Topic E.2 — Quantum Physics — is the most conceptually demanding HL-only topic in the IB Physics syllabus, and it returns every year on Paper 1 and Paper 2. The three pillars are the photoelectric effect (which proves light is quantised), the de Broglie wavelength (which proves matter is wavelike) and Compton scattering (which proves photons carry momentum). Together they killed the classical wave model of light and replaced it with wave–particle duality.
This cheatsheet condenses every formula, graph interpretation, derivation and trap from Topic E.2 HL onto a single page. Scroll to the bottom for the printable PDF, the full notes pack, and the gated tutorial library used by Photon Academy students in Singapore.
Cross-reference with the official IB Physics subject page for the full syllabus.
§1 — Photoelectric Effect E.2 HL
Key concepts
- Classical theory fails: below the threshold frequency $f_0$ no emission occurs, regardless of intensity. Emission is instantaneous. KE depends on frequency, not intensity.
- Photon model (Einstein): light consists of photons, each with energy $E = hf$. One photon ejects one electron. $\Phi$ is the work function — minimum energy needed to escape the surface.
Classical vs photon theory comparison
| Property | Classical predicts | Observed (photon) |
|---|---|---|
| Effect of intensity | KE increases | More electrons; KE unchanged |
| Threshold freq. $f_0$ | None | Sharp threshold exists |
| Time delay | Yes (energy builds up) | Instantaneous |
| Effect of frequency | No effect on KE | $E_{\max} = hf - \Phi$ |
§2 — Wave–Particle Duality & de Broglie Wavelength E.2 HL
Wave–particle duality summary
| Entity | Wave evidence | Particle evidence |
|---|---|---|
| Light | Diffraction, interference, polarisation | Photoelectric effect, Compton scattering |
| Electrons | Electron diffraction (Davisson-Germer) | Definite mass, momentum, charge |
§3 — Compton Scattering E.2 HL
Compton shift at key angles
| Angle $\theta$ | $\cos\theta$ | $1 - \cos\theta$ | $\Delta\lambda$ |
|---|---|---|---|
| $0^\circ$ | $1$ | $0$ | $0$ |
| $90^\circ$ | $0$ | $1$ | $h/(m_e c) = 2.43 \times 10^{-12}$ m |
| $180^\circ$ | $-1$ | $2$ | $2 h/(m_e c) = 4.86 \times 10^{-12}$ m (max) |
§4 — Constants & Formula Summary E.2 HL
All formulae at a glance
- $E = hf = hc/\lambda$ (photon energy)
- $\Phi = h f_0$ (work function)
- $E_{\max} = hf - \Phi$ (Einstein's equation)
- $E_{\max} = e V_s$ (stopping voltage)
- $\lambda = h/p = h/(mv)$ (de Broglie)
- $\lambda = h/\sqrt{2 m q V}$ (accelerated particle)
- $p = h/\lambda$ (photon momentum)
- $\Delta\lambda = (h/m_e c)(1 - \cos\theta)$ (Compton)
Constants: $h = 6.63 \times 10^{-34}$ J s; $m_e = 9.11 \times 10^{-31}$ kg; $e = 1.60 \times 10^{-19}$ C; 1 eV $= 1.60 \times 10^{-19}$ J.
Worked Example — Photoelectric Effect & de Broglie
Question (HL Paper 2 style — 7 marks)
A clean caesium surface (work function $\Phi = 2.10$ eV) is illuminated with monochromatic light of wavelength $\lambda = 400$ nm.
(a) Show that emission occurs and calculate the maximum KE of the photoelectrons in eV. [3]
(b) Find the stopping voltage $V_s$. [1]
(c) Find the de Broglie wavelength of the most energetic photoelectron. [3]
Solution
- Photon energy: $E = hc/\lambda = (6.63 \times 10^{-34})(3.00 \times 10^8)/(400 \times 10^{-9}) = 4.97 \times 10^{-19}$ J $= 3.11$ eV. (M1)
- Since $E = 3.11$ eV $> \Phi = 2.10$ eV, emission occurs. (R1)
- $E_{\max} = hf - \Phi = 3.11 - 2.10 = 1.01$ eV. (A1)
- Stopping voltage: $V_s = E_{\max}/e = 1.01$ V. (A1)
- Convert to joules: $E_{\max} = 1.01 \times 1.60 \times 10^{-19} = 1.62 \times 10^{-19}$ J. Then $p = \sqrt{2 m_e E_{\max}} = \sqrt{2(9.11 \times 10^{-31})(1.62 \times 10^{-19})} = 5.43 \times 10^{-25}$ kg m s$^{-1}$. (M1)(A1)
- $\lambda = h/p = (6.63 \times 10^{-34})/(5.43 \times 10^{-25}) \approx 1.22 \times 10^{-9}$ m $\approx 1.22$ nm. (A1)
Examiner's note: Two common errors here. (i) Forgetting to convert $\Phi$ from eV to joules (or vice versa) before mixing it with $hc/\lambda$ — keep the entire calculation in one unit system. (ii) Using $E_{\max}$ directly as $\tfrac{1}{2}mv^2$ to find $v$ then $p$ — this works but adds a needless step. Use $p = \sqrt{2 m E_k}$ in one line.
Common Student Questions
Does intensity affect the kinetic energy of photoelectrons?
How do I read the $E_{\max}$ vs frequency graph?
What is the difference between Compton scattering and the photoelectric effect?
What happens to the electron diffraction rings if I increase the accelerating voltage?
Does the Compton shift depend on the incident wavelength?
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