To solve the problem of finding the change in stopping potential when the wavelength of incident light is changed from 4000 Å to 3000 Å in a photoelectric cell, we can follow these steps:
### Step 1: Understand the Photoelectric Equation
The photoelectric effect is described by Einstein's photoelectric equation:
\[
\frac{hc}{\lambda} - \phi = eV_0
\]
where:
- \( h \) = Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \))
- \( c \) = speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda \) = wavelength of the incident light
- \( \phi \) = work function of the material
- \( e \) = charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \))
- \( V_0 \) = stopping potential
### Step 2: Set Up the Equations for Two Wavelengths
Let:
- \( \lambda_1 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} \)
Using the photoelectric equation for both wavelengths:
1. For \( \lambda_1 \):
\[
\frac{hc}{\lambda_1} - \phi = eV_{01}
\]
2. For \( \lambda_2 \):
\[
\frac{hc}{\lambda_2} - \phi = eV_{02}
\]
### Step 3: Subtract the Two Equations
Subtract the second equation from the first:
\[
\left(\frac{hc}{\lambda_1} - \phi\right) - \left(\frac{hc}{\lambda_2} - \phi\right) = eV_{01} - eV_{02}
\]
This simplifies to:
\[
\frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = e(V_{01} - V_{02})
\]
### Step 4: Rearrange for Change in Stopping Potential
Rearranging gives us:
\[
V_{01} - V_{02} = \frac{hc}{e} \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)
\]
### Step 5: Substitute Values
Now substitute the known values:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( \lambda_1 = 4000 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 3000 \times 10^{-10} \, \text{m} \)
Calculating:
\[
V_{01} - V_{02} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{1.6 \times 10^{-19}} \left(\frac{1}{4000 \times 10^{-10}} - \frac{1}{3000 \times 10^{-10}}\right)
\]
### Step 6: Calculate the Values
Calculate \( \frac{hc}{e} \):
\[
\frac{(6.63 \times 10^{-34})(3 \times 10^8)}{1.6 \times 10^{-19}} \approx 1.24 \, \text{V}
\]
Now calculate \( \left(\frac{1}{4000 \times 10^{-10}} - \frac{1}{3000 \times 10^{-10}}\right) \):
\[
\frac{1}{4000 \times 10^{-10}} - \frac{1}{3000 \times 10^{-10}} = \frac{3 - 4}{12000} = -\frac{1}{12000} \, \text{m}^{-1}
\]
### Step 7: Final Calculation
Putting it all together:
\[
V_{01} - V_{02} = 1.24 \times -\frac{1}{12000} \approx -0.103 \, \text{V} \approx -1.03 \, \text{V}
\]
Thus, the change in stopping potential is:
\[
V_{02} - V_{01} = 1.03 \, \text{V}
\]
### Conclusion
The change in stopping potential when the wavelength is changed from 4000 Å to 3000 Å is approximately \( 1.03 \, \text{V} \).
