Question

A light beam of power 1.5 mW and 400 nm wavelength incident on a cathode. If quantum efficiency is 0.1% then, find out obtained photo current and number of photoelectron per second.

Answer 1

No of photoelectrons can be find by the relation as:

$n=\frac{p\lambda }{hc}$

$p=1.5\times {10}^{-3}W$

$\lambda =400\times {10}^{-9}$

$\therefore n=\frac{1.5\times {10}^{-3}\times 400\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}$

$n=3\times {10}^{15}$

n is the number of photoelectrons per second.

Quantum efficiency is given as:

$I=ne\times QE$

$where,QEisthequantumefficiency$

$I=3\times {10}^{15}\times 1.6\times {10}^{-19}\times \frac{0.1}{100}$

$=0.48\mu A$

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