In photoelectric effect, the number of electrode emitted per second from a metallic surface is proportional to the

Step 1: Analyze the question The question asks what determines the number of electrons emitted per second (the photocurrent), not the energy of each electron.

Step 2: The concept In the photoelectric effect, light behaves as a stream of photons, each carrying energy E = hf.

• Frequency (f) decides whether emission happens at all (f must be at least the threshold frequency f₀) and sets the maximum kinetic energy of each electron: KE_max = hf − φ
• Intensity is the number of photons arriving per second per unit area. Each photon can free at most one electron, so more photons per second means more electrons emitted per second.

Step 3: Check each option

• A. Intensity of the incident radiation: The number of photons per second is proportional to intensity, so the number of electrons emitted per second is proportional to intensity (provided f ≥ f₀). ✔
• B. Frequency: This affects the kinetic energy of the electrons, not how many are emitted. ✘
• C. Energy of the incident radiation: The energy of each photon depends on frequency, so this also affects electron energy rather than the number. ✘
• D. Work function: This is a property of the metal that sets the threshold. It is not something the emission rate is proportional to. ✘

Step 4: Answer The number of electrons emitted per second is proportional to the intensity of the incident radiation.

Exam tip: Remember it this way:

• Intensity → Number of electrons (photocurrent)
• Frequency → Kinetic energy of electrons (and whether emission happens)