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.
§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.
Photon energy:$E = hf = \dfrac{hc}{\lambda}$
Work function:$\Phi = h f_0$
Einstein's eqn:$E_{\max} = hf - \Phi$
Stopping voltage:$E_{\max} = e V_s \;\Rightarrow\; V_s = \dfrac{hf - \Phi}{e}$
TrickOn an $E_{\max}$ vs $f$ graph: gradient $= h$ (same for all metals); x-intercept $= f_0$ (metal-specific); y-intercept (extrapolated) $= -\Phi$. Changing intensity changes the photocurrent, NOT this graph.
TrapMaximum KE depends on frequency only, not intensity. Intensity controls the number of electrons (i.e. the photocurrent). Never say "more intense = faster electrons."
Note$h = 6.63 \times 10^{-34}$ J s, $c = 3.00 \times 10^8$ m s$^{-1}$, $e = 1.60 \times 10^{-19}$ C.
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
de Broglie:$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$
Accelerated through V:$\lambda = \dfrac{h}{\sqrt{2 m q V}}$
Photon momentum:$p = \dfrac{h}{\lambda} = \dfrac{E}{c}$
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 |
TrickTo compare wavelengths at equal kinetic energy, use $\lambda = h/\sqrt{2 m E_k}$ — heavier particle $\Rightarrow$ shorter $\lambda$. At equal momentum, $\lambda = h/p$ applies directly.
TrapFor electron diffraction, increasing the accelerating voltage decreases $\lambda$ (more momentum), so the diffraction rings move inward (less spreading), not outward. Many students get this the wrong way round.
NoteElectron diffraction rings: bright rings = constructive interference; dark rings = minimum intensity (destructive). The first minimum defines the edge of the central maximum, just like single-slit light diffraction.
§3 — Compton Scattering E.2 HL
Compton shift:$\Delta\lambda = \lambda_f - \lambda_i = \dfrac{h}{m_e c}(1 - \cos\theta)$
Compton wavelength:$\dfrac{h}{m_e c} = 2.43 \times 10^{-12}$ m
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) |
Trick$\Delta\lambda$ depends only on $\theta$, not on the incident wavelength $\lambda_i$. The scattered photon always has a longer wavelength (lower energy). Because the shift is tiny ($\sim$ pm), Compton scattering is significant only for X-rays / gamma rays — not visible light.
TrapDo NOT confuse Compton scattering with the photoelectric effect. In Compton: photon survives with a longer wavelength, electron recoils. In photoelectric: photon is absorbed, electron is ejected from the surface.
NoteDerivation of the Compton formula is NOT required by IB. The formula is given in the data booklet. You must know how to apply it and explain why it proves photons carry momentum.
§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?
No. Maximum kinetic energy depends on frequency only: $E_{\max} = hf - \Phi$. Intensity (photons per second) controls how many electrons are emitted — the photocurrent — not their KE. Saying "more intense light = faster electrons" is the single most common mark loss on E.2.
How do I read the $E_{\max}$ vs frequency graph?
The graph is a straight line above the threshold frequency. Gradient $= h$ (Planck's constant — same for every metal); x-intercept $= f_0$ (threshold frequency, metal-specific); y-intercept (extrapolated below $f_0$) $= -\Phi$ (the work function, negative). Changing intensity does NOT change this graph at all.
What is the difference between Compton scattering and the photoelectric effect?
In Compton scattering the incident photon survives but with a longer (shifted) wavelength, and a free electron recoils. In the photoelectric effect the photon is fully absorbed and a bound electron is ejected from a metal surface. Compton needs X-rays or gamma rays to be measurable; the photoelectric effect happens with UV / visible light.
What happens to the electron diffraction rings if I increase the accelerating voltage?
Increasing $V$ increases electron momentum, so $\lambda = h/\sqrt{2 m q V}$ decreases. A smaller $\lambda$ means less diffraction spreading, so the rings move inward (smaller radius) — not outward. This is a frequent IB Paper 1 trap, often paired with an "intensity changes" distractor (intensity changes brightness, not ring size).
Does the Compton shift depend on the incident wavelength?
No. $\Delta\lambda = (h/m_e c)(1 - \cos\theta)$ depends only on the scattering angle $\theta$, not on the incident wavelength. Maximum shift occurs at $\theta = 180^\circ$ (back-scattering): $\Delta\lambda = 2h/(m_e c) \approx 4.86$ pm. Because the shift is so small, Compton scattering is only significant for X-rays and gamma rays — not visible light.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy E.2 Quantum Physics library (only available to enrolled students or via the resource library subscription) adds:
- 40-page Notes PDF — every concept worked through in full, including derivations of the photoelectric and de Broglie equations.
- Tutorial booklet — 30+ IB-style questions sequenced from foundation to AHL difficulty.
- Tutorial Solutions — full mark-scheme-style worked solutions with M1/A1/R1 annotations.
- Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
- Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
Get the printable PDF version
Same cheatsheet, formatted for A4 print — keep it next to your study desk. Free for signed-in users.
