When ultraviolet radiation of a certain frequency falls on a potassium target, the photoelectrons released can be stopped completely by a retarding potential of 0.5 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

Explanation

To find the work function of potassium based on the given information, we can follow these steps:

### Step 1: Understand the photoelectric effect equation
The maximum kinetic energy (K.E.) of the photoelectrons can be expressed using the equation:
\[ K.E. = hf - \phi \]
where:
- \( K.E. \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident radiation,
- \( \phi \) is the work function of the material (potassium in this case).

### Step 2: Relate stopping potential to kinetic energy
The maximum kinetic energy can also be related to the stopping potential \( V \) by the equation:
\[ K.E. = eV \]
where \( e \) is the charge of an electron. Thus, we can rewrite the equation as:
\[ eV = hf - \phi \]

### Step 3: Set up equations for the two scenarios
1. For the first case (stopping potential \( V_1 = 0.5 \, V \)):
\[ e(0.5) = hf - \phi \tag{1} \]

2. For the second case, where the frequency is increased by 10% (new frequency \( f' = 1.1f \)) and the stopping potential \( V_2 = 0.9 \, V \):
\[ e(0.9) = h(1.1f) - \phi \tag{2} \]

### Step 4: Substitute and simplify the equations
From equation (1):
\[ \phi = hf - e(0.5) \tag{3} \]

Substituting equation (3) into equation (2):
\[ e(0.9) = h(1.1f) - (hf - e(0.5)) \]

This simplifies to:
\[ e(0.9) = h(1.1f) - hf + e(0.5) \]
\[ e(0.9) = hf(1.1 - 1) + e(0.5) \]
\[ e(0.9) = hf(0.1) + e(0.5) \]

### Step 5: Rearrange to find \( hf \)
Rearranging gives:
\[ hf(0.1) = e(0.9) - e(0.5) \]
\[ hf(0.1) = e(0.4) \]
\[ hf = \frac{e(0.4)}{0.1} = 4e \]

### Step 6: Substitute \( hf \) back to find \( \phi \)
Now substituting \( hf = 4e \) back into equation (3):
\[ \phi = 4e - e(0.5) \]
\[ \phi = 4e - 0.5e \]
\[ \phi = (4 - 0.5)e \]
\[ \phi = 3.5e \]

### Step 7: Calculate the work function in electron volts
Using \( e \approx 1 \, eV \):
\[ \phi \approx 3.5 \, eV \]

### Conclusion
The work function of potassium is approximately \( 3.5 \, eV \).

---