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Subatomic Physics

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11.15. References 381<br />

11.41. For nucleons composed of point quarks,<br />

(a) show that the total cross section for high-energy neutrino scattering<br />

varies linearly with the laboratory energy E.<br />

(b) How is the total cross section modified as the center-of-mass energy<br />

approaches mW c 2 ?<br />

(c) What is the laboratory energy of neutrinos for which the center-of-mass<br />

energy is equal to mW c 2 ?<br />

11.42. Verify Eq. (11.97).<br />

11.43. Electron neutrinos can interact with electrons via charged and neutral weal<br />

currents.<br />

(a) Draw Feynman diagrams showing these possibilities.<br />

(b) Explain why muon neutrinos can interact with electrons only via neutral<br />

currents and not charged currents.<br />

11.44. Electron neutrinos produced in beta decays are actually a mixture of two mass<br />

eigenstates, ν1 and ν2, | νe〉 = cosθ | ν1〉 + sinθ | ν2〉 .<br />

(a) Deduce the equation that gives the probability of observing the neutrino<br />

as a muon neutrino for vacuum oscillations.<br />

(b) Consider electron neutrinos from the decays of 8 B in the sun. Assume<br />

θ =45degandm 2 2 − m2 1 =5× 10−5 eV 2 . Plot the electron neutrino<br />

energy spectrum on earth assuming vacuum oscillations.<br />

11.45. (a) Explain why the total cross section for high energy muon neutrinos to<br />

scatter off hadronic targets is three times larger than for antineutrinos.<br />

(b) Would you expect electron neutrinos to have approximately the same,<br />

larger, or smaller cross sections than those for muon neutrinos?<br />

(c) Repeat for electron antineutrinos.<br />

11.46. Assume that the Sun obtains its energy from the transformation of 4 protons<br />

into a doubly ionized He atom, liberating ≈ 26 MeV: 4p → 4 He ++ +2e + +2νe<br />

and use the solar luminosity on Earth, 1.4 kWatt/m 2 , to derive the expected<br />

intensity of neutrinos (number per unit time and area) on Earth.<br />

11.47. Estimate Vud using the ft values for the decays Λ0 → pe − νe and n → pe − νe<br />

from Tables 11.2 and 11.3.

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