Practice a College Board-style free response question on Modern Physics. Write your response, then reveal the model answer to see exactly what earns each point.
Free Response Question · Unit 7 · The Photoelectric Effect
A metal surface has a work function of 2.3 eV. Light of frequency 8.0 × 10¹⁴ Hz shines on the metal, ejecting photoelectrons. Use Planck's constant h = 6.63 × 10⁻³⁴ J·s = 4.14 × 10⁻¹⁵ eV·s.
Quantity
Value
Work function, φ
2.3 eV
Light frequency, f
8.0 × 10¹⁴ Hz
Planck's constant, h
4.14 × 10⁻¹⁵ eV·s
A
Calculate the energy of a single photon of this light, in eV.
✓ Model answer (earns the point)
Using E = hf: E = (4.14 × 10⁻¹⁵ eV·s)(8.0 × 10¹⁴ Hz) ≈ 3.3 eV.
Why it scores: Correctly states and applies E = hf, uses consistent units (eV·s with Hz, giving eV directly), AND arrives at the correct numerical value.
B
Calculate the maximum kinetic energy of the ejected photoelectrons.
✓ Model answer (earns the point)
Using KE_max = hf − φ: KE_max = 3.3 eV − 2.3 eV = 1.0 eV.
Why it scores: Correctly states and applies KE_max = hf − φ, uses the photon energy from Part A, AND arrives at the correct numerical value.
C
If the intensity of the light is doubled (with frequency unchanged), explain what happens to (1) the maximum kinetic energy of the ejected electrons, and (2) the number of electrons ejected per second.
✓ Model answer (earns the point)
(1) The maximum kinetic energy of the ejected electrons stays the same (1.0 eV), because KE_max depends only on the frequency of the light (through hf) and the work function — both unchanged here. Intensity does not appear in the KE_max equation.
(2) The number of electrons ejected per second doubles (approximately), because intensity corresponds to the number of photons arriving per second. Doubling intensity doubles the photon arrival rate, and since each photon above threshold frequency can eject one electron, the ejection rate increases correspondingly.
Why it scores: Correctly states that KE_max is unaffected by intensity (with the reasoning that intensity doesn't appear in hf − φ), AND correctly states that the ejection rate increases with intensity (with the reasoning connecting intensity to photon arrival rate) — both parts of this two-part conceptual question must be addressed for full credit.
How to score points on AP Physics 2 FRQs
State the relevant equation (E = hf, KE_max = hf − φ) explicitly before substituting values. Graders need to see the relationship, not just a final number.
Keep units consistent. Using h in eV·s when frequencies are in Hz gives energy directly in eV — saves a conversion step and avoids errors.
For "explain" prompts about intensity vs. frequency, always separate the two effects clearly. Frequency/photon energy affects KE_max; intensity/photon number affects the rate of ejection.
Justify with the underlying equation, not just a memorized rule. "KE_max = hf − φ has no intensity term, so KE_max doesn't change" earns more than just stating the conclusion.
Multi-part conceptual questions usually require BOTH parts to be addressed explicitly — don't let one half of the explanation crowd out the other.