The maximum kinetic energy of photoelectrons emitted from a metal surface increses from 0.4 eV to 1.2 eV when the frequency of the incident radiation is increased by 40% . What is the work function (in eV) of the metal surface?

Answer

To solve the problem, we will use the photoelectric effect equation, which relates the maximum kinetic energy (K.E.) of emitted photoelectrons to the frequency of the incident radiation and the work function (Φ) of the metal.

### Step-by-Step Solution:

1. **Understand the Photoelectric Equation**:
The maximum kinetic energy of photoelectrons is given by:
\[ K.E. = h \nu - \Phi \]
where:
- \( K.E. \) is the maximum kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident radiation,
- \( \Phi \) is the work function of the metal.

2. **Set Up the Initial Conditions**:
Initially, the maximum kinetic energy is given as:
\[ K.E._1 = 0.4 \, \text{eV} = h \nu - \Phi \quad \text{(1)} \]

3. **Calculate the New Frequency**:
When the frequency is increased by 40%, the new frequency becomes:
\[ \nu_2 = 1.4 \nu \]

4. **Set Up the New Conditions**:
The new maximum kinetic energy is given as:
\[ K.E._2 = 1.2 \, \text{eV} = h (1.4 \nu) - \Phi \quad \text{(2)} \]

5. **Substitute and Rearrange Equations**:
From equation (1):
\[ \Phi = h \nu - 0.4 \quad \text{(3)} \]
From equation (2):
\[ K.E._2 = 1.2 = 1.4 h \nu - \Phi \]
Substituting equation (3) into equation (2):
\[ 1.2 = 1.4 h \nu - (h \nu - 0.4) \]
Simplifying this gives:
\[ 1.2 = 1.4 h \nu - h \nu + 0.4 \]
\[ 1.2 = 0.4 h \nu + 0.4 \]

6. **Isolate \( h \nu \)**:
Rearranging the equation:
\[ 1.2 - 0.4 = 0.4 h \nu \]
\[ 0.8 = 0.4 h \nu \]
\[ h \nu = \frac{0.8}{0.4} = 2 \, \text{eV} \quad \text{(4)} \]

7. **Calculate the Work Function**:
Now substitute \( h \nu \) back into equation (3):
\[ \Phi = h \nu - 0.4 = 2 - 0.4 = 1.6 \, \text{eV} \]

### Final Answer:
The work function (Φ) of the metal surface is:
\[ \Phi = 1.6 \, \text{eV} \]