JEE Main 2021 · 27 August · morning · Single correct · +4 / −1
In a photoelectric experiment, increasing the intensity of incident light :
- (a) increases the number of photons incident and also increases the K.E. of the ejected electrons
- (b) increases the frequency of photons incident and increases the K.E. of the ejected electrons
- (c) increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged
- (d) increases the number of photons incident and the K.E. of the ejected electrons remains unchanged
Answer and solution
Answer (d) increases the number of photons incident and the K.E. of the ejected electrons…
Solution
Physics sir · 4 steps
Intensity decides how many photons arrive; frequency alone decides how much energy each one carries.
- What intensity means. For light of a fixed frequency, intensity is the energy delivered per unit area per second. Each photon carries hν, so a brighter beam means more photons per second, each with the same hν.
Kₘₐₓ = hν - φ₀
- One photon, one electron. Each electron absorbs exactly one photon. Its maximum kinetic energy depends only on hν and the work function φ₀, and intensity appears nowhere in this equation.
- Effect of brighter light. More photons per second eject more electrons per second, so the photocurrent rises. The kinetic energy of each ejected electron stays the same.
- So. Increasing intensity increases the number of incident photons while the K.E. of the ejected electrons remains unchanged, which is option (d).
The idea to keep. In the photoelectric effect, intensity controls the number of electrons (the current), and frequency controls their energy. Never mix the two.
Answer from the NTA final answer key. Solution by PrepWiser.