Step 1: Understanding the Question:
The stopping potential is the negative voltage required to halt the fastest emitted photoelectrons. It depends on the energy of the incident light and the work function of the metal. We need to find how a change in wavelength affects this potential.
Step 2: Key Formula or Approach:
Einstein’s Photoelectric Equation:
\[ E_{\text{photon}} = \phi + K.E._{\text{max}} \]
\[ \frac{hc}{\lambda} = \phi + eV_s \]
where \(V_s\) is the stopping potential. A helpful value for entrance exams is \(hc \approx 12400 \text{ eV}\cdot\text{\AA}\) or \(12420 \text{ eV}\cdot\text{\AA}\).
Step 3: Detailed Explanation:
Step A: Calculate photon energy (\(E_1\)) for 4000 \(\text{\AA}\).
\[ E_1 = \frac{12420}{4000} = 3.105 \text{ eV} \]
Step B: Find the work function (\(\phi\)) of the metal using the first case. Given \(V_s = 2 V\), the kinetic energy is 2 eV.
\[ \phi = E_1 - K.E. = 3.105 - 2 = 1.105 \text{ eV} \]
Step C: Calculate photon energy (\(E_2\)) for 3000 \(\text{\AA}\).
\[ E_2 = \frac{12420}{3000} = 4.14 \text{ eV} \]
Step D: Calculate the new stopping potential (\(V_{s2}\)) using the work function calculated in Step B.
\[ K.E._{\text{new}} = E_2 - \phi \]
\[ K.E._{\text{new}} = 4.14 - 1.105 = 3.035 \text{ eV} \]
Since the stopping potential in volts is numerically equal to the kinetic energy in eV, the potential is approximately 3.035 V.
Qualitatively, decreasing the wavelength increases the frequency and energy of the incident photons. Since the "admission fee" (work function) for the metal remains the same, all extra energy is converted into kinetic energy of the electrons, thus requiring a higher potential to stop them.
Step 4: Final Answer:
The new stopping potential will be approximately 3.03 V.