Answer: a) 102.04 nA; b) yes, it is possible to estimate the work by changing the applied anade voltage to the photocell and obtain zero of photocurrent.
Explanation: In order to explain this questios it is necessary to use the following relationship for the power and energy of photon and its number per second.
Power=Energy/time= h*ν/ time= h*ν* number of electrons per seconds
Energy of photon is: h*c/λ= where h and c is the Planck constant and c the speed of light. For λ=632.8 nm the energy is: 1240/632.8=1.96 eV
(1eV= 1.6 *10 ^-19 J)
so, n=Power/h*ν= 2 *10^-3/(1.96*1.6*10^-19)=6.38*10^15 electrons/s
if the efficiency is 10^-4 we have
6.38*10^15*10^-4= 6.38*10^11 electrons/s
we know that the charge of electron is 1.6 *10^-19 C so
we have; 1.6 *10^-19 C *6.38*10^11 electrons/s=
102.08 *10 ^-9 C/s=102.08 nA
Finally if we applied voltage to the anode of the photocell we change the current of protoelectrons and can obtain a voltage ( stopping voltage) for zero current so in this case we have,
h*ν-eV= W where W is the work function so