Photons Have Momentum — Question 9

Section 38-3 · Physics pages 6308–6309

Question 9 Two Compton deflections can restore direction without restoring energy A photon initially traveling along +x has wavelength λC. It scatters by 90◦from one resting free electron into +y, then by 90◦from another resting free electron back into +x. For scattering from an initially resting free electron, use E′ = E/[1 + (E/mec2)(1 −cos θ)], λ′ −λ = λC(1 −cos θ), and λC = h/(mec). 1 2 Schematic; labels, not drawing lengths, specify the data. (a) Find the final wavelength and photon energy. (b) Find the kinetic energy delivered to each electron, in units of mec2.
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Question 9 - Solution Two Compton deflections can restore direction without restoring energy 1 2 (a) Method and calculation λ1 = λC + λC = 2λC, λ2 = 3λC . (E0, E1, E2) = mec2(1, 1/2, 1/3). (b) Interpretation and further calculation K1 = E0 −E1 = 1 2mec2 , K2 = E1 −E2 = 1 6mec2 . K1 + K2 = 2 3mec2 = E0 −E2 . Physical check Net photon direction is not enough to identify the number of interactions. Wavelength shifts add across distinct resting targets.

Question 9

Two Compton deflections can restore direction without restoring energy

A photon initially traveling along +x has wavelength λC\lambda_C. It scatters by 90∘90^\circ from one resting free electron into +y, then by 90∘90^\circ from another resting free electron back into +x. For scattering from an initially resting free electron, use E′=E/[1+(E/mec2)(1−cos⁡θ)]E\prime=E/[1+(E/m_ec^2)(1-\cos\theta)], λ′−λ=λC(1−cos⁡θ)\lambda\prime-\lambda=\lambda_C(1-\cos\theta), and λC=h/(mec)\lambda_C=h/(m_ec).

Diagram shown in the original SVG worksheet.

  1. Find the final wavelength and photon energy.

  2. Find the kinetic energy delivered to each electron, in units of mec2m_ec^2.

Question 9 - Solution

Two Compton deflections can restore direction without restoring energy

Diagram shown in the original SVG worksheet.

(a) Method and calculation

λ1=λC+λC=2λC,λ2=3λC.\lambda_1=\lambda_C+\lambda_C=2\lambda_C,\quad\boxed{\lambda_2=3\lambda_C}.(E0,E1,E2)=mec2(1,1/2,1/3).(E_0,E_1,E_2)=m_ec^2(1,1/2,1/3).

(b) Interpretation and further calculation

K1=E0−E1=12mec2,K2=E1−E2=16mec2.\boxed{K_1=E_0-E_1=\tfrac12m_ec^2},\qquad\boxed{K_2=E_1-E_2=\tfrac16m_ec^2}.K1+K2=23mec2=E0−E2.\boxed{K_1+K_2=\tfrac23m_ec^2=E_0-E_2}.

Physical check

Net photon direction is not enough to identify the number of interactions. Wavelength shifts add across distinct resting targets.