When a metal surface is illuminated by light of wavelength λ, the stopping potential is 8 V. When the same surface is illuminated by light of wavelength 3 λ, stopping potential is 2 V. The threshold wavelength for this surface is:

Solution

Physics sir · 4 steps

Write Einstein's equation for both wavelengths and subtract: the work function drops out first.

1. Einstein's equation. The fastest photoelectron has energy eV₀, which equals the photon energy minus the work function φ. Write it once for each wavelength.

   8e = hc/λ - φ, 2e = hc/3λ - φ

2. Subtract. Subtracting removes φ: 6e = 2/3hc/λ, so hc/λ = 9e, which is 9 eV. Put this back into the first equation: φ = 9e - 8e = 1 eV.
3. Threshold wavelength. At threshold the photon energy just equals φ, so hc/λ₀ = φ. The photon energy at λ is 9 times φ, and energy goes as 1/λ, so λ₀ = 9λ.
4. So. The threshold wavelength is 9λ, option (b).

The idea to keep. Two stopping potentials for two wavelengths means two Einstein equations. Subtract to find hc/λ, get φ, then use λ₀ = hc/φ. Working in eV saves you all the constants.

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