An Interlude — Question 6

Section 44-2 · Physics pages 7322–7323

Question 6 A recoil partner slightly raises the threshold for photon pair creation A photon strikes a stationary body of mass M and produces an electron-positron pair while the body survives unchanged. Let mec2 = .511 MeV and Mc2 = 1000 MeV. Use the threshold invariant mass M + 2me. γ M e+e− M Schematic; labels, not drawing lengths, specify the data. (a) Derive the exact photon threshold energy. (b) Evaluate it and compare with 2mec2.
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Question 6 - Solution A recoil partner slightly raises the threshold for photon pair creation γ M e+e− M (a) Method and calculation sc4 = M2c4 + 2Mc2Eγ, sthc4 = (M + 2me)2c4. Eγ,th = 2mec2 ( 1 + me M ) . (b) Interpretation and further calculation Eγ,th = 1.022(1 + .511/1000) = 1.02252224 MeV . ∆Erecoil = .000522242 MeV = 522.242 eV . Physical check At threshold all final particles share a common velocity in the lab. Pair rest energy alone is insufficient because incident momentum must be retained.

Question 6

A recoil partner slightly raises the threshold for photon pair creation

A photon strikes a stationary body of mass M and produces an electron-positron pair while the body survives unchanged. Let mec2=.511m_ec^2=.511 MeV and Mc2=1000Mc^2=1000 MeV. Use the threshold invariant mass M+2meM+2m_e.

Diagram shown in the original SVG worksheet.

  1. Derive the exact photon threshold energy.

  2. Evaluate it and compare with 2mec22m_ec^2.

Question 6 - Solution

A recoil partner slightly raises the threshold for photon pair creation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

sc4=M2c4+2Mc2Eγ,sthc4=(M+2me)2c4.sc^4=M^2c^4+2Mc^2E_\gamma,\quad s_{\mathrm{th}}c^4=(M+2m_e)^2c^4.Eγ,th=2mec2(1+meM).\boxed{E_{\gamma,\mathrm{th}}=2m_ec^2\left(1+\frac{m_e}{M}\right)}.

(b) Interpretation and further calculation

Eγ,th=1.022(1+.511/1000)=1.02252224MeV.\boxed{E_{\gamma,\mathrm{th}}=1.022(1+.511/1000)=1.02252224\,\mathrm{MeV}}.ΔErecoil=.000522242MeV=522.242eV.\boxed{\Delta E_{\mathrm{recoil}}=.000522242\,\mathrm{MeV}=522.242\,\mathrm{eV}}.

Physical check

At threshold all final particles share a common velocity in the lab. Pair rest energy alone is insufficient because incident momentum must be retained.