When ultraviolet lightofenergy6.2 eV incidents on a aluminimum surface, it emits photoelectrons. If work function for aluminium surface is 4.2 eV, then kinetic energy of emitted electrons is

Options

A. `3.2 xx 10^(-19) J`
B. `3.2 xx 10^(-17) J`
C. `3.2 xx 10^(-16) J`
D. `3.2 xx 10^(-11) J`

Explanation

To solve the problem, we will use Einstein's photoelectric equation, which relates the energy of the incident light to the work function of the material and the kinetic energy of the emitted photoelectrons. The equation is given by:

\[ KE = E - \phi \]

Where:
- \( KE \) is the kinetic energy of the emitted electrons.
- \( E \) is the energy of the incident light.
- \( \phi \) is the work function of the material.

### Step-by-Step Solution:

1. **Identify the energy of the incident light (E)**:
The energy of the ultraviolet light is given as \( E = 6.2 \, \text{eV} \).

2. **Identify the work function (φ)**:
The work function for the aluminium surface is given as \( \phi = 4.2 \, \text{eV} \).

3. **Substitute the values into the photoelectric equation**:
Using the equation \( KE = E - \phi \):
\[ KE = 6.2 \, \text{eV} - 4.2 \, \text{eV} \]

4. **Calculate the kinetic energy (KE)**:
\[ KE = 6.2 \, \text{eV} - 4.2 \, \text{eV} = 2.0 \, \text{eV} \]

5. **Conclusion**:
The kinetic energy of the emitted electrons is \( 2.0 \, \text{eV} \).

### Final Answer:
The kinetic energy of the emitted electrons is \( 2.0 \, \text{eV} \).

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