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 of
photoelectron increases to two times. Assuming efficiency of photoelectron generation per incident to be same for both the cases, calculate,
a. Threshold wavelength for the cell
b. efficiency of photoelectron generation.[( No. of photoelectrons emitted per incident photon) ×100]
c. Saturation current in the second case
Answers
Answered by
22
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
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
Nikki57:
Didn't understand anything, but Superb :D
Answered by
0
Explanation:
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 of
photoelectron increases to two times. Assuming efficiency of photoelectron generation per incident to be same for both the cases, calculate,
a. Threshold wavelength for the cell
b. efficiency of photoelectron generation.[( No. of photoelectrons emitted per incident photon) ×100]
c. Saturation current in the second case
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