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Q. Let a light beam of total intensity $1\, \mu W\, cm { }^{-2}$ falls on a clean iron sample piece with $J$ work function $4.5\, eV$ and of $1.0\, cm ^{2}$ area. Assume that iron sample reflects $96 \%$ of light and that only $3 \%$ of the absorbed energy lies in the violet region which can cause photoemission (of wavelength $250\, nm$ ). Number of electrons emitted will be

Dual Nature of Radiation and Matter

Solution:

Only $4 \%$ of energy is absorbed and only $3 \%$ of this can cause photoemission, part of intensity available for photoemission is
$I'=0.03 \times 0.04\, I_{0}=1.2\, n\,W\, cm^{-2}$
$\therefore$ Number of electrons/second
$=\frac{1.2 \times 10^{-9} W }{h f}=1.5 \times 10^{9}$ electrons per second