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

Options

A. `3.2 xx 10^(-21)`
B. `3.2 xx 10^(-19)`
C. `3.2 xx 10^(-17)`
D. `3.2 xx 10^(-15)`

Answer

To solve the problem, we need to determine the kinetic energy of the fastest electron emitted when ultraviolet radiation of energy 6.2 eV falls on an aluminum surface with a work function of 4.2 eV.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Energy of ultraviolet radiation (E) = 6.2 eV
- Work function of aluminum (Φ) = 4.2 eV

2. **Use the Photoelectric Effect Equation:**
The kinetic energy (K.E.) of the emitted electron can be calculated using the formula:
\[ K.E. = E - \Phi \]
where \(E\) is the energy of the incident radiation and \(\Phi\) is the work function.

3. **Substitute the Values:**
Substitute the given values into the equation:
\[ K.E. = 6.2 \, \text{eV} - 4.2 \, \text{eV} \]
\[ K.E. = 2.0 \, \text{eV} \]

4. **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, the kinetic energy in joules is:
\[ K.E. = 2.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
\[ K.E. = 3.2 \times 10^{-19} \, \text{J} \]

5. **Final Answer:**
The kinetic energy of the fastest electron emitted is approximately:
\[ K.E. \approx 3.2 \times 10^{-19} \, \text{J} \]