The threshold frequency of a metal is f₀. When the light of frequency 2 f₀ is incident on the metal plate, the maximum velocity of photoelectrons is v₁. When the frequency of incident radiation is increased to 5 f₀, the maximum velocity of photoelectrons emitted is v₂. The ratio of v₁ to v₂ is :

Solution

Physics sir · 4 steps

Subtract the work function first. The energy left over is what sets the speed, not the full photon energy.

1. Einstein's equation. The work function is φ = hf₀. The fastest photoelectron keeps whatever photon energy is left after paying φ.

   1/2mvₘₐₓ² = hf - hf₀

2. Both frequencies. At 2f₀: 1/2mv₁² = 2hf₀ - hf₀ = hf₀. At 5f₀: 1/2mv₂² = 5hf₀ - hf₀ = 4hf₀.
3. Take the ratio. Divide the two equations: v₁²/v₂² = hf₀/4hf₀ = 1/4. Speed goes as the square root of kinetic energy, so v₁/v₂ = √(1/4) = 1/2.
4. So. v₁/v₂ = 1/2, which is option (a).

The idea to keep. In the photoelectric effect, KEₘₐₓ is proportional to (f - f₀), not to f. Velocity goes as the square root of that, so take the root at the very end.

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