When ultraviolet radiation of a certain frequency falls on a potassium target, the photoelectrons released can be stopped completely by a retarding potential of 0.6 V. If the frequency of the radiation is increased by 10%, this stopping potential rises to 0.9 V. The work function of potassium is

Options

A. 2.0 eV
B. 2.4 eV
C. 3.0 eV
D. 2.8 eV

Answer

To solve the problem, we will use the concepts of photoelectric effect and the equations relating to the kinetic energy of photoelectrons and the work function of the material.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a potassium target exposed to ultraviolet (UV) radiation of a certain frequency. The stopping potential required to stop the emitted photoelectrons is given as 0.6 V. When the frequency of the radiation is increased by 10%, the stopping potential increases to 0.9 V. We need to find the work function (φ) of potassium.

2. **Using the Photoelectric Equation**:
The energy of the incident photons can be expressed using the equation:
\[ E = hf \]
where \( h \) is Planck's constant and \( f \) is the frequency of the radiation.

3. **Relating Kinetic Energy and Stopping Potential**:
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be expressed as:
\[ K.E. = eV_s \]
where \( e \) is the charge of the electron and \( V_s \) is the stopping potential.

For the first case (stopping potential = 0.6 V):
\[ K.E. = e \cdot 0.6 \text{ V} \]
Thus, we can write:
\[ hf = e \cdot 0.6 + φ \quad \text{(1)} \]

4. **Increasing the Frequency by 10%**:
When the frequency is increased by 10%, the new frequency \( f' \) is:
\[ f' = 1.1f \]
The energy of the photons at this new frequency is:
\[ E' = hf' = h(1.1f) = 1.1hf \]

The new stopping potential is 0.9 V, so:
\[ K.E. = e \cdot 0.9 \text{ V} \]
This gives us the second equation:
\[ 1.1hf = e \cdot 0.9 + φ \quad \text{(2)} \]

5. **Setting Up the Equations**:
We now have two equations:
- From equation (1): \( hf = e \cdot 0.6 + φ \)
- From equation (2): \( 1.1hf = e \cdot 0.9 + φ \)

6. **Substituting Equation (1) into Equation (2)**:
We can express \( hf \) from equation (1):
\[ hf = e \cdot 0.6 + φ \]
Substitute this into equation (2):
\[ 1.1(e \cdot 0.6 + φ) = e \cdot 0.9 + φ \]

7. **Expanding and Rearranging**:
Expanding the left side:
\[ 1.1e \cdot 0.6 + 1.1φ = e \cdot 0.9 + φ \]
Rearranging gives:
\[ 1.1e \cdot 0.6 - e \cdot 0.9 = φ - 1.1φ \]
Simplifying:
\[ 0.66e - 0.9e = -0.1φ \]
\[ -0.24e = -0.1φ \]
Thus:
\[ φ = \frac{0.24e}{0.1} = 2.4 \text{ eV} \]

8. **Conclusion**:
The work function of potassium is:
\[ φ = 2.4 \text{ eV} \]