1.

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 nearlyA. `9.61 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» Correct Answer - C
Intensity of light
`I = ( "Watt")/("Area") = ( nhc)/( A lambda) rArr "Number of photon" , n = (I A lambda)/(hc)`
`:.` Number of photo electron ` = (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)`


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