Two identical photocathodes receive the light of frequencies f₁ and f₂ respectively. If the velocities of the photo-electrons coming out are v₁ and v₂ respectively, then

Solution

Physics sir · 4 steps

Same cathode means same work function, so subtract the two Einstein equations and φ drops out.

1. Einstein equation. For each cathode, the photon energy pays the work function φ and the rest becomes kinetic energy of the electron. Identical cathodes have the same φ.
2. Write both. hf₁ = φ + 1/2mv₁² and hf₂ = φ + 1/2mv₂².
3. Subtract. Taking the second from the first removes φ, leaving h(f₁ - f₂) = 1/2m(v₁² - v₂²). Multiply both sides by 2/m.

   v₁² - v₂² = 2h/m(f₁ - f₂)

4. So. The relation links the squares of the speeds, not the speeds themselves, and it is a difference. That is option (d).

The idea to keep. Two experiments on the same metal? Subtract the Einstein equations. The work function cancels, and kinetic energy brings in v², never v alone.

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