JEE Main 2025 · 29 January · evening · Single correct · +4 / −1
In an experiment with photoelectric effect, the stopping potential,
- (a) is (1/e) times the maximum kinetic energy of the emitted photoelectrons
- (b) increases with increase in the intensity of the incident light
- (c) decreases with increase in the intensity of the incident light
- (d) increases with increase in the wavelength of the incident light
Answer and solution
Answer (a) is (1/e) times the maximum kinetic energy of the emitted photoelectrons
Solution
Physics sir · 3 steps
The stopping potential is just the fastest electron's kinetic energy measured in volts.
- What stops electrons. The stopping potential V₀ is the reverse voltage that just stops even the fastest photoelectron. The field does work eV₀ on it, and that work must use up all of its kinetic energy.
eV₀ = Kₘₐₓ = hc/λ - φ
- Read the equation. So V₀ = 1/eKₘₐₓ. Intensity does not appear anywhere: it changes how many electrons come out, not how fast each one is. A longer λ means less energy per photon, so V₀ goes down, not up.
- So. The stopping potential is (1/e) times the maximum kinetic energy of the photoelectrons, which is option (a).
The idea to keep. Intensity controls the number of photoelectrons. Frequency controls their energy. Stopping potential follows Kₘₐₓ, so only frequency (or wavelength) can change it.
Answer from the NTA final answer key. Solution by PrepWiser.