A beam of light of wavelength 400nm and power 1.55 mW is directed at the cathode of a photoelectric cell. If only 10% of the incident photons effectively produce photoelectrons, then find current due to these electrons. (given `hc=1240 eV-nm, e=1.6xx10^(-19)C`)

Options

A. `5 muA`
B. `40 muA`
C. `50 muA`
D. `114 muA`

Answer

c

Explanation

Energy of the electric photon of wavelength `400nm` is : `E=(hc)/lambda=(1240 eVnm)/400=3.1 eV`
No. of such photons in a beam of light of power `1.55mW`
`n=(1.55xx10^(-3))/(3.1xx1.6xx10^(-19))=3.125xx10^(15)s^(-1)`
No. of photoelectrons used to produce photoelectrons per sec `=10%` of n
`=10/100xx3.125xx10^(15)=3.125xx10^(14)s^(-1)`
Current due to such photoelectrons per second
`=(3.125xx10^(14))xx(1.6xx10^(-19))=5xx10^(-15)A`
`=50 muA`