Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work - function = 4.2 eV). The kinetic energy in joule of the fastest electrons amitted is

Options

A. `3.2xx10^(-21)`
B. `3.2xx10^(-19)`
C. `3.2xx10^(-17)`
D. `3.2xx10^(-15)`

Explanation

To solve the problem, we need to find the kinetic energy of the fastest electrons emitted when ultraviolet radiation falls on an aluminum surface. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Energy of ultraviolet radiation (E) = 6.2 eV
- Work function of aluminum (Φ) = 4.2 eV

### Step 2: Use the photoelectric equation
The kinetic energy (K_max) of the emitted electrons can be calculated using the photoelectric equation:
\[ K_{max} = E - \Phi \]
Where:
- \( K_{max} \) is the maximum kinetic energy of the emitted electrons,
- \( E \) is the energy of the incident radiation,
- \( \Phi \) is the work function of the material.

### Step 3: Substitute the values
Now, substitute the values into the equation:
\[ K_{max} = 6.2 \, \text{eV} - 4.2 \, \text{eV} \]
\[ K_{max} = 2.0 \, \text{eV} \]

### Step 4: Convert kinetic energy from eV to joules
To convert the kinetic energy from electronvolts (eV) to joules (J), use the conversion factor:
1 eV = \( 1.6 \times 10^{-19} \) J.

Thus, we calculate:
\[ K_{max} = 2.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
\[ K_{max} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 5: Final answer
The kinetic energy of the fastest electrons emitted is:
\[ K_{max} = 3.2 \times 10^{-19} \, \text{J} \]