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AP Course

AP Physics 2

Study AP Physics 2 Units 9–15, with the Fall 2026 radioactive-decay clarification applied and official exam updates linked separately.

Study Units 9–15 through original models, derivations, experiments, quizzes, and practice sets.

Lessons
Particle Motion Behind Temperature and PressureConnecting Gas Variables with the Ideal-Gas ModelHeat Flow and the Approach to Thermal BalanceEnergy Accounting for Thermodynamic ProcessesMaterial Response to Heating and Heat ConductionEntropy, Probability, and the Direction of Thermal ChangeCharge Interactions and Coulomb-Force ModelsTracking Charge During Contact and InductionMapping Electric Fields from Source ChargesEnergy of Configurations of ChargesElectric Potential and Equipotential ReasoningCapacitance and Energy Stored in Electric FieldsConserving Energy in Charged-Particle MotionCharge Flow and Conventional CurrentModeling Sources, Wires, and Loads in Simple NetworksHow Geometry and Material Set ResistanceElectrical Energy Transfer and PowerReducing Series-Parallel DC NetworksLoop Equations from Energy ConservationJunction Equations from Charge ConservationTransient Charging and Discharging in RC NetworksSources, Direction, and Strength of Magnetic FieldsMagnetic Forces and Curved Paths of ChargesForces Between Fields and Current-Carrying ConductorsChanging Magnetic Flux and Induced EMFReflection and Absorption at Material BoundariesLocating Images with Mirror Ray ModelsRefraction, Index, and Total Internal ReflectionLocating Images with Thin-Lens ModelsPulses, Wave Types, and Propagation SpeedPeriodic-Wave Measures and GraphsBoundary Changes and PolarizationElectromagnetic-Wave BehaviorFrequency Shifts from Relative MotionSuperposition, Interference, and Standing WavesDiffraction from Openings and EdgesTwo-Slit and Grating Pattern GeometryPhase Change and Thin-Film ColorQuantum Models and Dual Wave-Particle EvidenceEnergy Levels in a Hydrogen-Like AtomConnecting Spectral Lines to Energy TransitionsThermal Spectra and Blackbody CurvesPhoton Thresholds in the Photoelectric EffectPhoton-Electron Scattering and MomentumMass-Energy Accounting in Fission and FusionRandom Nuclear Decay, Activity, and Half-Life
Quizzes
Practice Problems Formula notes, diagrams, and practice sets will be added here.

AP Physics 2 · Unit 15 · Topic 15.6

Photon-Electron Scattering and Momentum

Quantization, probability, nuclear processes, and relativity extend classical models at atomic scales.

1. Topic Lens

Photon-Electron Scattering and Momentum is studied through modern physics. Connect the system boundary, interacting parts, and measurable evidence before applying a formula.

\[E=hf\]

2. Why the Formula Works

The relationship is built from definitions and conservation reasoning:

  1. A photon carries discrete energy hf.
  2. The material requires work function φ to release an electron.
  3. Energy conservation leaves the remainder as the maximum electron kinetic energy.
\[K_{\text{max}}=hf-\phi\]

3. Detailed Visual Model

Pixel diagram for Photon-Electron Scattering and MomentumOriginal schematic connecting Photon-Electron Scattering and Momentum to Modern Physics.
Photon-Electron Scattering and Momentum: an original pixel-style model. Use it as a schematic, not a literal scale drawing.

4. Worked Example and Lab Link

A 5 eV photon strikes a surface with work function 2 eV. Find maximum electron kinetic energy.

Answer: Kmax=5-2=3 eV.

Investigation idea: Analyze original stopping-potential data versus frequency and infer threshold frequency and Planck-slope meaning.

Common trap: Light intensity changes photon count; below threshold frequency it does not release electrons in the ideal model.

Expanded Topic 15.6 Lesson

5. Learning Targets

  • Relate photon energy, wavelength, and momentum using \(E=pc=hf=hc/\lambda\).
  • Apply vector momentum and relativistic energy conservation to photon–electron scattering.
  • Use the Compton-shift equation to predict scattered wavelength from angle.
  • Calculate the scattered photon energy and the recoil electron's kinetic energy and direction.
  • Explain why the observed angle-dependent shift supports a particle-momentum model of light.

6. A Massless Photon Still Carries Momentum

The relativistic energy–momentum relation is

\[E^2=p^2c^2+m_0^2c^4\]

For a photon, \(m_0=0\), so \(E=pc\). Combining this with \(E=hf=hc/\lambda\) gives

\[\boxed{p_\gamma=\frac{E_\gamma}{c}=\frac{h}{\lambda}}\]

The formula gives magnitude. The momentum vector points in the photon's direction of travel.

7. The Experimental Puzzle

When monochromatic X-rays scatter from targets such as graphite, detectors record a component with a wavelength longer than the incident wavelength. The shift changes systematically with scattering angle.

Longer \(\lambda'\)The scattered photon has smaller energy and momentum.
Recoil electronThe missing photon energy and momentum appear in the electron.
Angle dependenceThe measured shift follows collision geometry, not light intensity.

