Photons Have Momentum — Question 6

Section 38-3 · Physics pages 6302–6303

Question 6 A required recoil fraction imposes a minimum incident photon energy A photon scatters from a resting free electron. Let ϵ = E/(mec2). 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). ϵ Ke/E .8 Schematic; labels, not drawing lengths, specify the data. (a) Find the largest possible fraction of incident energy transferred to the electron. (b) Find the minimum incident energy that permits at least 80 percent transfer, using mec2 = 511 keV.
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Question 6 - Solution A required recoil fraction imposes a minimum incident photon energy ϵ Ke/E .8 (a) Method and calculation Ke E = 1 − 1 1 + ϵ(1 −cos θ). θ = π : (Ke/E)max = 2ϵ 1 + 2ϵ . (b) Interpretation and further calculation 2ϵ 1 + 2ϵ ≥.800 ⇒ ϵ ≥2. Emin = 1022 keV , attained at backscattering. Physical check At finite incident energy the scattered photon retains positive energy; complete transfer is only an asymptotic limit.

Question 6

A required recoil fraction imposes a minimum incident photon energy

A photon scatters from a resting free electron. Let ϵ=E/(mec2)\epsilon=E/(m_ec^2). 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 largest possible fraction of incident energy transferred to the electron.

  2. Find the minimum incident energy that permits at least 80 percent transfer, using mec2=511keVm_ec^2=511\,\mathrm{keV}.

Question 6 - Solution

A required recoil fraction imposes a minimum incident photon energy

Diagram shown in the original SVG worksheet.

(a) Method and calculation

KeE=1−11+ϵ(1−cos⁡θ).\frac{K_e}{E}=1-\frac1{1+\epsilon(1-\cos\theta)}.θ=π:(Ke/E)max=2ϵ1+2ϵ.\theta=\pi:\quad\boxed{(K_e/E)_{\max}=\frac{2\epsilon}{1+2\epsilon}}.

(b) Interpretation and further calculation

2ϵ1+2ϵ≥.800⇒ϵ≥2.\frac{2\epsilon}{1+2\epsilon}\ge.800\quad\Rightarrow\quad\epsilon\ge2.Emin=1022keV,attained at backscattering.\boxed{E_{\min}=1022\,\mathrm{keV}},\quad\text{attained at backscattering}.

Physical check

At finite incident energy the scattered photon retains positive energy; complete transfer is only an asymptotic limit.