When a beam of $10.6eV$ photon of intensity $2.0W{m}^{-2}$falls on a platinum surface of area $1.0c{m}^2$ and work function $5.6eV,0.53\%$ of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum kinetic energies (in eV). Take $1eV=1.6\times {10}^{-19}J$.

 

Since intensity of the photon beam is defined as the energy incident per second per unit area of a surface

$I=\frac{E}{At}=\frac{P}{A}$

$\Rightarrow P=\left|A\right|$

$\Rightarrow P=\left(2.0W{m}^{-2}\right)\left(1.0\times {10}^{-4}{m}^2\right)=2\times {10}^{-4}W$

According to the problem, energy carried by each photon is

$E_{\text{single photon }}=10.6eV=(10.6)\left(1.6\times {10}^{-19}\right)J$

$\Rightarrow E_{\text{single photon }}=16.96\times {10}^{-19}J$

So, the number of photons striking the metal surface per second is

$n=\frac{P_{\text{photon beam }}}{E_{\text{single photon }}}=\frac{2.0\times {10}^{-4}}{16.96\times {10}^{-19}}$

$\Rightarrow n=1.18\times {10}^{14}\text{ photons }/s$

Since, only $0.53\%$ of the incident photons are able to eject photoelectrons, so the number of photoelectrons ejected per second is

$n_{\text{photoelectrons }}=\left(\frac{0.53}{100}\right)n=\left(\frac{0.53}{100}\right)\left(1.18\times {10}^{14}\right)$

$\Rightarrow n_{\text{photoelectrons }}=6.254\times {10}^{11}$

So, minimum kinetic energy is zero and the maximum energy of the emitted photoelectron is

$E_{\max }=10.6eV-5.6eV=5.0eV$