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Topic: Einstein's photoelectric equation, particle nature of light· Dual Nature of Matter and Radiation

When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. The threshold wavelength for the metallic surface i…

When a metallic surface is illuminated with radiation of wavelength \(\lambda\), the stopping potential is V. If the same surface is illuminated with radiation of wavelength \(2\lambda\), the stopping potential is \(\frac{V}{4}\). The threshold wavelength for the metallic surface is :

1. \(4\lambda\)

2. \(5\lambda\)

3. \(\frac{5}{2}\lambda\)

4. \(3\lambda\)

1 answer

Best answer

Answer: 4. \(3\lambda\)

Key idea: \(eV=\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_0}\).

  1. \(eV=\dfrac{hc}{\lambda}-\phi\) and \(\dfrac{eV}{4}=\dfrac{hc}{2\lambda}-\phi\).
  2. Subtracting: \(\tfrac34eV=\dfrac{hc}{2\lambda}\Rightarrow eV=\dfrac{2hc}{3\lambda}\).
  3. \(\phi=\dfrac{hc}{\lambda}-\dfrac{2hc}{3\lambda}=\dfrac{hc}{3\lambda}\), so \(\lambda_0=3\lambda\).
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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