To solve the problem, we will use the relationship between the stopping potential (V₀), the energy of the incident photons, and the work function of the material. The stopping potential is related to the maximum kinetic energy of the emitted electrons, which can be expressed as:
1. **Understanding the relationship**:
The stopping potential \( V_0 \) can be expressed as:
\[
eV_0 = E_{photon} - \phi
\]
where \( E_{photon} \) is the energy of the incident photon and \( \phi \) is the work function of the material.
2. **Photon energy**:
The energy of a photon can be calculated using the formula:
\[
E_{photon} = \frac{hc}{\lambda}
\]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.
3. **Calculating for 4000 Å**:
Given that for \( \lambda = 4000 \, \text{Å} \) (which is \( 4000 \times 10^{-10} \, \text{m} \)), the stopping potential \( V_0 = 2 \, \text{V} \):
\[
eV_0 = \frac{hc}{4000 \times 10^{-10}} - \phi
\]
Rearranging gives:
\[
\phi = \frac{hc}{4000 \times 10^{-10}} - eV_0
\]
4. **Calculating for 3000 Å**:
Now, for \( \lambda = 3000 \, \text{Å} \) (which is \( 3000 \times 10^{-10} \, \text{m} \)), we can express the new stopping potential \( V_0' \):
\[
eV_0' = \frac{hc}{3000 \times 10^{-10}} - \phi
\]
5. **Substituting the work function**:
We can substitute the expression for \( \phi \) from step 3 into this equation:
\[
eV_0' = \frac{hc}{3000 \times 10^{-10}} - \left(\frac{hc}{4000 \times 10^{-10}} - eV_0\right)
\]
Simplifying this gives:
\[
eV_0' = \frac{hc}{3000 \times 10^{-10}} - \frac{hc}{4000 \times 10^{-10}} + eV_0
\]
6. **Finding a common denominator**:
The common denominator for the fractions involving \( hc \) is \( 12000 \times 10^{-10} \):
\[
eV_0' = \left(\frac{4hc - 3hc}{12000 \times 10^{-10}}\right) + eV_0
\]
\[
eV_0' = \frac{hc}{12000 \times 10^{-10}} + eV_0
\]
7. **Calculating the change in stopping potential**:
Since \( V_0 = 2 \, \text{V} \), we can see that \( V_0' \) will be greater than \( V_0 \) because the first term \( \frac{hc}{12000 \times 10^{-10}} \) is positive. Thus:
\[
V_0' > 2 \, \text{V}
\]
8. **Conclusion**:
Therefore, the potential required to stop the ejection of electrons when the incident light is changed to \( 3000 \, \text{Å} \) will be greater than \( 2 \, \text{V} \).
