The problems are on the photoelectric effect worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Light comes in photons of energy E = hf. In the photoelectric effect a photon ejects an electron only if hf is at least the work function φ, and the electron then has maximum kinetic energy Kmax = hf − φ. These worked answers correct the 20 problems of the Physics 11-12 worksheet on photoelectric effect, from energy from frequency to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. Keep track of whether energies are in joules or electronvolts, and convert before subtracting. Small rounding differences in the last digit are acceptable.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB E.2
Problem 1 Energy from frequency
1. E = hf = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.98 × 10⁻¹⁹ J.2. 3.98 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.49 eV.
Problem 2 Laser pointer photon
E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 650 × 10⁻⁹ = 3.06 × 10⁻¹⁹ J = 1.91 eV.Problem 3 Frequency of ultraviolet
f = c/λ = 3.00 × 10⁸ / 250 × 10⁻⁹ = 1.2 × 10¹⁵ Hz.Problem 4 Caesium threshold
1. φ = 2.1 × 1.60 × 10⁻¹⁹ = 3.36 × 10⁻¹⁹ J.2. f₀ = φ/h = 3.36 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.07 × 10¹⁴ Hz.
Problem 5 Photon sharing its energy
1. Kmax = hf − φ = 3.2 − 2.4 = 0.8 eV.2. 0.8 × 1.60 × 10⁻¹⁹ = 1.3 × 10⁻¹⁹ J.
Problem 6 Bright red light
Each electron absorbs one photon and needs at least the work function. Red photons have less energy than φ, so none can free an electron, however many arrive. Each ultraviolet photon has enough energy, so even a weak beam releases electrons.Problem 7 Stopping potential
eVs = Kmax, so Vs = 2.4 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.5 V.Problem 8 Green laser photons
1. E = hc/λ = 1.99 × 10⁻²⁵ / 532 × 10⁻⁹ = 3.74 × 10⁻¹⁹ J.2. N = P/E = 5.0 × 10⁻³ / 3.74 × 10⁻¹⁹ = 1.3 × 10¹⁶ photons/s.
Problem 9 Potassium test
1. 500 nm: 2.49 eV. 600 nm: 2.07 eV.2. Only 500 nm, since 2.49 eV is greater than 2.3 eV, and Kmax = 2.49 − 2.3 = 0.19 eV. At 600 nm the photon energy is below φ, so nothing is emitted.
Problem 10 Reading a Kmax graph
1. The line meets the axis at f₀ = 5.5 × 10¹⁴ Hz.2. Gradient = 1.87 / (4.5 × 10¹⁴) = 4.2 × 10⁻¹⁵ eV s.
3. φ = gradient × f₀ = 4.16 × 10⁻¹⁵ × 5.5 × 10¹⁴ = 2.29 eV. As a check, gradient × e = 6.6 × 10⁻³⁴ J s, which agrees with Planck’s constant.
Problem 11 Doubling the brightness
1. The photocurrent doubles.2. Kmax is unchanged.
3. Doubling the intensity doubles the number of photons per second, so twice as many electrons leave. Each photon still has energy hf, so each electron still has the same maximum kinetic energy.
Problem 12 Finding the work function
1. E = hc/λ = 6.63 × 10⁻¹⁹ J = 4.14 eV.2. Kmax = 1.74 eV, so φ = 4.14 − 1.74 = 2.40 eV.
3. λ₀ = hc/φ = 1.99 × 10⁻²⁵ / (2.40 × 1.60 × 10⁻¹⁹) = 517 nm.
Problem 13 Wavelength of an electron
1. λ = h/(mv) = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 2.0 × 10⁶) = 3.6 × 10⁻¹⁰ m (0.36 nm).2. It is of the same order as atomic spacings, so crystals diffract such electrons.
Problem 14 Cricket ball wavelength
1. λ = h/(mv) = 6.63 × 10⁻³⁴ / (0.16 × 38) = 1.1 × 10⁻³⁴ m.2. This is about 10¹⁹ times smaller than a nucleus, so no gap or obstacle is small enough to show diffraction.
Problem 15 Current–voltage graph
1. Current falls to zero at −1.2 V, so the stopping potential is 1.2 V and Kmax = 1.2 eV.2. The bright beam gives twice the saturation current (6 µA against 3 µA).
3. Both beams have the same frequency, so each photon has the same energy and Kmax is the same.
Problem 16 Photon against electron
1. E = hc/λ = 1.99 × 10⁻²⁵ / 1.0 × 10⁻¹⁰ = 1.99 × 10⁻¹⁵ J = 12.4 keV.2. p = h/λ = 6.63 × 10⁻²⁴ kg m/s. K = p²/2m = 2.41 × 10⁻¹⁷ J = 151 eV.
3. The electron has the same wavelength (the same resolution) but only about 151 eV, so it is far less damaging to a delicate sample than a 12 keV X-ray photon.
Problem 17 Electron diffraction
1. eV = p²/2m, so p = √(2meV) = √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 150) = 6.61 × 10⁻²⁴ kg m/s.2. λ = h/p = 6.63 × 10⁻³⁴ / 6.61 × 10⁻²⁴ = 0.100 nm.
3. sin θ = λ/d = 0.100 / 0.215 = 0.466, so θ = 27.8°.
Problem 18 Two wavelengths
1. 400 nm: E = 3.108 eV, Kmax = 0.21 eV, Vs = 0.21 V. 300 nm: E = 4.144 eV, Kmax = 1.24 eV, Vs = 1.24 V.2. v = √(2K/m) = √(2 × 1.244 × 1.60 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = 6.6 × 10⁵ m/s.
3. λ₀ = hc/φ = 1.99 × 10⁻²⁵ / (2.9 × 1.60 × 10⁻¹⁹) = 429 nm.
Problem 19 Photocell current
1. E = hc/λ = 4.97 × 10⁻¹⁹ J, so N = 2.0 × 10⁻³ / 4.97 × 10⁻¹⁹ = 4.02 × 10¹⁵ s⁻¹.2. Electrons per second = 0.010 × 4.02 × 10¹⁵ = 4.02 × 10¹³ s⁻¹.
3. I = (electrons per second) × e = 4.02 × 10¹³ × 1.60 × 10⁻¹⁹ = 6.4 × 10⁻⁶ A (6.4 µA).
Problem 20 Spot the errors
1. The denominator is 9.11 × 10⁻³¹ × 3.0 × 10⁶ = 2.73 × 10⁻²⁴, so λ = 2.4 × 10⁻¹⁰ m (0.24 nm), ten times smaller than the student’s value.2. Doubling the intensity doubles the photocurrent; Kmax depends on frequency only and is unchanged.
3. 10⁻³⁴ ÷ 10⁻²⁴ is about 10⁻¹⁰ m, so a result near 10⁻⁹ m would be suspect.
