Ultraviolet light of 6.2eV falls on Aluminium surface (work function = 5.2eV). The kinetic energy in joules of the fastest electron emitted is approximately:

Options

A. 3.2×`10^(−19)`J
B. 4.2×`10^(−19)`J
C. 1.6×`10^(−19)`J
D. 6.1×`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) = 6.2 eV
- Work function of aluminum (W) = 5.2 eV

2. **Use the photoelectric equation:**
The kinetic energy (KE) of the emitted electron can be calculated using the equation:
\[ KE = E - W \]

3. **Substitute the values into the equation:**
\[ KE = 6.2 \, \text{eV} - 5.2 \, \text{eV} \]

4. **Perform the subtraction:**
\[ KE = 1.0 \, \text{eV} \]

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

6. **Calculate the kinetic energy in joules:**
\[ KE = 1.6 \times 10^{-19} \, \text{J} \]

### Final Answer:
The kinetic energy of the fastest electron emitted is approximately \( 1.6 \times 10^{-19} \, \text{J} \).