Question:
A metallic surface is illuminated with radiation of wavelength \(\lambda\), the stopping potential is \(V_0\). If the same surface is illuminated with radiation of wavelength \(2\lambda\), the stopping potential becomes \(\frac{V_0}{4}\). The threshold wavelength for this metallic surface will be
A metallic surface is illuminated with radiation of wavelength \(\lambda\), the stopping potential is \(V_0\). If the same surface is illuminated with radiation of wavelength \(2\lambda\), the stopping potential becomes \(\frac{V_0}{4}\). The threshold wavelength for this metallic surface will be
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Apply Einstein photoelectric equation \(\frac{hc}{\lambda}=\phi+eV_s\) twice and solve for the work function \(\phi=\frac{hc}{\lambda_0}\).
Updated On: Nov 2, 2026
- \(3\lambda\)
- \(4\lambda\)
- \(\frac{3}{2}\lambda\)
- \(\frac{\lambda}{4}\)
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The Correct Option is A
Solution and Explanation
Let \(E=\frac{hc}{\lambda}\). Then
\(E=\phi+eV_0\) ... (i)
\(\frac{E}{2}=\phi+\frac{eV_0}{4}\) ... (ii)
Multiply (ii) by 4: \(2E=4\phi+eV_0\). Subtract (i): \(E=3\phi\), so \(\phi=\frac{E}{3}=\frac{hc}{3\lambda}\).
Threshold wavelength: \(\lambda_0=\frac{hc}{\phi}=3\lambda\).
Answer: (1) \(3\lambda\)
\(E=\phi+eV_0\) ... (i)
\(\frac{E}{2}=\phi+\frac{eV_0}{4}\) ... (ii)
Multiply (ii) by 4: \(2E=4\phi+eV_0\). Subtract (i): \(E=3\phi\), so \(\phi=\frac{E}{3}=\frac{hc}{3\lambda}\).
Threshold wavelength: \(\lambda_0=\frac{hc}{\phi}=3\lambda\).
Answer: (1) \(3\lambda\)
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