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Q. Find the number of photon emitted per second by a $25$ watt source of monochromatic light of wavelength $ 6600\, \mathring{A}$. What is the photoelectric current assuming $3\%$ efficiency for photoelectric effect ?

BITSATBITSAT 2016

Solution:

$ P _{\text {in }}=25 \,W , \lambda=6600 \,\mathring{A}=6600 \times 10^{-10} m$
$ nhv = P$
$\Rightarrow $ Number of photons emitted / sec,
$n =\frac{ P }{\frac{ hc }{\lambda}}=\frac{ P \lambda}{ hc }=\frac{25 \times 6600 \times 10^{-10}}{6.64 \times 10^{-34} \times 3 \times 10^{8}}$
$=8.28 \times 10^{19}=\frac{25}{3} \times 10^{19}$
$3 \%$ of emitted photons are producing current
$\therefore I =\frac{3}{100} \times ne $
$=\frac{3}{100} \times \frac{25}{3} \times 10^{19} \times 1.6 \times 10^{-19}=0.4 A$.