In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function φ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 × 10⁻³⁴ Js, c = 3 × 10⁸ ms⁻¹)

Solution

Physics sir · 4 steps

The work function is the same for both wavelengths, so it cancels when you subtract.

1. Einstein's equation. For each wavelength, eV₀ = hc/λ - φ. Subtract the two equations and φ drops out, because the cathode is the same.

   e Δ V₀ = hc(1/λ₁ - 1/λ₂)

2. Handy value of hc. hc = 6.63 × 10⁻³⁴ × 3 × 10⁸ = 1.989 × 10⁻²⁵ J m. Dividing by 1.6 × 10⁻¹⁹ J eV⁻¹ gives hc ≈ 1243 eV nm.
3. Put in numbers. 1243/280 = 4.44 eV and 1243/400 = 3.11 eV. The photon energy drops by 4.44 - 3.11 = 1.33 eV.
4. So. The stopping potential falls by about 1.3 V, from 1.94 V to 0.61 V. The answer is (a).

The idea to keep. Same metal, two wavelengths: the change in stopping potential equals the change in photon energy in eV. Keep hc ≈ 1240 eV nm in your pocket for quick work.

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