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 step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Calculate the Maximum Kinetic Energy of Emitted Electrons

The maximum kinetic energy (K.E.) of the emitted electrons can be calculated using the photoelectric equation:

\[ K.E. = E - \phi \]

where:
- \(E\) is the energy of each photon,
- \(\phi\) is the work function of the metal.

Given:
- \(E = 10 \, \text{eV}\)
- \(\phi = 5 \, \text{eV}\)

Substituting the values:

\[ K.E. = 10 \, \text{eV} - 5 \, \text{eV} = 5 \, \text{eV} \]

### Step 2: Calculate the Number of Photoelectrons Emitted Per Second

To find the number of photoelectrons emitted per second, we can use the formula:

\[ I = n \cdot E \cdot \text{Efficiency} \]

where:
- \(I\) is the intensity of the light,
- \(n\) is the number of photoelectrons emitted per second,
- \(E\) is the energy of one photon (in joules),
- Efficiency is given as a fraction.

First, we need to convert the intensity from mW to watts:

\[ I = 16 \, \text{mW} = 16 \times 10^{-3} \, \text{W} = 0.016 \, \text{W} \]

Next, convert the energy of one photon from eV to joules:

\[ E = 10 \, \text{eV} = 10 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-18} \, \text{J} \]

Now, using the efficiency of \(10\%\) or \(0.1\):

Substituting into the formula:

\[ 0.016 = n \cdot (1.6 \times 10^{-18}) \cdot 0.1 \]

Rearranging to solve for \(n\):

\[ n = \frac{0.016}{(1.6 \times 10^{-18}) \cdot 0.1} \]

Calculating \(n\):

\[ n = \frac{0.016}{1.6 \times 10^{-19}} = 10^{11} \, \text{photoelectrons per second} \]

### Final Answers

1. The maximum kinetic energy of emitted electrons: **5 eV**
2. The number of photoelectrons emitted per second: **\(10^{11}\) photoelectrons/s**