A light of intensity 16 mW and energy of each photon 10 eV incident on a metal plate of work function 5 eV and area `10^(-4)m^(2)` then find the maximum kinetic energy of emitted electrons and the number of photoelectrons emitted per second if photon efficiency is `10%`.

Options

A. 5 eV, `10^(11)`
B. 10 eV, `10^(12)`
C. 5 eV, `10^(13)`
D. 10 eV, `10^(14)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Maximum Kinetic Energy of Emitted Electrons
The maximum kinetic energy (K.E.) of emitted electrons can be calculated using the photoelectric equation:
\[ \text{K.E.} = E_{\text{photon}} - \phi \]
where:
- \(E_{\text{photon}} = 10 \, \text{eV}\) (energy of each photon)
- \(\phi = 5 \, \text{eV}\) (work function of the metal)

Substituting the values:
\[ \text{K.E.} = 10 \, \text{eV} - 5 \, \text{eV} = 5 \, \text{eV} \]

### Step 2: Calculate the Number of Photons Incident per Second
The intensity \(I\) of the light is given as \(16 \, \text{mW} = 16 \times 10^{-3} \, \text{W}\). The energy of each photon in joules can be calculated as:
\[ E_{\text{photon}} = 10 \, \text{eV} = 10 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-18} \, \text{J} \]

The number of photons incident per second (\(N\)) can be calculated using the formula:
\[ N = \frac{I}{E_{\text{photon}}} \]
Substituting the values:
\[ N = \frac{16 \times 10^{-3} \, \text{W}}{1.6 \times 10^{-18} \, \text{J}} = 10^{15} \, \text{photons/second} \]

### Step 3: Calculate the Number of Photoelectrons Emitted per Second
Given that the photon efficiency is \(10\%\), the number of photoelectrons emitted per second (\(N_e\)) can be calculated as:
\[ N_e = \text{Efficiency} \times N \]
Substituting the values:
\[ N_e = 0.1 \times 10^{15} = 10^{14} \, \text{photoelectrons/second} \]

### Final Answers
- Maximum Kinetic Energy of emitted electrons: \(5 \, \text{eV}\)
- Number of photoelectrons emitted per second: \(10^{14} \, \text{photoelectrons/second}\)