Ultraviolet light of 4.2eV falls on Aluminium surface (work function = 3.2eV). The kinetic energy in joules of the fastest electron emitted is approximately:

Options

A. 4.2×`10^(−19)`J
B. 1.6×`10^(−19)`J
C. 2×`10^(−19)`J
D. 3.2×`10^(−19)`J

Explanation

To solve the problem of finding the kinetic energy of the fastest electron emitted when ultraviolet light falls on an aluminum surface, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Energy of the ultraviolet light (E) = 4.2 eV
- Work function of aluminum (W) = 3.2 eV

2. **Use the Photoelectric Equation**:
The kinetic energy (KE) of the emitted electrons can be calculated using the equation:
\[ KE = E - W \]
where:
- KE is the kinetic energy of the emitted electron,
- E is the energy of the incident photon,
- W is the work function of the material.

3. **Substitute the Values**:
Substitute the given values into the equation:
\[ KE = 4.2 \, \text{eV} - 3.2 \, \text{eV} \]
\[ KE = 1.0 \, \text{eV} \]

4. **Convert Electron Volts to Joules**:
To convert the kinetic energy from electron volts to joules, use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ KE = 1.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
\[ KE = 1.6 \times 10^{-19} \, \text{J} \]

5. **Final Answer**:
The kinetic energy of the fastest electron emitted is approximately:
\[ KE \approx 1.6 \times 10^{-19} \, \text{J} \]