IB Diploma Physics · topic E.2
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.
Same light energy, different photon counts
A source emits monochromatic pulses of equal energy 6.626×10⁻⁶ J at two frequencies: 5.0×10¹⁴ Hz and 1.0×10¹⁵ Hz. Use h = 6.626×10⁻³⁴ J s and c = 3.0×10⁸ m s⁻¹. The graph provides photon energy as a function of frequency. Supplied model: Use photon energy E = hf, wavelength λ = c/f and photon count N = Epulse/Ephoton. Supplied data (Frequency / 10¹⁴ Hz; Photon energy / 10⁻¹⁹ J): (0, 0), (5, 3.313), (10, 6.626). (a) Calculate each wavelength, photon energy and photon count. Identify the approximate spectral region of each pulse. [3 marks] (b) Explain why interference supports a wave model while photon-counting detection needs a particle-like model. Evaluate the claim that all electromagnetic radiation has identical biological effects. [3 marks]
Photon-energy model
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Photon energy / 10⁻¹⁹ J
Horizontal axis (x): Frequency / 10¹⁴ Hz
Read the exact data as a table
| Frequency / 10¹⁴ Hz | Photon energy / 10⁻¹⁹ J |
|---|---|
| 0 | 0 |
| 5 | 3.313 |
| 10 | 6.626 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Calculate each wavelength, photon energy and photon count. Identify the approximate spectral region of each pulse.
Reveal the worked solution
- Use λ = c/f: the lower frequency has wavelength 6.0×10⁻⁷ m and the higher 3.0×10⁻⁷ m.
- Ephoton = hf gives 3.313×10⁻¹⁹ and 6.626×10⁻¹⁹ J. Dividing equal total pulse energy by these values gives 2.0×10¹³ and 1.0×10¹³ photons.
- 600 nm is visible light and 300 nm is ultraviolet. Both propagate at c in vacuum; the higher-frequency pulse contains fewer photons for the same total energy.
- Spatial interference patterns follow phase and superposition, which a simple stream of classical particles does not explain.
- Localized detector events with frequency-dependent energy are described by photons. These are complementary descriptions of quantum light rather than literal tiny balls riding classical waves.
- Absorption, penetration, dose and photon energy depend on wavelength and material. Ultraviolet can cause photochemical damage; exposure assessment cannot be inferred from vacuum speed alone.
Answer (a) 600 nm, 3.313×10⁻¹⁹ J, 2.0×10¹³ photons; 300 nm, 6.626×10⁻¹⁹ J, 1.0×10¹³ photons. (b) Different models describe different observations; common vacuum speed does not imply identical interaction or hazard.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.