Question

A metal plate of area $1\times {10}^{-4}{m}^2$ is illuminated by a radiation of intensity $16mW/m^2$.The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be :
$[1eV=1.6\times {10}^{-19}]$

• A. ${10}^{10}$ and 5 eV
• B. ${10}^{12}$ and 5 eV
• C. ${10}^{14}$ and 10 eV
• D. ${10}^{11}$ and 5 eV

Answer 1

The correct option is B ${10}^{12}$ and 5 eV

Maximum Kinetic Energy $K.E{.}_{max}=E-\varphi =(10-5)eV=5eV$

$I=\frac{nE}{At}$
$16\times {10}^{-3}={\left(\frac{n}{t}\right)}_{Photon}\frac{10\times 1.6\times {10}^{-19}}{{10}^{-4}}={10}^{12}$