Question:
When radiation of wavelength \(\lambda\) is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength \(3\lambda\), the stopping potential is \(\frac{V}{4}\). If the threshold wavelength for the metallic surface is \(n\lambda\) then value of n will be ________.
When radiation of wavelength \(\lambda\) is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength \(3\lambda\), the stopping potential is \(\frac{V}{4}\). If the threshold wavelength for the metallic surface is \(n\lambda\) then value of n will be ________.
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Write Einstein's equation \(eV_0 = hc/\lambda - \phi\) for the two wavelengths and subtract to eliminate \(\phi\).
Updated On: Nov 5, 2026
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Correct Answer: 9
Solution and Explanation
Einstein's equation: \(\dfrac{hc}{\lambda} = \phi + eV\) ...(1) and \(\dfrac{hc}{3\lambda} = \phi + \dfrac{eV}{4}\) ...(2).
Subtract (2) from (1): \(\dfrac{hc}{\lambda}\left(1-\dfrac{1}{3}\right) = \dfrac{3eV}{4}\), so \(eV = \dfrac{8hc}{9\lambda}\).
Put in (1): \(\phi = \dfrac{hc}{\lambda} - \dfrac{8hc}{9\lambda} = \dfrac{hc}{9\lambda}\).
Threshold wavelength \(\lambda_0 = \dfrac{hc}{\phi} = 9\lambda\), so \(n = 9\).
Answer: 9
Subtract (2) from (1): \(\dfrac{hc}{\lambda}\left(1-\dfrac{1}{3}\right) = \dfrac{3eV}{4}\), so \(eV = \dfrac{8hc}{9\lambda}\).
Put in (1): \(\phi = \dfrac{hc}{\lambda} - \dfrac{8hc}{9\lambda} = \dfrac{hc}{9\lambda}\).
Threshold wavelength \(\lambda_0 = \dfrac{hc}{\phi} = 9\lambda\), so \(n = 9\).
Answer: 9
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