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_(1)` so , `K_(2)=4K_(1)` (iii)
Solving above equations, we get
`W=3 eV`
`:.` Threshold wavelength `lambda_(0)=(12400)/(3)=4.133A`
(b) Energy of photon in first case `=(12400)/(3000)=4.13 eV`
or `E_(1)=6.6xx10^(-19)J`
Rate of incident photons (number of photon per second)
`=(P_(1))/(E_(1))=(10^(-3))/(6.6xx10^(-19))=1.5xx10^(15)`per second
Number of electron ejected `=(4.8xx10^(-6))/(1.6xx10^(-19))` per second
`:.` Efficiency of phtoelectron generation
`(eta)=(3.0xx10^(13))/(1.5xx10^(15))xx100=2%`
(c ) Energy of photon in second case
`E_(2)=(12400)/(1650)=7.51 eV=12xx10^(-19)J`
Therefore, number of photons incident per second
`n_(2)=(P_(2))/(E_(2))=(5.0xx10^(-3))/(12xx10^(-19))=4.17xx10^(15)` per second
Number of electron emitted per second `=(2)/(100)xx4.7xx10^(15)`
`=9.4xx10^(13)` per second
`:.` Saturation current in second case `i=(9.4xx10^(13))(1.6xx10^(-19))amp`
`=15 mu A`