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Question:

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:

5λ

5/2λ

4λ

3λ

Solution:

We know that: eV = hc/λ - φ (1) eV/4 = hc/2λ - φ (2) From equations (1) and (2), we get: => 4 = 1/λ - φ/hc 2/λ - φ/hc => φ/hc = 1/λ - 4/2λ => φ/hc = 1/λ - 2/λ => φ/hc = -1/λ => λ₀ = 3λ