When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v₁. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v₂. If v₂ = x v₁, the value of x will be __________.

Solution

Physics sir · 4 steps

Subtract the work function first, then take the square root. Speed goes as the root of kinetic energy.

1. Einstein's equation. The maximum kinetic energy is the photon energy minus the work function, and the work function is hν₀, where ν₀ is the threshold frequency.

   1/2mvₘₐₓ² = hν - hν₀

2. Both cases. At ν = 2ν₀: 1/2mv₁² = 2hν₀ - hν₀ = hν₀. At ν = 5ν₀: 1/2mv₂² = 5hν₀ - hν₀ = 4hν₀.
3. Take the ratio. Dividing gives v₂²/v₁² = 4hν₀/hν₀ = 4, so v₂/v₁ = √4 = 2.
4. So. v₂ = 2v₁, which gives x = 2.

The common slip: 4. Taking the ratio of the kinetic energies, 4hν₀ to hν₀, and stopping there gives 4. That ratio is v₂²/v₁², so its square root gives 2.

The idea to keep. In photoelectric questions, kinetic energy depends on ν - ν₀, not on ν alone. Find the energies first, then remember speed goes as the square root of kinetic energy.

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