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Unformatted text preview: FROM QUARKS TO THE COSMOS E XERCISES Section 39.1 Particles and Forces 17. I NTERPRET This problem is about finding the lifetime of a virtual photon by applying the uncertainty principle. D EVELOP In order to test the conservation of energy in a process involving one virtual photon, a measurement of energy with uncertainty less than the photons energy ( / ), E hc < must be performed in a time interval less than the virtual photons lifetime ( ). t < Thus, / . E t hc < But Heisenbergs uncertainty principle limits the product of these uncertainties to , E t h so / . hc h Knowing the wavelength allows us to find the upper bound on the lifetime of the virtual photon. E VALUATE With 633 nm, = we get 9 16 8 633 10 m 3 10 s 2 2 (3 10 m/s) c -- = for the given wavelength. A SSESS If the lifetime of a virtual photon of wavelength 633 nm were less than 16 3 10 s,- no measurement showing a violation of conservation of energy would be possible. 18. Yukawas argument (see the solution to Problem 34) shows that / (197.3 eV nm/ )/ m c x c = h . 2 ( 100 m) 2 neV/ . c c . (The longer the range of the force, the smaller the mass of the mediating particle.) Section 39.2 Particles and More Particles 19. I NTERPRET Were asked about the decay of a positive pion to a muon and a neutrino. D EVELOP The decay process must conserve both charge and muon-lepton number L (see Table 39.1). E VALUATE The only decay of the positive pion consistent with conservation laws is + + + A SSESS The muon-lepton numbers of , , and + + are 0, - 1, and +1, respectively. Therefore, we see that L is conserved. Similarly, the charges of , , and + + are +1, +1, and 0, respectively, so we see that charge is also conserved. 20. The has strangeness 1,- while the p and - both have zero strangeness. Thus 1 S = for this decay (final minus initial strangeness). Since strangeness is conserved in strong and electromagnetic interactions, the decay must be a weak interaction. 21. I NTERPRET Were asked to verify the conservation laws for the decay of into a positive, a negative, and a neutral pion. D EVELOP The decay process +- + + must conserve charge, strangeness, and baryon number (see Table 39.1). E VALUATE The decay conserves charge (0 1 1 0), =- + and all the particles have zero baryon number and strangeness. A SSESS The problem illustrates how conservation laws restrict the possible decay modes of a particle. 22. Reference to Table 39.1 shows that both decays satisfy conservation of charge, electron-lepton number, and tau- lepton number (the negatively charged leptons are particles). None of the particles are baryons, weak decays dont 39.1 39 39.2 Chapter 39 have to conserve strangeness, and the taus rest energy is greater than that of either final state. Therefore, both decays are possible....
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