Photons Have Momentum — Question 7

Section 38-3 · Physics pages 6304–6305

Question 7 The heavy-target limit suppresses Compton energy transfer A photon scatters from a resting target of mass M, with E′/E = [1 + a]−1 and a = [E/(Mc2)](1 −cos θ). Assume the target retains its internal state. a ∆E/E Schematic; labels, not drawing lengths, specify the data. (a) Expand the fractional energy loss for a ≪1. (b) At fixed photon energy and angle, compare recoil from targets of masses m and 1836m in this limit and find the relative error of using a instead of the exact loss.
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Question 7 - Solution The heavy-target limit suppresses Compton energy transfer a ∆E/E (a) Method and calculation E −E′ E = a 1 + a = a −a2 + O(a3). ∆E ≃E2 Mc2 (1 −cos θ) . (b) Interpretation and further calculation ∆E1836m/∆Em ≃1/1836 . a −a/(1 + a) a/(1 + a) = a . Physical check The approximation is controlled by a, including its angular factor, rather than photon energy alone.

Question 7

The heavy-target limit suppresses Compton energy transfer

A photon scatters from a resting target of mass M, with E′/E=[1+a]−1E\prime/E=[1+a]^{-1} and a=[E/(Mc2)](1−cos⁡θ)a=[E/(Mc^2)](1-\cos\theta). Assume the target retains its internal state.

Diagram shown in the original SVG worksheet.

  1. Expand the fractional energy loss for a≪1a\ll1.

  2. At fixed photon energy and angle, compare recoil from targets of masses m and 1836m in this limit and find the relative error of using a instead of the exact loss.

Question 7 - Solution

The heavy-target limit suppresses Compton energy transfer

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E−E′E=a1+a=a−a2+O(a3).\frac{E-E\prime}{E}=\frac a{1+a}=a-a^2+O(a^3).ΔE≃E2Mc2(1−cos⁡θ).\boxed{\Delta E\simeq\frac{E^2}{Mc^2}(1-\cos\theta)}.

(b) Interpretation and further calculation

ΔE1836m/ΔEm≃1/1836.\boxed{\Delta E_{1836m}/\Delta E_m\simeq1/1836}.a−a/(1+a)a/(1+a)=a.\boxed{\frac{a-a/(1+a)}{a/(1+a)}=a}.

Physical check

The approximation is controlled by a, including its angular factor, rather than photon energy alone.