## Sapling #4

Kayko Lee 1C
Posts: 79
Joined: Wed Sep 30, 2020 9:39 pm

### Sapling #4

When a metal was exposed to photons at a frequency of 1.30×1015 s−1, electrons were emitted with a maximum kinetic energy of 3.30×10−19 J.
A) Calculate the work function, Φ, of this metal.
B) What is the maximum number of electrons that could be ejected from this metal by a burst of photons (at some other frequency) with a total energy of 6.24×10−7 J?
I was able to get part A, 5.3 x 10^-19, but I'm not sure on how to get the answer for B. I thought I would have to subtract the work function from total energy given in part B but I guess that isn't the case. Thank you for any help!

Kamille Kibria 2A
Posts: 65
Joined: Wed Sep 30, 2020 9:52 pm

### Re: Sapling #4

for part b you actually have to divide the total energy by the energy of one photon. since you are already given the frequency, to find the energy of one photon you have to use the equations c=lamda(v) and E=hv. once you find the energy of one photon, take the total energy of 6.24x10^-7 J (given) and divide it by the energy which you have just solved for. this will give you the number of photons. hope that helps :)