`10^-3` W of `5000A` light is directed on a photoelectric cell. If the current in the cell is `0.16muA`, the percentage of incident photons which produce photoelectrons, is

Options

A. `40%`
B. `0.04%`
C. `20%`
D. `10%`

Answer

B

Explanation

Number of photons falling per second :
`N_p=(10^-3)/((6.6xx10^-34xx3xx10^8)/(5000xx10^-10))=2.5xx10^15`
Let `N_e` is the number of photoelectrons emitted per second.
`I=(q)/(t)=(N_ee)/(1)impliesN-e=(I)/(e)=(0.16xx10^-6)/(1.6xx10^-19)=10^12`
percentage of photons producing photoelectrons
`=(N_e)/(N_p)xx100=(10^12)/(2.5xx10^15)xx100=0.04%`