A2 · Practice questions · The photoelectric effect

One photon, one electron.

Six original Cambridge-style questions on the work function and threshold frequency, Einstein's photoelectric equation hf = Φ + KE_max, stopping potentials, and why intensity and frequency do different jobs.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Frequency sets the energy, intensity sets the rate.

Work function Φ: the minimum energy needed to remove an electron from the surface of the metal; the threshold frequency is f₀ = Φ / h.
Einstein's equation: hf = Φ + KE_max, so KE_max = hf − Φ depends only on the frequency and the metal; the stopping potential V_s satisfies eV_s = KE_max.
Intensity: the number of photons per second; above the threshold it sets the number of photoelectrons per second (the current), never their maximum energy.

Analysis

[3 marks]

Define the work function of a metal and its threshold frequency. Light of frequency below the threshold frequency is shone on the metal and its intensity is then increased greatly. Explain why no photoelectrons are emitted.

The work function Φ is the minimum energy needed to remove an electron from the surface of the metal. ✓
The threshold frequency f₀ is the lowest frequency of radiation that causes photoelectric emission from that metal, where hf₀ = Φ. ✓
Each electron is freed by a single photon; below f₀ every photon carries less than Φ, and increasing the intensity only sends more of these too-small photons, so none can free an electron. ✓

Calculation

[3 marks]

Caesium has a work function of 1.9 eV. Blue light of wavelength 450 nm falls on a caesium surface. Calculate the energy of a photon of this light in electronvolts, the maximum kinetic energy of the emitted electrons in joules, and their maximum speed. (Electron mass = 9.11 × 10⁻³¹ kg.)

E = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (450 × 10⁻⁹) = 4.42 × 10⁻¹⁹ J = 2.76 eV. ✓
KE_max = hf − Φ = 2.76 − 1.9 = 0.86 eV = 0.86 × 1.60 × 10⁻¹⁹ = 1.4 × 10⁻¹⁹ J. ✓
½mv² = 1.38 × 10⁻¹⁹, so v = √(2 × 1.38 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = 5.5 × 10⁵ m s⁻¹. ✓

Calculation

[3 marks]

The longest wavelength of light that causes photoelectric emission from potassium is 540 nm. Calculate the work function of potassium in joules and in electronvolts. Light of frequency 7.5 × 10¹⁴ Hz is then used; calculate the maximum kinetic energy of the photoelectrons, in electronvolts.

Φ = hc / λ₀ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (540 × 10⁻⁹) = 3.68 × 10⁻¹⁹ J. ✓
Φ = 3.68 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.30 eV. ✓
hf = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.97 × 10⁻¹⁹ J = 3.11 eV, so KE_max = 3.11 − 2.30 = 0.81 eV. ✓

Analysis

[2 marks]

State two observations from photoelectric experiments that cannot be explained if light is treated purely as a wave, and explain how the photon model accounts for one of them.

Any two of: there is a threshold frequency below which no electrons are emitted however intense the light; the maximum kinetic energy of the electrons depends on frequency and not on intensity; emission is instantaneous even for very faint light. ✓
In the photon model the energy arrives in packets of hf absorbed one at a time by single electrons, so an electron gains at most hf: below Φ it cannot escape (threshold), and above it the surplus hf − Φ sets KE_max whatever the intensity. ✓

Calculation

[3 marks]

In a photoelectric experiment the stopping potential V_s is measured for two frequencies of light on the same metal: at f = 6.0 × 10¹⁴ Hz, V_s = 0.60 V; at f = 8.0 × 10¹⁴ Hz, V_s = 1.43 V. Use these results to determine a value for the Planck constant, the threshold frequency of the metal and its work function in electronvolts.

eV_s = hf − Φ, so h = eΔV_s / Δf = 1.60 × 10⁻¹⁹ × (1.43 − 0.60) / (2.0 × 10¹⁴) = 6.6 × 10⁻³⁴ J s. ✓
The line reaches V_s = 0 at f₀ = 6.0 × 10¹⁴ − 0.60 × 1.60 × 10⁻¹⁹ / 6.64 × 10⁻³⁴ = 4.6 × 10¹⁴ Hz. ✓
Φ = hf₀ = 6.64 × 10⁻³⁴ × 4.55 × 10¹⁴ = 3.0 × 10⁻¹⁹ J = 1.9 eV. ✓

Calculation

[4 marks]

Zinc has a work function of 4.3 eV. Ultraviolet light of wavelength 200 nm falls on a zinc plate; the power of the light reaching the plate is 3.0 mW, and on average one photon in every 1000 releases a photoelectron.

(a) Calculate the threshold frequency of zinc. (b) Calculate the maximum kinetic energy of the photoelectrons, in electronvolts. (c) Calculate the photoelectric current.

(a) f₀ = Φ / h = (4.3 × 1.60 × 10⁻¹⁹) / 6.63 × 10⁻³⁴ = 1.0 × 10¹⁵ Hz. ✓
(b) E = hc / λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (200 × 10⁻⁹) = 9.95 × 10⁻¹⁹ J = 6.22 eV, so KE_max = 6.22 − 4.3 = 1.9 eV. ✓
(c) Photons per second = P / E = 3.0 × 10⁻³ / 9.95 × 10⁻¹⁹ = 3.0 × 10¹⁵ s⁻¹, so electrons per second = 3.0 × 10¹² s⁻¹. ✓
I = 3.0 × 10¹² × 1.60 × 10⁻¹⁹ = 4.8 × 10⁻⁷ A (0.48 μA). ✓

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