A beam of light has three wavelength $4144Å, 4972Å$, and $6216Å$  with a total intensity of $3.6\times {10}^{-3}W{m}^{-2}$ equally distributed amongst the three wavelength The beam falls normally on $1.0c{m}^2$ area of a clear metallic surface of work function $2.3eV$ . Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons emitted in two second

Threshold wavelength for the metal having a work function of $2.3eV$ is

${\lambda }_{th}=\frac{12375}{2.3}Å=5380Å$

So, only the wavelengths $4144Å$and $4972Å$ will be able to emit electrons from the metal surface because they are less than the threshold wavelength.

Since the intensity is equally distributed among the three incident wavelengths, intensity corresponding to each wavelength

$I=\left(\frac{I_{\text{total }}}{3}\right)=1.2\times {10}^{-3}W{m}^{-2}$

The energy incident per second (i.e. power incident on the surface for each wavelength is)

$P=IA$

$\Rightarrow P=\left(1.2\times {10}^{-3}W{m}^{-2}\right)\left(1.0c{m}^2\right)$

$\Rightarrow P=\left(1.2\times {10}^{-3}\right)\times \left({10}^{-4}\right)W=1.2\times {10}^{-7}W$

Energy incident on surface for each wavelength in interval of 2 seconds is

$E=Pt=\left(1.2\times {10}^{-7}\right)\left(2\right)=2.4\times {10}^{-7}J$

The number of photons $\left(N\right)$ in a light beam of energy $E$ having photons of wavelength $\lambda$ are given by

$N=\frac{E_{\text{light beam }}}{E_{\text{single photon }}}=\frac{E}{\left(\frac{hc}{\lambda }\right)}=\frac{E\lambda }{hc}$

Number of photons $N_1$ due to wavelength $4144Å$

$N_1=\frac{\left(2.4\times {10}^{-7}\right)\left(4144\times {10}^{-10}\right)}{\left(6.63\times {10}^{-34}\right)\left(3\times {10}^8\right)}=0.5\times {10}^{12}$

Number of photons $N_2$ due to wavelength $4972Å$ is

$N_2=\frac{\left(2.4\times {10}^{-7}\right)\left(4972\times {10}^{-10}\right)}{\left(6.63\times {10}^{-34}\right)\left(3\times {10}^8\right)}=0.6\times {10}^{12}$

So, total number of photons $\left(N\right)$ is given by

$N=N_1+N_2=0.5\times {10}^{12}+0.6\times {10}^{12}$

$N=1.1\times {10}^{12}$ = Number of photoelectrons emitted.