Which of the following statements about photoelectrons is correct?

Analyze the question: It asks which statement about photoelectrons (the photoelectric effect) is correct, comparing green light and red light.

Concept: Einstein's photoelectric equation:

hf = φ + KE_max

• The maximum kinetic energy of photoelectrons depends only on the frequency of the light (and the metal's work function φ). It does not depend on intensity (brightness).
• The number of photoelectrons emitted per second depends on the intensity of the light (provided f ≥ threshold frequency).
• Green light has a higher frequency than red light (green ≈ 5.5 × 10¹⁴ Hz, red ≈ 4.3 × 10¹⁴ Hz), so its photons carry more energy.

Step by step:

• A. Says faint green gives less KE than bright red. Wrong. Brightness doesn't affect KE, and green has the higher frequency, so its electrons have more KE.
• B. Says red releases fewer electrons than green. Not necessarily true. The number depends on intensity, not colour. Red may not even release any electrons if its frequency is below the threshold, but the statement isn't a general rule.
• C. Says faint green gives photoelectrons with greater KE than bright red. Correct. Green has the higher frequency, so each photon has more energy (E = hf). Faintness only means fewer photons (fewer electrons), not less energy per electron.
• D. Says red produces more photoelectrons than green. Wrong as a general statement. The number depends on intensity, and red's lower frequency may be below the threshold.

Answer: C

Exam tip:

• Frequency → energy (KE) of each electron
• Intensity → number of electrons

Whenever you see "bright vs faint" and "different colours", compare the frequencies first. Remember the colour order of increasing frequency: Red < Orange < Yellow < Green < Blue < Violet.