The problems are on the photoelectric effect worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Light is absorbed and emitted in photons, each carrying energy E = hf. These worked answers correct the 20 problems of the AP Physics 2 worksheet on photoelectric effect, from photon energy from frequency to measure h with LEDs: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the relationship, the substitution and the result with its unit. Use hc = 1240 eV·nm and 1 eV = 1.60 × 10⁻¹⁹ J. Check whether you reasoned with the energy of one photon, not the energy of the whole beam.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 15
Problem 1 Photon energy from frequency
1. E = hf = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.3 × 10⁻¹⁹ J.2. 3.3 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.07 eV.
Problem 2 Converting energy units
1. 3.20 × 1.60 × 10⁻¹⁹ = 5.12 × 10⁻¹⁹ J.2. 2.0 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.25 eV.
Problem 3 Energy from wavelength
1. E = 1240 / 620 = 2.00 eV.2. Energy is inversely proportional to wavelength, so 310 nm photons have twice the energy, 4.00 eV.
Problem 4 Bright light below the threshold
1. Each red photon has less energy than the work function of zinc. An electron absorbs one photon at a time, so no electron receives enough energy, and extra photons cannot add up.2. Each ultraviolet photon has more energy than the work function, so a single photon can free an electron, even if there are few photons. Frequency, not brightness, decides whether emission occurs.
Problem 5 Threshold of a metal
1. λ₀ = hc / φ = 1240 / 2.10 = 590 nm.2. f₀ = c / λ₀ = 3.00 × 10⁸ / 5.90 × 10⁻⁷ = 5.1 × 10¹⁴ Hz.
Problem 6 Maximum kinetic energy
1. E = 1240 / 350 = 3.54 eV.2. K max = 3.54 − 2.30 = 1.24 eV.
3. V s = K max / e = 1.24 V.
Problem 7 Wavelength of a cricket ball
1. p = mv = 0.16 × 30 = 4.8 kg·m/s.2. λ = h / p = 6.63 × 10⁻³⁴ / 4.8 = 1.4 × 10⁻³⁴ m.
3. This is far smaller than any object or gap, so no diffraction or interference is possible.
Problem 8 Reading a K max graph
1. The line meets the axis at 4.0 × 10¹⁴ Hz.2. Gradient = (1.66 − 0) / (8.0 − 4.0) × 10¹⁴ = 4.15 × 10⁻¹⁵ eV·s.
3. 4.15 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.64 × 10⁻³⁴ J·s, matching h = 6.63 × 10⁻³⁴ J·s.
4. φ = h f₀ = 4.14 × 10⁻¹⁵ × 4.0 × 10¹⁴ = 1.66 eV.
Problem 9 Photons per second
1. E = hc / λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 5.32 × 10⁻⁷ = 3.74 × 10⁻¹⁹ J.2. N = P / E = 5.0 × 10⁻³ / 3.74 × 10⁻¹⁹ = 1.3 × 10¹⁶ photons per second.
Problem 10 Momentum of an electron wave
1. p = h / λ = 6.63 × 10⁻³⁴ / 2.0 × 10⁻¹⁰ = 3.32 × 10⁻²⁴ kg·m/s.2. v = p / m = 3.32 × 10⁻²⁴ / 9.11 × 10⁻³¹ = 3.64 × 10⁶ m/s.
3. K = p² / 2m = 6.03 × 10⁻¹⁸ J = 37.7 eV.
Problem 11 Predict the effect of changes
1. Current doubles; K max and V s are unchanged, because the number of photons doubles but each keeps the same energy.2. K max and V s increase since each photon has more energy; the current may change slightly because fewer photons arrive per watt.
3. K max and V s decrease because φ is larger, and emission may stop if f falls below the new threshold.
Problem 12 Which light ejects electrons?
1. UV: 1240 / 250 = 4.96 eV. Red: 1240 / 700 = 1.77 eV.2. Only ultraviolet works, since 1.77 eV is below 2.0 eV.
3. K max = 4.96 − 2.0 = 2.96 eV.
Problem 13 Stopping potential data
1. Energies: 1240/450 = 2.76, 1240/400 = 3.10, 1240/350 = 3.54, 1240/300 = 4.13 eV.2. E − eV s: 2.20, 2.20, 2.20, 2.20 eV.
3. The result is constant, so the work function is 2.20 eV.
Problem 14 Spot the error
1. The student treats the beam energy as shared by all electrons. In the photon model an electron absorbs one photon, so what matters is the energy per photon, not the total energy.2. Brighter red light delivers more red photons, each with too little energy, so no electrons are ejected at any intensity.
Problem 15 Sunscreen and bond breaking
1. λ = 1240 / 3.8 = 326 nm.2. 300 nm has 4.13 eV and can break the bond; 600 nm has 2.07 eV and cannot.
Problem 16 Unknown frequency from a stopping potential
1. E = φ + eV s = 2.40 + 1.10 = 3.50 eV.2. f = E / h = 3.50 × 1.60 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 8.45 × 10¹⁴ Hz.
3. λ = 1240 / 3.50 = 354 nm.
Problem 17 Electrons through a voltage
1. K = 400 × 1.60 × 10⁻¹⁹ = 6.4 × 10⁻¹⁷ J.2. p = √(2mK) = √(2 × 9.11 × 10⁻³¹ × 6.4 × 10⁻¹⁷) = 1.08 × 10⁻²³ kg·m/s.
3. λ = h / p = 6.63 × 10⁻³⁴ / 1.08 × 10⁻²³ = 6.1 × 10⁻¹¹ m.
4. The wavelength is the same order of magnitude as the atom spacing, so the lattice acts as a diffraction grating.
Problem 18 Single electrons through slits
1. The spots form bright and dark fringes as in the double-slit pattern for light.2. Δy = λL / d = 5.0 × 10⁻¹¹ × 0.50 / 1.0 × 10⁻⁶ = 2.5 × 10⁻⁵ m.
3. Each electron arrives at one point like a particle, but the distribution of arrivals follows an interference pattern like a wave. Electrons show both particle and wave behaviour.
Problem 19 Photon against electron
1. p = h / λ is the same for both: 6.63 × 10⁻³⁴ / 1.0 × 10⁻¹⁰ = 6.63 × 10⁻²⁴ kg·m/s.2. E = pc = 6.63 × 10⁻²⁴ × 3.00 × 10⁸ = 1.99 × 10⁻¹⁵ J = 12.4 keV.
3. K = p² / 2m = 2.41 × 10⁻¹⁷ J = 151 eV.
4. The photon has about 80 times more energy because it moves at c, while a massive electron needs far less kinetic energy for the same momentum.
Problem 20 Measure h with LEDs
1. Measure the voltage at which each LED just glows (use a dark tube) and plot V against 1/λ for several LEDs; the gradient is hc / e.2. 1/λ is 1.587 × 10⁶ m⁻¹ and 2.128 × 10⁶ m⁻¹, so the gradient = 0.66 / 0.541 × 10⁶ = 1.22 × 10⁻⁶ V·m.
3. h = gradient × e / c = 1.22 × 10⁻⁶ × 1.60 × 10⁻¹⁹ / 3.00 × 10⁸ = 6.5 × 10⁻³⁴ J·s.
