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Compton Scattering Reference cheat sheet - grade 11-12

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Physics Grade 11-12

Compton Scattering Reference Cheat Sheet

A printable reference covering Compton wavelength shift, photon energy, momentum, scattering angle, and electron recoil for grades 11-12.

TopicsQuantum PhysicsLightEnergyMomentumPhysicsStudy Guide
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Compton scattering describes how high-energy photons, such as X-rays, collide with electrons and leave with a longer wavelength. This cheat sheet helps students connect wave behavior, particle momentum, and conservation laws in one reference. It is useful for solving photon scattering problems and understanding evidence for the particle nature of light.

The most important result is the Compton shift equation, Δλ=λ′−λ=hmec(1−cos⁡θ)\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos \theta). Photon energy is related to frequency by E=hfE = hf and to wavelength by E=hcλE = \frac{hc}{\lambda}. Momentum conservation explains why the scattered photon loses energy while the electron gains kinetic energy.

Key Facts

  • The Compton wavelength shift is Δλ=λ′−λ=hmec(1−cos⁡θ)\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos \theta).
  • The electron Compton wavelength is λC=hmec≈2.43×10−12 m\lambda_C = \frac{h}{m_e c} \approx 2.43 \times 10^{-12}\,\text{m}.
  • A photon has energy E=hf=hcλE = hf = \frac{hc}{\lambda}.
  • A photon has momentum p=hλ=Ecp = \frac{h}{\lambda} = \frac{E}{c}.
  • The wavelength shift is zero at θ=0∘\theta = 0^\circ because 1−cos⁡0∘=01 - \cos 0^\circ = 0.
  • The maximum wavelength shift occurs at θ=180∘\theta = 180^\circ and equals Δλmax⁡=2λC\Delta \lambda_{\max} = 2\lambda_C.
  • The scattered photon has a longer wavelength, so λ′>λ\lambda' > \lambda and E′<EE' < E.
  • Energy conservation gives the recoiling electron kinetic energy as Ke=E−E′K_e = E - E'.

Vocabulary

Compton scattering
The scattering of a photon by an electron in which the photon loses energy and its wavelength increases.
Compton shift
The increase in photon wavelength after scattering, written as Δλ=λ′−λ\Delta \lambda = \lambda' - \lambda.
Scattering angle
The angle θ\theta between the incoming photon direction and the scattered photon direction.
Photon momentum
The momentum carried by a photon, given by p=hλp = \frac{h}{\lambda}.
Recoil electron
The electron that gains kinetic energy and momentum after being struck by the photon.
Compton wavelength
The constant λC=hmec\lambda_C = \frac{h}{m_e c} that sets the scale of wavelength shifts for electron scattering.

Common Mistakes to Avoid

  • Using the wrong angle in Δλ=hmec(1−cos⁡θ)\Delta \lambda = \frac{h}{m_e c}(1 - \cos \theta) is wrong because θ\theta must be the photon scattering angle, not the electron recoil angle.
  • Forgetting that the scattered wavelength is longer is wrong because the photon transfers energy to the electron, so λ′>λ\lambda' > \lambda and E′<EE' < E.
  • Using electron mass instead of photon momentum directly is wrong because photons have no rest mass but still carry momentum p=hλp = \frac{h}{\lambda}.
  • Leaving wavelengths in nanometers or picometers without conversion can give wrong units because hh, mem_e, and cc in SI require meters.
  • Assuming the shift depends on the original wavelength is wrong because Δλ\Delta \lambda depends only on θ\theta and λC\lambda_C for scattering from a free electron.

Practice Questions

  1. 1 An X-ray photon scatters from an electron at 90∘90^\circ. Calculate Δλ\Delta \lambda using λC=2.43×10−12 m\lambda_C = 2.43 \times 10^{-12}\,\text{m}.
  2. 2 A photon with initial wavelength 7.00×10−11 m7.00 \times 10^{-11}\,\text{m} scatters at 180∘180^\circ. Find the scattered wavelength λ′\lambda'.
  3. 3 A photon changes from λ=5.00×10−11 m\lambda = 5.00 \times 10^{-11}\,\text{m} to λ′=5.20×10−11 m\lambda' = 5.20 \times 10^{-11}\,\text{m}. Find the energy lost by the photon using E=hcλE = \frac{hc}{\lambda}.
  4. 4 Explain why Compton scattering supports the idea that light has particle-like momentum as well as wave-like wavelength.