In a photoelectric cell, the wavelength of incident light is chaged from `4000A` to `3600A`. The change in stopping potential will be

Options

A. 0.14 V
B. 0.24 V
C. 0.35 V
D. 0.44 V

Explanation

To solve the problem of finding the change in stopping potential when the wavelength of incident light changes from 4000 Å to 3600 Å in a photoelectric cell, we will use the photoelectric effect equation derived from Einstein's theory. Here’s a step-by-step solution:

### Step 1: Understand the Photoelectric Effect Equation
The energy of the incident photons can be expressed using the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the incident light.

The stopping potential \( V_0 \) is related to the energy of the photon and the work function \( W \) of the material:
\[ eV_0 = \frac{hc}{\lambda} - W \]
Where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

### Step 2: Set Up the Equations
For two different wavelengths \( \lambda_1 = 4000 \, \text{Å} \) and \( \lambda_2 = 3600 \, \text{Å} \), we can write two equations:
1. For \( \lambda_1 \):
\[ eV_{0_1} = \frac{hc}{\lambda_1} - W \]
2. For \( \lambda_2 \):
\[ eV_{0_2} = \frac{hc}{\lambda_2} - W \]

### Step 3: Find the Change in Stopping Potential
Subtract the first equation from the second:
\[ eV_{0_2} - eV_{0_1} = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \]
This simplifies to:
\[ e(V_{0_2} - V_{0_1}) = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) \]

### Step 4: Solve for the Change in Stopping Potential
Rearranging gives:
\[ V_{0_2} - V_{0_1} = \frac{hc}{e} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) \]

### Step 5: Substitute the Values
Substituting the known values:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- Convert wavelengths from Ångströms to meters:
- \( \lambda_1 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 3600 \, \text{Å} = 3600 \times 10^{-10} \, \text{m} \)

Now, plug in the values:
\[ V_{0_2} - V_{0_1} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.6 \times 10^{-19}} \left( \frac{1}{3600 \times 10^{-10}} - \frac{1}{4000 \times 10^{-10}} \right) \]

### Step 6: Calculate the Values
Calculating \( \frac{1}{3600 \times 10^{-10}} - \frac{1}{4000 \times 10^{-10}} \):
\[ = \frac{1}{3.6 \times 10^{-7}} - \frac{1}{4.0 \times 10^{-7}} = \frac{4 - 3.6}{14.4} \times 10^{7} = \frac{0.4}{14.4} \times 10^{7} = \frac{1}{36} \times 10^{7} \]

Now substituting back:
\[ V_{0_2} - V_{0_1} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.6 \times 10^{-19}} \times \frac{1}{36} \times 10^{7} \]

Calculating the numerical values yields:
\[ V_{0_2} - V_{0_1} \approx 0.35 \, \text{V} \]

### Final Answer
The change in stopping potential is approximately \( 0.35 \, \text{V} \).

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