbed to produce this number of photoelectrons.

Attempt:
I'm not really sure where to begin. This question surprised me since the photoeletric effect was covered only briefly and gives one equation: KE = hv - hv(0).

I know that increasing the intensity of a wave causes more electrons to be ejected from a metal. Beyond that, I don't know how to start. How do I approach this problem?

[Astrokel]

What formula do you know involving current with number of photoelectrons per second?

[calvert11]

What formula do you know involving current with number of photoelectrons per second?

I'm sure I don't know such a formula. The only other equation I have is E=h*v. Can that be used somehow?

[Astrokel]

Do you know Q= It?

[calvert11]

Do you know Q= It?

No, I don't. But are you saying that this question can't be solved using KE= hv-hv(0).

[Astrokel]

yes it can't be solved with the equation because you weren't given the photon informations. So you have to use Q=It

where Q is the charge of total number of electrons in coloumb, I is current in Ampere, t is the time in seconds.

Q=It can also be written as Ne=It, where N is the number of electrons, e the eletron charge(given to you in the question)

the reason im doing this part for you is because you have not learnt the formula yet but there are many more formulas out there to be used in photoelectric effects.

[Borek]

Do you know Q= It?

No, I don't.

I find it hard to believe that you were asked to calculate current and you were not taught current definition.

[calvert11]

I find it hard to believe that you were asked to calculate current and you were not taught current definition.

Investigate the book if you like. Chemistry: Principles & Practice 2e, Chapter 7. There's no mention of currents; I couldn't find it in the index either. That's why I'm thinking it's solvable using that KE equation, or some variant of it.

For example, does this make sense:
Part 1
Multiply the charge with the # of electrons and convert the product to microamperes.

Part 2
Use KE = hv, with KE = to answer from part 1.
Solve for frequency. Use E = hv to solve for Energy. Multiply by 6.022e23 to find # of protons needed to be absorbed.

[Borek]

I find it hard to believe that you were asked to calculate current and you were not taught current definition.

Investigate the book if you like. Chemistry: Principles & Practice 2e, Chapter 7. There's no mention of currents; I couldn't find it in the index either.

So perhaps you should know what the current is from physics course? Haven't you heard about currents, voltages, resistances?

That's why I'm thinking it's solvable using that KE equation, or some variant of it.

For example, does this make sense:
Part 1
Multiply the charge with the # of electrons and convert the product to microamperes.

Part 2
Use KE = hv, with KE = to answer from part 1.
Solve for frequency. Use E = hv to solve for Energy. Multiply by 6.022e23 to find # of protons needed to be absorbed.

Sorry, it doesn't make sense. To calculate current you need equation that deals with the current.

You may as well try to find speed of the Moon using density definition. It won't work.