How Does the Work Function of Cesium Impact Photoelectric Effect Calculations?

[MarkusNaslund19]

Homework Statement

http://www.astro.uvic.ca/~jwillis/teaching/phys215/phys215_assign_2.pdf

It's question 2.

Homework Equations

Work = U/t
E = hf
f = c/lambda
I=Q/t

The Attempt at a Solution

f = c/lambda = (3.00x10^8 m/s)/(530x10^-9 m) = 5.66x10^14 Hz

E = hf = (6.626x10^-34 Js)(5.66x10^14Hz) = 3.75x10^-19 J/photon

Work = 2.00x10^-3 J/s ----> in one second U = 2.00x10^-3 J

#photons = U / E = 5.33 x 10^15 photons

#photo-electrons produced = (5.33 x 10^15)/(10^5) = 5.33x10^10 photo-electrons

Q = (5.33x10^10)(1.60x10^-19 C) = 8.54x10^-9 C

I = 8.54x10^-9 C/s

The thing is I did not use the work function of cesium to solve this. Hence I am unsure if this is right... Any help would be appreciated.

THANKS!

 

[mjsd]

btw, when you say "work", I think you mean "Watts" or more precisely "power".
your method looks good to me.
work function only comes in when you want to know the energy of the photo-electron..but you are only concerned with the amount of them (ie. current) the most you could do is to check that the incoming photon energy can indeed overcome the work function: E_\text{photon} > \phi. I don't think there is any problem with that