Ultraviolet light of `6.2 eV` falls on an aluminium surface (work function `= 4.2 eV`). The kinetic energy (in joule) of the fastest electron emitted is approximately.

Options

A. `3 xx 10^-21`
B. `3 xx 10^-19`
C. `3 xx 10^-17`
D. `3 xx 10^-15`

Explanation

To find 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 ultraviolet light (E) = 6.2 eV
- Work function of aluminum (Φ) = 4.2 eV

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

3. **Substitute the Values:**
Substitute the given values into the equation:
\[ KE = 6.2 \, \text{eV} - 4.2 \, \text{eV} \]

4. **Calculate the Kinetic Energy:**
\[ KE = 2.0 \, \text{eV} \]

5. **Convert Kinetic Energy from eV to Joules:**
To convert electron volts to joules, use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Therefore,
\[ KE = 2.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
\[ KE = 3.2 \times 10^{-19} \, \text{J} \]

### Final Answer:
The kinetic energy of the fastest electron emitted is approximately \( 3.2 \times 10^{-19} \, \text{J} \).

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