QJEEM19Ph2fa4: JEE Main 2019 Physics - Class 12 JEEM19Ph2fa4 Solution: A metal plate of area 1 10 –4 m 2 is illuminated by a radiation …

Given that the area of the metal plate $A = 1 \times 10^{-4} m^2$ , and the intensity of the radiation

$I = 16 mW/m^2 = 16 \times 10^{-3} W/m^2$ .

The power $P$ falling on the metal plate is given by the product of the intensity of the radiation and the area:

$P = IA = (16 \times 10^{-3} W/m^2)(1 \times 10^{-4} m^2) = 1.6 \times 10^{-6} W$ .

The energy of each incident photon is given as 10 eV. Converting this to joules (since 1 eV = $1.6 \times 10^{-19} J$ ):

$E_{\text{photon}} = 10 eV = 10 \times 1.6 \times 10^{-19} J = 1.6 \times 10^{-18} J$ .

The number of incident photons per second $n_{\text{photon}}$ is then given by the total power divided by the energy per photon:

$n_{\text{photon}} = P / E_{\text{photon}} = 1.6 \times 10^{-6} W / 1.6 \times 10^{-18} J = 10^{12} \text{ photons/s}$ .

Given that only 10% of these photons actually produce photoelectrons, the number of photoelectrons produced per second $n_{\text{electron}}$ is:

$n_{\text{electron}} = 0.1 \times n_{\text{photon}} = 0.1 \times 10^{12} = 10^{11} \text{ electrons/s}$ .

The work function of the metal is given as 5 eV, which is the energy required to remove an electron from the metal. Therefore, the maximum energy of the photoelectrons $E_{\text{max}}$ is given by the energy of the incident photons minus the work function:

$E_{\text{max}} = E_{\text{photon}} - \text{Work function} = 10 eV - 5 eV = 5 eV$ .