Light from a he-ne laser (wavelength 632.8 nm) falls on a photo-emissive cell with a quantumefficiency of 10-4(the number of electrons emitted per incident photon). if the laser power is 2 mw, and all liberated electrons reach the anode, how large is the current? could you estimate the work function of the cathode material by varying the anodevoltage of the photocell?

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

its a 5.5 n

glad that