It's question 2.
Work = U/t
E = hf
f = c/lambda
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.