The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

Solution

Physics sir · 4 steps

Find the kinetic energy first; speed goes as the square root of it, so the ratio is the root of the KE ratio.

1. Einstein's equation. Each photon spends φ = 0.6 eV to free the electron. Whatever is left becomes the electron's maximum kinetic energy.

   1/2mvₘₐₓ² = hν - φ

2. Kinetic energies. For the first light, K₁ = 3.8 - 0.6 = 3.2 eV. For the second, K₂ = 1.4 - 0.6 = 0.8 eV. So K₁ : K₂ = 4 : 1.
3. Speed from energy. The electron mass is the same in both cases, so vₘₐₓ ∝ √K. That gives v₁/v₂ = √(3.2/0.8) = √4 = 2.
4. So. The ratio of maximum speeds is 2 : 1, which is option (b).

The idea to keep. Subtract the work function before taking any ratio, and remember speed goes as √K. A kinetic energy ratio of 4 means a speed ratio of only 2.

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