Suppose the wavelength of the incident light is increased from `3000 Å` to `3040Å`. Find the corresponding change in the stopping potential. [Take the product `hc=12.4xx10^(-7)eVm)`]

Explanation

To solve the problem of finding the change in stopping potential when the wavelength of the incident light is increased from 3000 Å to 3040 Å, we can follow these steps:

### Step 1: Understand the relationship between stopping potential and wavelength
The stopping potential \( V \) is related to the energy of the incident photons. The energy of a photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the light.

The stopping potential \( V \) is given by the energy of the photon divided by the charge of the electron \( e \):
\[ V = \frac{E}{e} = \frac{hc}{e\lambda} \]

### Step 2: Calculate the stopping potential for both wavelengths
We can calculate the stopping potential for both wavelengths \( \lambda_1 = 3000 \, \text{Å} \) and \( \lambda_2 = 3040 \, \text{Å} \).

Given:
- \( hc = 12.4 \times 10^{-7} \, \text{eV m} \)
- Convert wavelengths from Ångströms to meters:
- \( \lambda_1 = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 3040 \, \text{Å} = 3040 \times 10^{-10} \, \text{m} \)

Now, calculate the stopping potential for both wavelengths:
\[ V_1 = \frac{hc}{e\lambda_1} = \frac{12.4 \times 10^{-7}}{1.6 \times 10^{-19} \times 3000 \times 10^{-10}} \]
\[ V_2 = \frac{hc}{e\lambda_2} = \frac{12.4 \times 10^{-7}}{1.6 \times 10^{-19} \times 3040 \times 10^{-10}} \]

### Step 3: Calculate the change in stopping potential
The change in stopping potential \( \Delta V \) is given by:
\[ \Delta V = V_1 - V_2 \]

### Step 4: Substitute the values and calculate
1. Calculate \( V_1 \):
\[ V_1 = \frac{12.4 \times 10^{-7}}{1.6 \times 10^{-19} \times 3000 \times 10^{-10}} \approx 4.133 \, \text{V} \]

2. Calculate \( V_2 \):
\[ V_2 = \frac{12.4 \times 10^{-7}}{1.6 \times 10^{-19} \times 3040 \times 10^{-10}} \approx 4.078 \, \text{V} \]

3. Now, find \( \Delta V \):
\[ \Delta V = 4.133 - 4.078 = 0.055 \, \text{V} = 5.5 \times 10^{-2} \, \text{V} \]

### Final Answer
The change in stopping potential is:
\[ \Delta V = 5.5 \times 10^{-2} \, \text{V} \]

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