When radiation of wavelength λ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3λ, the stopping potential is V/4. If the threshold wavelength for the metallic surface is nλ then value of n will be __________.

Solution

Physics sir · 4 steps

Write Einstein's equation for both wavelengths, then remove V. The work function is all that remains.

1. Two equations. Use eV₀ = hc/λ - φ for each light. With λ you get eV = hc/λ - φ. With 3λ you get eV/4 = hc/3λ - φ.
2. Eliminate V. Multiply the second equation by 4 to get eV = 4hc/3λ - 4φ. Set it equal to the first: hc/λ - φ = 4hc/3λ - 4φ.

   3φ = 4hc/3λ - hc/λ = hc/3λ ⇒ φ = hc/9λ

3. Threshold wavelength. At threshold the photon energy just equals the work function, so hc/λ₀ = φ = hc/9λ. This gives λ₀ = 9λ.
4. So. The threshold wavelength is 9λ, so n = 9.

The idea to keep. Two stopping potentials for one metal means two Einstein equations with one shared φ. Eliminate V first, then use λ₀ = hc/φ.

Answer from the NTA final answer key. Solution by PrepWiser.