Ultraviolet light of 5.2eV falls on Aluminium surface (work function = 3.2eV). The kinetic energy in joules of the fastest electron emitted is approximately:

Options

A. 17.2×`10^(−19)`J
B. 4.2×`10^(−19)`J
C. 3.2×`10^(−19)`J
D. 2.2×`10^(−19)`J

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between energy, work function, and kinetic energy
The energy of the incident ultraviolet light (E) can be expressed as the sum of the work function (W) of the material and the kinetic energy (KE) of the emitted electrons:
\[ E = W + KE \]

### Step 2: Rearrange the equation to find kinetic energy
We can rearrange the equation to solve for the kinetic energy:
\[ KE = E - W \]

### Step 3: Substitute the given values
From the problem, we know:
- Energy of the ultraviolet light, \( E = 5.2 \, \text{eV} \)
- Work function of aluminum, \( W = 3.2 \, \text{eV} \)

Now we substitute these values into the equation:
\[ KE = 5.2 \, \text{eV} - 3.2 \, \text{eV} \]
\[ KE = 2.0 \, \text{eV} \]

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

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

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

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