A photocell is operating in saturation mode with a photocurrent 4.8 mA when a monochromatic radiation of wavelength `3000 Å`and power of 1wM is incident. When another monochromatic radiation of wavelength `1650 Å` and power 5wM is indcident, 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.

Answer

A, B, C, D

Explanation

(a) `K_1 = (12375)/(3000) -W …..(i)`
`K_2 = (12375)/(1650) - W….(ii)`
`v_2 = 2_(v1) :. K_2 - 4K_1……(iii)`
Solving these equations, we get
W = 3 eV
:. Threshold wavelength,
`lambda_0 = (12375)/(3)`
=4125 Å
(b) `E_2 = (12375)/(1650) = 7.5 eV = 12 xx10^(-19) J`
Therefore, number of photons incident per
second
`n_2 = (P^2)/(E_2) = (5.0xx10^(-3)/(12xx10^9-19)`
`4.17xx10^(15) per second`
Number of electrons emitted per second
`(eta = 5.1%)`
`=(5.1)/(100)xx 4.17xx10^(15)`
`=2.13 xx10^(14)` per second
`:.` Saturation current in second case
`i = (2.13xx10^(14)(1.6xx10^(-19) A`
`= 3.4 xx10^(-5) A`
`=34 etaA`
(c ) Energy of photon in first case,
`=(12375)/(3000)`
=4.125 eV
or `E_1 = 6.6xx10^(-19) J`
Rate of incident photons
`=(P_1)/(E_1) = (10^(-3))/(6.6xx10^(-19))`
`=1.52xx10^(15)` per second
Number of electorns ejected
`=(4.8xx10)^(-3)/(1.6xx10)^(-19)` per second
`=3.0 xx10^(16)` per second
:. Efficiency of photoelectrons generation
`=(1.52xx10)^(15)/(3.0xx10)^(16) xx100`
= 5.1%.