A metal exposed to light of wavelength 800 nm and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The workfunction of the metal is : (Take hc =1230 eV-nm ).

Solution

Physics sir · 4 steps

Write Einstein's equation for both wavelengths, then use K₂ = 2K₁ to solve for φ.

1. Photon energies. Use E = hc/λ with hc = 1230 eV nm. At 800 nm, E₁ = 1.5375 eV. At 500 nm, E₂ = 2.46 eV.
2. Einstein's equation. For each light, Kₘₐₓ = E - φ. So K₁ = 1.5375 - φ and K₂ = 2.46 - φ, both in eV.

   2.46 - φ = 2(1.5375 - φ)

3. Solve. The right side is 3.075 - 2φ, so φ = 3.075 - 2.46 = 0.615 eV. Check: K₁ = 0.9225 eV and K₂ = 1.845 eV, which is exactly double.
4. So. The work function is 0.615 eV, option (c).

The idea to keep. When a question links two kinetic energies, write K = hc/λ - φ for each case and plug in the given relation. The work function is the same in both equations, so it is the only unknown.

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