Question

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

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

Answer 1

The correct option is C ${10}^{11}\text{and}5\text{eV}$

maximum Kinetic energy $K{E}_{max}=E-\Phi =10-5=5\text{eV}$

Using intensity formula $I=\frac{nE}{At}$
$n=$ no. of photoelectrons
$\Rightarrow 16\times {10}^{-3}=\left(\frac{n}{t}\right)\times \frac{10\times 1.6\times {10}^{-19}}{{10}^{-4}},\Rightarrow \frac{n}{t}={10}^{12}$
So, effective number of photoelectrons ejected per unit time$={10}^{12}\times \frac{10}{100}={10}^{11}$.