Cambridge International A-Level Physics · topic 22
Quantum physics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A photoelectric threshold measured with stopping voltage
A metal's stopping-potential measurements against incident-light frequency are supplied. Use electron charge e = 1.60×10⁻¹⁹ C and Kmax = eVs. The stopping-potential fit extrapolates to Vs = 0 at 5.0×10¹⁴ Hz. Assume one-photon photoemission and no contact-potential correction. Supplied data (Frequency / 10¹⁴ Hz; Stopping potential / V): (5, 0), (6, 0.414), (7, 0.828), (8, 1.242). (a) Estimate Planck's constant and the work function from the graph; find maximum electron kinetic energy at 8.0×10¹⁴ Hz. [3 marks] (b) Predict the effect of doubling intensity at fixed frequency above threshold, and of using intense light below threshold. [3 marks]
Photoelectric stopping potential
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Stopping potential / V
Horizontal axis (x): Frequency / 10¹⁴ Hz
Read the exact data as a table
| Frequency / 10¹⁴ Hz | Stopping potential / V |
|---|---|
| 5 | 0 |
| 6 | 0.414 |
| 7 | 0.828 |
| 8 | 1.242 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Estimate Planck's constant and the work function from the graph; find maximum electron kinetic energy at 8.0×10¹⁴ Hz.
Reveal the worked solution
- Einstein's relation gives Vs = (h/e)f−φ/e. Graph slope is 0.414/(1.0×10¹⁴) = 4.14×10⁻¹⁵ V s.
- Thus h = e×slope = 6.624×10⁻³⁴ J s. At threshold, φ = hf0 = 3.312×10⁻¹⁹ J, or 2.07 eV.
- At 8.0×10¹⁴ Hz, Kmax = eVs = (1.60×10⁻¹⁹)(1.242)≈1.987×10⁻¹⁹ J. The graph intercept is an energy-per-charge condition, not a delay.
- At fixed frequency, each photon carries the same hf. Doubling intensity supplies more photons per second, which can increase photocurrent if collection and material conditions remain unchanged.
- Stopping potential depends on the maximum energy per emitted electron, so it does not double with intensity.
- Below threshold each photon lacks the work-function energy. Increasing photon arrival rate does not fix that within the stated single-photon model; multiphoton regimes are outside this approximation.
Answer (a) h≈6.624×10⁻³⁴ J s; work function≈3.312×10⁻¹⁹ J = 2.07 eV; Kmax≈1.987×10⁻¹⁹ J = 1.242 eV. (b) Above threshold, emission rate can rise without changing Kmax; below threshold the ideal one-photon model predicts no photoemission.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Inferring discrete levels from a line spectrum
An idealised atom has the supplied energy levels relative to ionisation at 0 eV. Use h = 4.14×10⁻¹⁵ eV s and c = 3.0×10⁸ m s⁻¹. The atom initially occupies the −1.5 eV level; allowed transitions between all listed levels may emit photons. Supplied data (Level index; Energy / eV): (1, -6), (2, -3), (3, -1.5). (a) List the possible emitted photon energies in cascades and find the shortest emitted wavelength. [3 marks] (b) From the −6.0 eV ground level, can isolated 2.0 eV and 6.5 eV photons be absorbed? Explain what line spectra establish about the model. [3 marks]
Idealised energy levels
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Energy / eV
Horizontal axis (x): Level index
Read the exact data as a table
| Level index | Energy / eV |
|---|---|
| 1 | -6 |
| 2 | -3 |
| 3 | -1.5 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): List the possible emitted photon energies in cascades and find the shortest emitted wavelength.
Reveal the worked solution
- Energy drops are (−1.5)−(−3.0)=1.5 eV, (−3.0)−(−6.0)=3.0 eV and (−1.5)−(−6.0)=4.5 eV.
- A direct transition emits one 4.5 eV photon; a two-step cascade emits separate 1.5 and 3.0 eV photons, not one of each plus 4.5 eV in the same decay.
- The greatest photon energy gives shortest wavelength: λ = hc/E = (4.14×10⁻¹⁵)(3.0×10⁸)/4.5≈2.76×10⁻⁷ m.
- A bound–bound absorption must match an allowed energy difference; the listed ground-state excitation gaps are 3.0 and 4.5 eV, not 2.0 eV.
- Ground-state ionisation needs 6.0 eV. Absorbing 6.5 eV can free the electron with 0.5 eV excess kinetic energy, neglecting recoil.
- Discrete lines support quantised energy changes. They do not show electrons following literal tiny planetary trajectories, and a real atom may have selection rules absent from this idealised model.
Answer (a) Possible photons: 1.5, 3.0 and 4.5 eV; shortest wavelength≈276 nm. (b) A 2.0 eV photon matches no listed excitation; a 6.5 eV photon can ionise the atom, leaving 0.5 eV kinetic energy.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.