A light source of wavelength λ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a light source of wavelength λ/2, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV )

Solution

Physics sir · 3 steps

Find the photon energy first, double it for half the wavelength, then subtract the work function again.

1. First photon energy. Einstein's equation is hν = φ + Kₘₐₓ. With φ = 1 eV and Kₘₐₓ = 2 eV, the photon of wavelength λ carries hc/λ = 3 eV.
2. Halve the wavelength. Photon energy is hc/λ, so it varies as 1/λ. At wavelength λ/2 each photon carries twice as much: hc/(λ/2) = 2 × 3 = 6 eV.

   K'ₘₐₓ = 2hc/λ - φ = 6 eV - 1 eV = 5 eV

3. So. The same metal still takes 1 eV to free an electron, so the maximum kinetic energy is 5 eV, option (a).

The idea to keep. Changing the wavelength scales the photon energy, not the kinetic energy. Rebuild hν first, scale it, then subtract the same work function.

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