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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 is

1. 4λ
2. 5λ
3. λ/2
4. 3λ

Correct answer: 5λ

Solution

Using the photoelectric equation, we have eV = hc/λ - hc/λ₀ for the first case and e(V/4) = hc/(2λ) - hc/λ₀ for the second case. Solving these equations simultaneously gives the threshold wavelength λ₀ = 5λ.

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