The Leptons — Question 5

Section 44-3 · Physics pages 7340–7341

Question 5 Inverse beta decay has a recoil threshold beyond the mass difference Consider ¯νe + p →n + e+ with the proton initially at rest and negligible neutrino mass. Use proton, neutron and electron rest energies 938.27, 939.57 and .511 MeV. ¯νe + p e+ n Schematic; labels, not drawing lengths, specify the data. (a) Derive the antineutrino threshold from invariant energy. (b) Evaluate the threshold and compare with the final-minus-initial rest-energy difference.
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Question 5 - Solution Inverse beta decay has a recoil threshold beyond the mass difference ¯νe + p e+ n (a) Method and calculation sc4 = m2 pc4 + 2mpc2Eν, sthc4 = (mn + me)2c4. Eν,th = (mn + me)2 −m2 p 2mp c2 . (b) Interpretation and further calculation ∆= (939.57 + .511) −938.27 = 1.811 MeV, Eν,th = ∆+ ∆2 2(938.27) = 1.812748 MeV . Eν,th −∆= 1.748 keV > 0. Physical check Final particles must share a common center-of-mass motion at threshold. A stationary target cannot absorb incident momentum without a recoil-energy cost.

Question 5

Inverse beta decay has a recoil threshold beyond the mass difference

Consider ν‾e+p→n+e+\bar\nu_e+p\to n+e^+ with the proton initially at rest and negligible neutrino mass. Use proton, neutron and electron rest energies 938.27, 939.57 and .511 MeV.

Diagram shown in the original SVG worksheet.

  1. Derive the antineutrino threshold from invariant energy.

  2. Evaluate the threshold and compare with the final-minus-initial rest-energy difference.

Question 5 - Solution

Inverse beta decay has a recoil threshold beyond the mass difference

Diagram shown in the original SVG worksheet.

(a) Method and calculation

sc4=mp2c4+2mpc2Eν,sthc4=(mn+me)2c4.sc^4=m_p^2c^4+2m_pc^2E_\nu,\quad s_{\mathrm{th}}c^4=(m_n+m_e)^2c^4.Eν,th=(mn+me)2−mp22mpc2.\boxed{E_{\nu,\mathrm{th}}=\frac{(m_n+m_e)^2-m_p^2}{2m_p}c^2}.

(b) Interpretation and further calculation

Δ=(939.57+.511)−938.27=1.811MeV,Eν,th=Δ+Δ22(938.27)=1.812748MeV.\Delta=(939.57+.511)-938.27=1.811\,\mathrm{MeV},\quad\boxed{E_{\nu,\mathrm{th}}=\Delta+\frac{\Delta^2}{2(938.27)}=1.812748\,\mathrm{MeV}}.Eν,th−Δ=1.748keV>0.\boxed{E_{\nu,\mathrm{th}}-\Delta=1.748\,\mathrm{keV}}>0.

Physical check

Final particles must share a common center-of-mass motion at threshold. A stationary target cannot absorb incident momentum without a recoil-energy cost.