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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?

Respuesta :

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

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