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.

Options

A. `0.59muA`
B. `1.16 muA`
C. `0.48muA`
D. `0.79muA`

Answer

C

Explanation

Energy of photon
`E=hv(hc)/(lamda)=(6.6xx10^(-34)xx3xx10^(8))/(4000xx10^(-10))`
`approx5xx10^(-19)J`
So, number of photons emitted per second by light source,
`n_p=(P)/(E)=(1.5xx10^-3)/(5xx10^(-19))=3xx10^(15)`
Now, as only 0.1% of the photons emit electrons, number of photoelectrons emitted per second,
`n_e=(0.1)/(100)xx3xx10^(15)=3xx10^(12)`
But current is the rate of flow of charge i.e.,
`(I)=(q)/(t)=(Ne)/(t)-n_exxe[as(N)/(t)=n_e]`
So, `I=3xx10^(12)xx1.6xx10^(-19)=0.48muA`