A photocell is operating in saturation mode with a photocurrent `4.8 mA` when a monochromatic radiation of wavelength `3000 Å`and power of `1mW` is incident. When another monochromatic radiation of wavelength `1650 Å` and power `5mW` is incident, it is observed that maximum velocity of photoelectron increases to two times. Assuming efficiency of photoelectron generation per incident photon to be same for both the cases, calculate.
(a) the threshold wavelength for the cell
(b) the saturation current in second case
(c ) the efficiency of photoelectron generation per incident photon.

Explanation

a. `K_1=(12400)/(3000)-W=4.13 - W `....(i)
`K_2=(12400)/(1650) - W=7.51 - W` ....(ii)
Since `V_2=2V` so `K_2=4K_1` ….(iii)
Solving above equations, we get
`W=3eV`
Therefore, threshold wavelength
`lamda_0=(12400)/(3)=4133 A`
b.Enerby of a photon in the first case `=(12400)/(3000)=4.13eV`
or `E_1=6.6xx10^(-19)J`
Rate of incident photons (number of photons per second)
`(P_1)/(E_1)=(10^(-3))/(6.6xx10^(-19))=1.52xx10^(15)s^(-1)`
Number of electrons ejected
`(4.8xx10^(-6))/(1.6xx10^(-19))xx100=1.97%`
c. Energy of photon in the second case,
`E_2=(12400)/(1650)=7.51eV=12xx10^(-19)J`
Therefore, number of photons incident per second,
`n_2=(P_2)/(E_2)=(5.0xx10^(-3))/(12xx10^(-19))=4.17xx10^(15)s^(-1)`
Number of electrons emitted per second is
`(1.97)/(100)xx4.7xx10^(15)=9.27xx10^(13)s^(-1)`
Therefore, saturation current in the second case
`i=(9.27xx10^(13))(1.6xx10^(-19))A`
`=14.8muA`