if the wavelength is chaged from`4000A. to 3000A`, then stopping potential will become

Options

A. 1.36 V
B. 3.40 V
C. 1.60 V
D. 1.97 V

Explanation

To solve the problem of how the stopping potential changes when the wavelength is changed from 4000 Å to 3000 Å, we can use the Einstein equation for the photoelectric effect. Let's break down the solution step by step.

### Step 1: Write the Einstein Equation
The Einstein equation for the photoelectric effect is given by:

\[ E \cdot V_0 = \frac{hc}{\lambda} - \phi \]

Where:
- \( E \) is the charge of the electron.
- \( V_0 \) is the stopping potential.
- \( h \) 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.
- \( \phi \) is the work function.

### Step 2: Set Up the Equations for Both Wavelengths
For the two different wavelengths, we can set up two equations:

1. For \( \lambda_1 = 4000 \, \text{Å} \):
\[ E \cdot V_{0_1} = \frac{hc}{\lambda_1} - \phi \]

2. For \( \lambda_2 = 3000 \, \text{Å} \):
\[ E \cdot V_{0_2} = \frac{hc}{\lambda_2} - \phi \]

### Step 3: Rearrange the Equations
Rearranging both equations to express the work function \( \phi \):

From equation 1:
\[ \phi = \frac{hc}{\lambda_1} - E \cdot V_{0_1} \]

From equation 2:
\[ \phi = \frac{hc}{\lambda_2} - E \cdot V_{0_2} \]

### Step 4: Equate the Two Expressions for \( \phi \)
Setting the two expressions for \( \phi \) equal to each other gives:

\[ \frac{hc}{\lambda_1} - E \cdot V_{0_1} = \frac{hc}{\lambda_2} - E \cdot V_{0_2} \]

### Step 5: Solve for the Change in Stopping Potential
Rearranging gives:

\[ E \cdot V_{0_2} - E \cdot V_{0_1} = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \]

Factoring out \( E \):

\[ E \cdot (V_{0_2} - V_{0_1}) = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) \]

Now, we can express the change in stopping potential:

\[ V_{0_2} - V_{0_1} = \frac{hc}{E} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) \]

### Step 6: Substitute Values
Now we substitute the known values:

- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( E = 1.6 \times 10^{-19} \, \text{C} \) (charge of an electron)
- \( \lambda_1 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} \)

Calculating:

\[ V_{0_2} - V_{0_1} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.6 \times 10^{-19}} \left( \frac{1}{3000 \times 10^{-10}} - \frac{1}{4000 \times 10^{-10}} \right) \]

### Step 7: Calculate the Values
1. Calculate \( \frac{hc}{E} \):
\[ \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.6 \times 10^{-19}} \approx 1.241 \times 10^{-6} \, \text{V} \]

2. Calculate \( \frac{1}{3000 \times 10^{-10}} - \frac{1}{4000 \times 10^{-10}} \):
\[ \frac{1}{3 \times 10^{-7}} - \frac{1}{4 \times 10^{-7}} = \frac{4 - 3}{12} = \frac{1}{12} \]

3. Combine the results:
\[ V_{0_2} - V_{0_1} = 1.241 \times 10^{-6} \times \frac{1}{12} \approx 0.1034 \, \text{V} \]

### Final Answer
Thus, the change in stopping potential is approximately \( 0.1034 \, \text{V} \).

---