1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1% of the incident photons produce photoelectrons, find the current in the cell.

Answer

A, D

Explanation

Number of photon incident per second
`= (Power)/(Energy of one photon)`
`(P)/((hc//lambda))`
Number of electrons emitted per second
=0.1 % of
`(Plambda)/(hc) = (Plambda)/(1000hc)`
`:. Current = charge (on photoelectrons per second)`
`=(Plambdae)/(1000hc)`
`=(1.5xx10^(-3) (400xx10^(-9)) (1.6xx10^(-19))/((1000)(6.63xx10^(-34))(3xx10^8))`
`= 0.48xx10^(-6) A = 0.48 muA.`