(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

It's question 2.

2. Relevant equations

Work = U/t

E = hf

f = c/lambda

I=Q/t

3. 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!

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# Homework Help: Photo-electric Effect

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