8. Original Photon–Electron Collision Map

Momentum vectors in Compton scattering An incident photon travels right and collides with an electron initially at rest. A lower-momentum photon scatters upward at angle theta while the electron recoils downward at angle phi. incident photonp = h/λscattered photonp′ = h/λ′recoil electron pₑθφelectron initially at restp⃗ = p⃗′ + p⃗ₑ
Momentum is conserved as a vector. The recoil electron supplies the vertical component needed to balance the scattered photon and takes the photon's lost energy.

9. State the Collision Model First

Initial electronat rest and sufficiently weakly bound to approximate a free electron
Systemincident photon plus electron, with negligible external impulse during collision
Final particlesone scattered photon and one recoil electron
Physicsrelativistic total energy and vector momentum are conserved

If binding or whole-atom recoil matters, the free-electron formula requires modification.

10. The Two Conservation Equations

For an electron initially at rest, total energy conservation gives

\[\frac{hc}{\lambda}+m_ec^2=\frac{hc}{\lambda'}+E_e\]

Momentum conservation gives

\[\frac{h}{\lambda}\hat{\mathbf x}=\vec p_{\gamma}' + \vec p_e\]

The final electron must obey \(E_e^2=p_e^2c^2+m_e^2c^4\). A nonrelativistic collision equation alone is not sufficient for an X-ray photon.

11. Derivation Roadmap

Rearrange the momentum equation and square its magnitude:

\[p_e^2=p_\gamma^2+p_\gamma'^2-2p_\gamma p_\gamma'\cos\theta\]

Use the energy equation to express \(E_e\), then substitute \(E_e^2=p_e^2c^2+m_e^2c^4\) and \(p_\gamma=h/\lambda\). After canceling common terms,

\[\lambda'-\lambda=\frac{h}{m_ec}(1-\cos\theta)\]

The measured angle dependence follows from combining scalar energy conservation with vector momentum geometry.

12. Compton Shift and Its Natural Scale

\[\boxed{\Delta\lambda=\lambda'-\lambda=\lambda_C(1-\cos\theta)}\]
\[\lambda_C=\frac{h}{m_ec}=2.426\,\mathrm{pm}\]

\(\lambda_C\) is the electron Compton wavelength. It sets the shift scale for scattering from a free electron; it is not the incident photon's wavelength.

13. Original Shift-versus-Angle Graph

Normalized Compton wavelength shift versus scattering angle The shift begins at zero for forward scattering, equals one Compton wavelength at ninety degrees, and reaches two Compton wavelengths for backscattering. Δλ/λC0°60°90°120°180°scattering angle θ0.511.52backscatter maximum
The curve is \(1-\cos\theta\): forward scattering gives no shift, \(90^\circ\) gives \(\lambda_C\), and backward scattering gives \(2\lambda_C\).

14. Angle Checkpoints

\(\theta\)\(1-\cos\theta\)\(\Delta\lambda\)Interpretation
\(0^\circ\)00Forward direction; no Compton shift
\(60^\circ\)0.5\(1.213\,\mathrm{pm}\)Intermediate transfer
\(90^\circ\)1\(2.426\,\mathrm{pm}\)Shift equals one electron Compton wavelength
\(180^\circ\)2\(4.852\,\mathrm{pm}\)Maximum wavelength shift

15. Worked Wavelength Example

A 71.0 pm X-ray photon scatters through \(60^\circ\):

\[\Delta\lambda=(2.426\,\mathrm{pm})(1-\cos60^\circ)=1.213\,\mathrm{pm}\]
\[\lambda'=71.0+1.213=72.213\,\mathrm{pm}\]

The scattered photon has longer wavelength, so its energy and momentum magnitudes are both smaller than before the collision.

16. Endpoints and Maximum Transfer

Because \(-1\le\cos\theta\le1\),

\[0\le\Delta\lambda\le2\lambda_C=4.852\,\mathrm{pm}\]

The maximum occurs at \(\theta=180^\circ\), when the photon reverses direction. A shift larger than \(2\lambda_C\) cannot result from the stated free electron initially at rest.

17. An Equivalent Energy Formula

Substitute \(\lambda=hc/E_\gamma\) into the shift relation:

\[\boxed{E_\gamma'=\frac{E_\gamma}{1+\dfrac{E_\gamma}{m_ec^2}(1-\cos\theta)}}\]

For an electron initially at rest, energy conservation then gives

\[K_e=E_\gamma-E_\gamma'\]

The electron rest energy \(m_ec^2\approx511\,\mathrm{keV}\) is the natural scale in the denominator.

18. Worked Energy-Transfer Example

A 100 keV photon scatters through \(90^\circ\). Since \(1-\cos90^\circ=1\),

\[E_\gamma'=\frac{100}{1+100/511}=83.6\,\mathrm{keV}\]
\[K_e=100-83.6=16.4\,\mathrm{keV}\]

Total energy includes electron rest energy on both sides; the 16.4 keV difference becomes recoil kinetic energy.

