Question

A beam of light has three wavelengths 4121 A, 4972 A and 6216 A with a total intensity of $3.6\times {10}^{-3}W{m}^{-2}$ equally distributed amongst the three wavelengths .The beam falls normally on an area $1.0c{m}^2$ of a clean metallic surface of work function 2.3 eV .Assume that there is no loss of lighy by reflection and that each energetically oapable photon eject one electron .Calculate the number of photoelectrons liberated in two seconds

Answer 1

The maximum wavelength capable of emitting photo-electron is,

${\lambda }_m=\frac{hc}{{\varphi }_o}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{2.3\times 1.6\times {10}^{-19}}=5404A$

So photons with wavelength of $4121Aand4972A$ would be able to eject photo-electrons.

So Intensity of photons of both the wavelengths is, $I=1.2\times {10}^{-3}W{m}^{-2}$

Power, $P=1.2\times {10}^{-3}\times {10}^{-4}{m}^2=1.2\times {10}^{-7}W$

Number of photo-electrons ejected per second by $4121A$ wavelength is, $N_1=\frac{P\lambda }{hc}=\frac{1.2\times {10}^{-7}\times 4121\times {10}^{-10}}{6.626\times {10}^{-34}\times 3\times {10}^8}=249\times {10}^9$

Similarly, Number of photo-electrons ejected per second by $4972A$ wavelength is, $N_2=\frac{P\lambda }{hc}=\frac{1.2\times {10}^{-7}\times 4972\times {10}^{-10}}{6.626\times {10}^{-34}\times 3\times {10}^8}=300\times {10}^9$

So total number of photo-electrons emitted per second is, $N=N_1+N_2=549\times {10}^9$

total number of photo-electrons emitted in two second is, $1098\times {10}^9$