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A photocell is operating in saturation mode with a photocurrent 4.8MuA when a monochromatic radiation of wavelength 3000Å and power 1 mW is incident. When another monochromatic radiation of wavelength 1650Å and power 5 mW is incident, it is observed that maximum velocity ofphotoelectron increases to two times. Assuming efficiency of photoelectron generation per incident to be same for both the cases, calculate,a. Threshold wavelength for the cellb. efficiency of photoelectron generation.[( No. of photoelectrons emitted per incident photon) ×100]c. Saturation current in the second case |
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Answer» KINETIC energy maximum = energy of photon - WORK function(threshold energy) THUS k.E Max = hc/λ -w k.e 1 = E1- w = (12400/3000)-w = 4.13-w k.E2 = E2-w = (12400/1650)- w = 7.51 -w since k.e directly proportional to (velocity)^2 v2 =2v1 thus ke2 =4 ke1 = > 7.51-w = 16.51-4w => 3w = 9 => w =3 threshold energy = hc/threshold wavelength thus threshold wavelength = 12400/3 = 4133 Å E1 = 4.13 EV = 6.6*10^-19 ws photon incident per second = power incident /E1 = 10^-3/6.6*10^-19 = 1.52 * 10 ^15 per second efficiency% = (photo current /photon incident*e)*100 = (4.8*10^-6/1.6*10^-19*1.52*10^15) *100 = (1.97/100)*100 = 1.97% photon incident per second in second case = power incident (P2) /E2 = 5*10^-3 /7.51*1.6*10^-19 =4.16*10^15 saturation current = efficiency*photon incident*e = (1.97/100) *4.16*10^15*1.6*10^-19 = 13.11*10^-6 ampere |
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