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 step by step, we will use the photoelectric effect formula to find the kinetic energy of the emitted electrons.

### Step 1: Understand the Problem
We are given:
- Energy of ultraviolet radiation (E) = 6.2 eV
- Work function of aluminium (φ) = 4.2 eV

We need to find the kinetic energy (KE) of the fastest electrons emitted in joules.

### Step 2: Use the Photoelectric Effect Formula
The maximum kinetic energy (KE) of the emitted electrons can be calculated using the formula:
\[ KE = E - φ \]
Where:
- \( KE \) = Kinetic energy of the emitted electrons
- \( E \) = Energy of the incident radiation
- \( φ \) = Work function of the material

### Step 3: Substitute the Values
Substituting the given values into the formula:
\[ KE = 6.2 \, \text{eV} - 4.2 \, \text{eV} \]
\[ KE = 2.0 \, \text{eV} \]

### Step 4: Convert Kinetic Energy from eV to Joules
To convert the kinetic energy from electron volts (eV) to joules (J), we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]

Now, we can convert:
\[ KE = 2.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
\[ KE = 3.2 \times 10^{-19} \, \text{J} \]

### Final Answer
The kinetic energy of the fastest electrons emitted is:
\[ KE = 3.2 \times 10^{-19} \, \text{J} \]

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