The questions are on the photoelectric effect test. Take it under real conditions before reading the mark scheme.
Quick answer: Light is absorbed and emitted in photons, each carrying energy E = hf. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics 2 test on photoelectric effect, from photon energy to LED colours, with the equivalent answers to accept.
Marks are shown for each step. Method marks are given for a correct relationship even if the arithmetic slips, but final answers need units.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 15
Question 1 Photon energy (2 marks)
a. E = 1240 / 480 = 2.58 eV.b. f = c / λ = 3.00 × 10⁸ / 4.80 × 10⁻⁷ = 6.25 × 10¹⁴ Hz.
Marking: one mark each part.
Question 2 Photoelectric calculation (3 marks)
a. E = 1240 / 450 = 2.76 eV, so K max = 2.76 − 2.10 = 0.66 eV.b. λ₀ = 1240 / 2.10 = 590 nm.
Marking: photon energy (1); subtraction (1); threshold (1).
Question 3 Failure of the wave model (2 marks)
a. A wave of high intensity should give any electron enough energy after a short time, whatever the frequency, but no emission is seen.b. Each photon has energy hf; a red photon has less than the work function, so no electron is freed, however many photons arrive.
Marking: one mark each part.
Question 4 Reading a current graph (3 marks)
a. The upper curve is brighter light, because more photons release more electrons per second, giving a larger saturation current.b. The current drops to zero at about −1.5 V, so V s = 1.5 V and K max = 1.5 eV.
c. Both curves use light of the same frequency, so each photon has the same energy and the electrons have the same maximum kinetic energy.
Marking: one mark each part.
Question 5 de Broglie wavelength of an electron (3 marks)
a. p = mv = 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.82 × 10⁻²⁴ kg·m/s.b. λ = h / p = 6.63 × 10⁻³⁴ / 1.82 × 10⁻²⁴ = 3.6 × 10⁻¹⁰ m.
c. Its wavelength, about 4 × 10⁻³⁵ m, is far too small to meet any gap of comparable size.
Marking: one mark each part.
Question 6 Evidence for the models (2 marks)
a. The photoelectric effect supports the particle nature of light.b. Electron diffraction supports the wave nature of matter.
Marking: one mark each part.
Question 7 Photons per second (3 marks)
a. E = hc / λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 6.00 × 10⁻⁷ = 3.3 × 10⁻¹⁹ J.b. N = P / E = 2.0 × 10⁻³ / 3.3 × 10⁻¹⁹ = 6.0 × 10¹⁵ per second.
Marking: energy (1); division (1); answer with unit (1).
Question 8 LED colours (2 marks)
a. A blue photon has a higher frequency and so a higher energy, E = hf, than a red photon. The electrons crossing the LED must gain that energy from the supply, eV = hf, so a larger voltage is needed to emit blue light.Marking: relation between colour and photon energy (1); link to voltage (1).
