(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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Photo-electric Effect

**Physics Forums | Science Articles, Homework Help, Discussion**