A photoelectric surface is illuminated successively by monochromatic light of wavelengths \(\lambda\) and \(\frac{\lambda}{2}\). If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :
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- \(\frac{hc}{3\lambda}\)
- \(\frac{hc}{2\lambda}\)
- \(\frac{hc}{\lambda}\)
- \(\frac{3hc}{\lambda}\)
The Correct Option is B
Solution and Explanation
Given \(K_2 = 3K_1\): \(\frac{2hc}{\lambda} - \phi = \frac{3hc}{\lambda} - 3\phi\).
So \(2\phi = \frac{hc}{\lambda}\), giving \(\phi = \frac{hc}{2\lambda}\).
Answer: (2) \(\frac{hc}{2\lambda}\)
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