PC Chapter 15

# PC Chapter 15 - 15 Particle Physics 15-1 The time for a...

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245 15 Particle Physics 15-1 The time for a particle traveling with the speed of light to travel a distance of - × 15 3 10 m is - - × = == × 15 23 8 3 10 m 10 s 3 10 ms d t v . 15-2 The ρ ππ -+ →+ 0 decay must occur via the strong interaction. The 0 K decay must occur via the weak interaction. 15-3 The minimum energy is released, and hence the minimum frequency photons are produced. The proton and antiproton are at rest when they annihilate. That is, = 0 EE and = 0 K . To conserve momentum, each photon must carry away one-half the energy. Thus, = = 0 min min 0 2 938.3 MeV 2 E E h fE . Thus, (29 ( 29 - - × = × 19 23 min 34 938.3 MeV 1.6 10 J MeV 2.26 10 Hz 6.63 10 J s f and λ - × = = × 8 15 max 23 min 1.32 10 m 2.26 10 Hz c f 15-4 E , p nucleus at rest M n = 100 u Before 100 u proton antiproton K E n , p n After ( 29 ( photon photon n E pp c

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246 CHAPTER 15 PARTICLE PHYSICS ( 29 ( == photon photon n E pp c , ( 29 = + =+ 2 photon 2 K 1 877 MeV K p nn E m c EE . Assuming that the kinetic energy of the recoiling nucleus, K n E , is small compared to 1 877 MeV, ( 2245 photon 1 877 MeV E , and 2245 1 877 MeV n pc . Since K n E is small we can use the classical expression = 2 K 2 n n n p E m : ( (29 ( ( λ - - = += = = × × = = × 2 2 23 15 8 16 23 1 877 MeV K 18.9 MeV 2 100 u 931.5 MeV u photon 1 877 MeV 19 MeV 1 896 MeV 1 896 MeV 4.58 10 Hz 4.14 10 eV s 3.00 10 m s 6.55 10 m 4.58 10 Hz n c E c E E f h c f 15-5 The rest energy of the 0 Z boson is = 0 96 GeV E . The maximum time a virtual 0 Z boson can exist is found from ∆∆= h Et . ( 29 - - - × = = × h 34 27 10 1.055 10 J s 6.87 10 s 96 GeV 1.6 10 J GeV t E . The maximum distance it can travel in this time is ( 29 ( 29( 29 -- =∆=× × = × 8 27 18 3 10 m s 6.87 10 s 2.06 10 m dct . The distance d is an approximate value for the range of the weak interaction. 15-6 μ + +→+ e vv muon-lepton number before reaction ( 29 (29 =-+ 10 electron-lepton number before reaction (29 (29 =+= 0 11 . Therefore, after the reaction, the muon-lepton number must be –1. Thus, one of the neutrinos must be the anti-neutrino associated with muons or μ v . Also, after the reaction, the electron-lepton number must be 1. Thus, one of the neutrinos must be the neutrino associated with electrons, or e v . Thus, μ μ + + →+ e evv . 15-7 Use Table 15.2 to find properties that can be conserved in the given reactions Reaction 1 Reaction 2 (a) Charge: π - -+ + → pK ππ - p (29 (29 (29 (29 -++→-++ 00 ǆ ǆ
MODERN PHYSICS 247 (b) Baryon number: ( 29(29 ( 29(29 ++ → ++ 0101 ( + →+ 11 ǆ ǆ (c) Strangeness: ( 29( 29 (29(29 + →+ +- 0 0 (29(29 (29( 29 + 0 001 00 ǆ (29 →- 01 X Thus, the second reaction is not allowed since it does not conserve strangeness. 15-8 (a) μ πμ -- →+ v μ L : 0 11 (b) μ μ Kv μ L : →-+ 0 (c) + +→+ pn e ve e L : -+→- 10 01 (d) - np e e L : (e) μ μ - v μ L : (f) μ μ - →++ e evv μ L : 1 001, and e L : +-+ 0110 15-9 (a) ππ - 0 p (Baryon number is violated: 1 00 ) (b) π +→++ 0 pp pp (This reaction can occur) (c) π + ppp (Baryon number is violated: 11 10 ) (d) μ v (This reaction can occur) (e) - e ev (This reaction can occur) (f) n (Violates baryon number: 0 01, and violates muon-lepton number: 0 10 .

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## This note was uploaded on 11/05/2009 for the course PHSX 313 taught by Professor Staff during the Fall '09 term at Kansas.

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PC Chapter 15 - 15 Particle Physics 15-1 The time for a...

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