The photoelectric cut off voltage in a certain experiment is 1.5V. What is the maximum kinetic energy of photoelectrons emitted? `e=1.6xx10^(-19)C`.

Options

A. 2.4 eV
B. 1.5 eV
C. 3.1 eV
D. 4.5 eV

Explanation

To find the maximum kinetic energy of photoelectrons emitted when the photoelectric cut-off voltage is given, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept**: The photoelectric cut-off voltage (also known as stopping potential) is the voltage at which the emitted photoelectrons have just enough energy to overcome the work function and stop. The maximum kinetic energy (KE) of the emitted photoelectrons can be calculated using the formula:
\[ KE = e \cdot V \]
where \( e \) is the charge of the electron and \( V \) is the cut-off voltage.

2. **Identify Given Values**:
- Cut-off voltage, \( V = 1.5 \, \text{V} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

3. **Calculate the Maximum Kinetic Energy**:
Using the formula for kinetic energy:
\[ KE = e \cdot V = (1.6 \times 10^{-19} \, \text{C}) \cdot (1.5 \, \text{V}) \]

4. **Perform the Calculation**:
\[ KE = 1.6 \times 10^{-19} \cdot 1.5 = 2.4 \times 10^{-19} \, \text{J} \]

5. **Convert Joules to Electron Volts**:
Since the energy is often expressed in electron volts (eV), we can convert joules to electron volts using the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE = \frac{2.4 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 1.5 \, \text{eV} \]

6. **Final Answer**:
The maximum kinetic energy of the photoelectrons emitted is \( 1.5 \, \text{eV} \).