On a photosensitive material, when frequency of incident radiation is increased by 30% kinetic energy of emitted photoelectrons increases from `0.4 eV` to `0.9 eV`. The work function of the surface is

Options

A. 1 eV
B. 1.267 Ev
C. 1.4 eV
D. 1.8 eV

Answer

To solve the problem, we will use the photoelectric effect equation, which relates the kinetic energy of emitted photoelectrons to the frequency of the incident radiation and the work function of the material.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a photosensitive material where the frequency of incident radiation is increased by 30%.
- The initial kinetic energy (KE1) of emitted photoelectrons is 0.4 eV.
- The final kinetic energy (KE2) after the increase in frequency is 0.9 eV.
- We need to find the work function (φ) of the surface.

2. **Define the Variables**:
- Let the initial frequency of the incident radiation be \( \nu \).
- The new frequency after a 30% increase is \( \nu' = 1.3 \nu \).

3. **Use the Photoelectric Equation**:
The kinetic energy of the emitted photoelectrons can be expressed as:
\[ KE = h\nu - \phi \]
where \( h \) is Planck's constant and \( \phi \) is the work function.

4. **Set Up the Equations**:
- For the initial condition (before frequency increase):
\[ KE_1 = h\nu - \phi \quad \text{(1)} \]
Substituting \( KE_1 = 0.4 \) eV:
\[ 0.4 = h\nu - \phi \quad \text{(1)} \]

- For the final condition (after frequency increase):
\[ KE_2 = h\nu' - \phi \quad \text{(2)} \]
Substituting \( KE_2 = 0.9 \) eV and \( \nu' = 1.3\nu \):
\[ 0.9 = h(1.3\nu) - \phi \quad \text{(2)} \]

5. **Subtract the Two Equations**:
- From equation (2) and (1):
\[ (0.9 = h(1.3\nu) - \phi) - (0.4 = h\nu - \phi) \]
This simplifies to:
\[ 0.9 - 0.4 = h(1.3\nu) - h\nu \]
\[ 0.5 = h(1.3\nu - \nu) \]
\[ 0.5 = h(0.3\nu) \]
\[ h\nu = \frac{0.5}{0.3} = \frac{5}{3} \text{ eV} \quad \text{(3)} \]

6. **Substitute \( h\nu \) Back to Find Work Function**:
- Substitute \( h\nu \) from equation (3) into equation (1):
\[ 0.4 = \frac{5}{3} - \phi \]
Rearranging gives:
\[ \phi = \frac{5}{3} - 0.4 \]
Converting \( 0.4 \) to a fraction:
\[ 0.4 = \frac{12}{30} = \frac{6}{15} \]
Thus,
\[ \phi = \frac{5}{3} - \frac{12}{30} \]
Converting \( \frac{5}{3} \) to a common denominator:
\[ \phi = \frac{50}{30} - \frac{12}{30} = \frac{38}{30} = 1.267 \text{ eV} \]

7. **Final Answer**:
The work function \( \phi \) of the surface is approximately \( 1.27 \text{ eV} \).