Ultraviolet light of wavelength 300nn and intensity `1.0Wm^-2` falls on the surface of a photosensitive material. If one per cent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of 1.0 `cm^2` of the surface is nearly

Options

A. `9.6 xx 10^(14) per sec`
B. `4.12 xx 10^(13) per sec`
C. `1.51 xx 10^(12) per sec`
D. `2.13 xx 10^(11) per sec`

Answer

C

Explanation

Intensiyt of light
`I = ("Watt")/(" Area") = (nhc)/(A lambda) rArr` "Number of photon" n = (IA lambda)/(hc)`
`:.` "Number of photoelectron" ` = `(1)/(100) xx (IA lambda)/(hc)`
`= (1)/(100) (1 xx 10^(-4) xx 300 xx 10^(-9))/(6.6 xx 10^(-34) xx 3 xx 10^(8)) = 1.5 xx 10^(12)`