Ultraviolet light of 6.2eV falls on Aluminium surface (work function = 2.2eV). The kinetic energy in joules of the fastest electron emitted is approximately:

Options

A. 1.2×`10^(−19)`J
B. 6.4×`10^(−19)`J
C. 4.8×`10^(−19)`J
D. 2.8×`10^(−19)`J

Explanation

To solve the problem, we need to calculate the kinetic energy of the fastest electron emitted when ultraviolet light of energy 6.2 eV falls on an aluminum surface with a work function of 2.2 eV.

### Step-by-Step Solution:

1. **Identify the given values:**
- Energy of the ultraviolet light (E) = 6.2 eV
- Work function of aluminum (W) = 2.2 eV

2. **Use the photoelectric equation:**
The kinetic energy (KE) of the emitted electron 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 incoming photon,
- W is the work function.

3. **Substitute the values into the equation:**
\[ KE = 6.2 \, \text{eV} - 2.2 \, \text{eV} \]

4. **Calculate the kinetic energy in electron volts:**
\[ KE = 4.0 \, \text{eV} \]

5. **Convert the kinetic energy from electron volts to joules:**
To convert electron volts to joules, use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Therefore, to convert 4.0 eV to joules:
\[ KE = 4.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]

6. **Perform the multiplication:**
\[ KE = 6.4 \times 10^{-19} \, \text{J} \]

### Final Answer:
The kinetic energy of the fastest electron emitted is approximately:
\[ KE \approx 6.4 \times 10^{-19} \, \text{J} \]