Step 1: Understanding the Concept:
First, find the maximum kinetic energy (\( K_{\text{max}} \)) of the emitted photoelectrons using Einstein's photoelectric equation. Then, calculate the de Broglie wavelength using the relationship between kinetic energy and wavelength.
Step 2: Key Formula or Approach:
1. Photoelectric equation: \( K_{\text{max}} = E - \phi \)
2. de Broglie wavelength: \( \lambda = \frac{12.27}{\sqrt{V}} \, \text{\AA} \) (where \( V \) is stopping potential or kinetic energy in eV). Alternatively, \( \lambda = \frac{h}{\sqrt{2mK}} \).
Step 3: Detailed Explanation:
Given:
Photon Energy \( E = 4.5 \) eV
Work Function \( \phi = 3 \) eV
Calculate \( K_{\text{max}} \):
\[ K_{\text{max}} = 4.5 - 3 = 1.5 \, \text{eV} \]
Using the shortcut formula for an electron's wavelength where energy is in eV:
\[ \lambda \approx \frac{12.27}{\sqrt{K_{\text{max}}(\text{in eV})}} \, \text{\AA} \]
\[ \lambda = \frac{12.27}{\sqrt{1.5}} \, \text{\AA} \]
Since \( \sqrt{1.5} \approx 1.225 \):
\[ \lambda \approx \frac{12.27}{1.225} \approx 10 \, \text{\AA} \]
Step 4: Final Answer:
The wavelength is nearly 10 \AA.