JEE Main 2024 Physics question
When a metal surface is illuminated by light of wavelength λ, the stopping potential is 8V. When the same surface is illuminated by light of wavelength 3λ, stopping potential is 2V. The threshold wavelength for this surface is:
Answer: (C)
Solution
8 eV = hc/λ − φ₀ and 2 eV = hc/(3λ) − φ₀. Subtracting, 6 eV = (2/3)hc/λ, so
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