19. Recoil Momentum from Components

Choose the incident photon direction as \(+x\) and let the scattered photon go above the axis:

\[p_{e,x}=p_\gamma-p_\gamma'\cos\theta,\qquad p_{e,y}=-p_\gamma'\sin\theta\]
\[p_e=\sqrt{p_\gamma^2+p_\gamma'^2-2p_\gamma p_\gamma'\cos\theta}\]

The negative \(y\)-component is essential: it cancels the scattered photon's positive vertical momentum.

20. Relativistic Electron Consistency Check

Once recoil kinetic energy is known, electron momentum must satisfy

\[(m_ec^2+K_e)^2=p_e^2c^2+m_e^2c^4\]
\[p_ec=\sqrt{K_e^2+2K_em_ec^2}\]

Using \(K=p^2/(2m)\) is only an approximation when \(K_e\ll m_ec^2\). The relativistic relation provides a robust check at X-ray and gamma-ray energies.

21. What Depends on Incident Energy?

Absolute wavelength shift\(\Delta\lambda=\lambda_C(1-\cos\theta)\) depends on angle and scatterer mass, not incident wavelength in the free-electron model.
Fractional wavelength change\(\Delta\lambda/\lambda\) is larger for a shorter incident wavelength.
Energy transferredDepends on both incident photon energy and scattering angle.

“The shift is independent of incident wavelength” refers to the absolute \(\Delta\lambda\), not the fraction or energy loss.

22. Shifted and Unshifted Peaks

Real target spectra can contain both a shifted component and an approximately unshifted component. Weakly bound electrons behave more like free electrons and show the usual Compton shift. For tightly bound electrons, momentum can be shared with the whole atom.

\[\lambda_{\mathrm{scatterer}}=\frac{h}{Mc}\]

Replacing \(m_e\) with the much larger atomic mass \(M\) makes the recoil wavelength scale far smaller, so that component may appear unshifted at available resolution.

23. Why X-Rays and Gamma Rays Reveal the Effect

The maximum electron Compton shift is only about 4.85 pm. That is a meaningful fraction of an X-ray wavelength but an extremely small fraction of a 500 nm visible wavelength.

Short-wavelength photons also carry larger momentum \(h/\lambda\), making electron recoil and energy transfer easier to resolve experimentally.

24. Do Not Mix Three Photon–Matter Processes

ProcessPhoton after interactionElectron outcomeKey evidence
Photoelectric effectabsorbedejected if \(hf\ge\phi\)threshold and \(K_{\max}=hf-\phi\)
Compton scatteringsurvives with lower energy for nonzero shiftrecoilsangle-dependent \(\lambda'-\lambda\)
Classical/Thomson limitapproximately unchanged frequencydriven oscillation or negligible recoil energyuseful when photon energy is small relative to rest-energy scale

25. Open-Data Modeling Investigation

A. Test the angular law
  1. Use \(0^\circ,30^\circ,\ldots,180^\circ\) and compute \(\Delta\lambda\).
  2. Plot \(\Delta\lambda\) against \(1-\cos\theta\).
  3. Fit a line and interpret slope and intercept.
  4. Compare the slope with \(h/(m_ec)\) using stated constants.
B. Audit conservation
  1. Choose an incident photon energy and scattering angle.
  2. Calculate \(E_\gamma'\), \(K_e\), and both momentum components.
  3. Verify energy and \(x,y\) momentum separately.
  4. Vary angle and explain which transfer becomes maximal.

For a source-backed comparison dataset, use the worked scattering cases in OpenStax 6.3. Spreadsheet calculations are sufficient; no radiation source is required.

26. Common Reasoning Traps

  • Treating photon momentum as \(mv\): a photon has zero rest mass and \(p=E/c=h/\lambda\).
  • Conserving momentum only in magnitude: momentum must balance in both \(x\) and \(y\).
  • Making the scattered wavelength shorter: for a resting free electron, \(\lambda'\ge\lambda\).
  • Calling \(90^\circ\) the maximum shift: the maximum is at \(180^\circ\).
  • Using ordinary electron kinetic energy at every energy: check against \(m_ec^2\).
  • Confusing absolute and fractional shift: only the absolute shift is incident-wavelength independent in the ideal model.

27. AP-Style Reasoning Checks

  1. A photon wavelength increases after scattering. Compare its final energy and momentum with the initial values.
  2. What scattering angle gives \(\Delta\lambda=\lambda_C\)?
  3. At what angle is electron energy transfer greatest?
  4. Why must the recoil electron have a transverse momentum component?
  5. Two photons of different initial wavelength scatter from resting electrons through the same angle. Compare their absolute wavelength shifts.
  6. What observation makes Compton scattering evidence for photon momentum?

Answers: both decrease; \(90^\circ\); \(180^\circ\); it balances the scattered photon's transverse component; the ideal absolute shifts are equal; the measured angle-dependent wavelength change agrees with a collision model conserving photon and electron momentum.

Checkpoint · Topic 15.6

Explain how photon-electron scattering and momentum supports or limits this conclusion: Kmax=5-2=3 eV.

Official curriculum reference: College Board AP Physics 2 course page. The explanation and worked example are independently written for this study site.