To solve the problem of how the change in stopping potential occurs when the wavelength of incident light is changed from 4000 Å to 3000 Å in a photoelectric cell, we can follow these steps:
### Step-by-Step Solution:
1. **Understand the Photoelectric Effect:**
The photoelectric effect is described by Einstein's photoelectric equation:
\[
E = \phi + eV_s
\]
where \( E \) is the energy of the incident photon, \( \phi \) is the work function of the material, \( e \) is the charge of the electron, and \( V_s \) is the stopping potential.
2. **Calculate Energy of Photons:**
The energy of a photon is given by:
\[
E = \frac{hc}{\lambda}
\]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)), \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), and \( \lambda \) is the wavelength in meters.
3. **Convert Wavelengths:**
Convert the wavelengths from Ångströms to meters:
- \( \lambda_1 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} = 4 \times 10^{-7} \, \text{m} \)
- \( \lambda_2 = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} = 3 \times 10^{-7} \, \text{m} \)
4. **Calculate Energies for Each Wavelength:**
- For \( \lambda_1 = 4000 \, \text{Å} \):
\[
E_1 = \frac{hc}{\lambda_1} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \, \text{J}
\]
- For \( \lambda_2 = 3000 \, \text{Å} \):
\[
E_2 = \frac{hc}{\lambda_2} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{3 \times 10^{-7}} = 6.63 \times 10^{-19} \, \text{J}
\]
5. **Calculate Stopping Potential for Each Wavelength:**
The stopping potential can be calculated using:
\[
eV_s = E - \phi
\]
Therefore, the change in stopping potential when changing the wavelength can be expressed as:
\[
e(V_{s2} - V_{s1}) = E_2 - E_1
\]
6. **Calculate Change in Stopping Potential:**
\[
e(V_{s2} - V_{s1}) = E_2 - E_1 = (6.63 \times 10^{-19} - 4.97 \times 10^{-19}) \, \text{J} = 1.66 \times 10^{-19} \, \text{J}
\]
Now, divide by the charge of an electron (\( e = 1.6 \times 10^{-19} \, \text{C} \)):
\[
V_{s2} - V_{s1} = \frac{1.66 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.0375 \, \text{V}
\]
### Final Answer:
The change in stopping potential when the wavelength is changed from 4000 Å to 3000 Å is approximately **1.04 V**.
