A source emits monochromatic light of frequency `5.5xx10^(14)` Hzat a rate of 0.1 W. Of the photons given out, 0.15% fall on the cathode of a photocell which gives a current of `6muA` in an extrnal circuit.
(a) Find the enrgy of a photon.
(b) Find the number of photons leaving the source per second.
(C) Find the percentage of the photons falling on the cathode which produce photoelectrons.

Answer

A, B, C

Explanation

(a)`E = hf= (6.6xx10^(-034))(5.5xx10^(14))`
`=36.3xx10^(-20) J`
=2.75 eV
(b) Number of photons leaving the source per
second,
n = (p)/(E ) = (0.1)/(36.3xx10^(-20)
`=2.75xx10^(-17)`
(c ) Number of photons falling on cathode per
second
`n_1 = (0.15)/(100)xx2.75xx10^(17)`
`=4.125x10^(-14)`
Number of photonelectrons emitting per second,
`n_2 = (6xx10^(-6))/(1.6xx10^(-19)) = 3.75xx10^(13)`
`:. % = (n_2)/(n_1)xx100 = (3.75)/(4.125x10^(14)xx100`
=9%.