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Chapter 44 - 44.1 m a K 9 109 mc 2 1 1 v 2 1 c 2 0 1547 mc...

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44.1: a) 2 2 2 2 1547 . 0 1 1 1 mc c v mc K = - - = J 10 27 . 1 so kg, 10 109 . 9 14 31 - - × = × = K m b) The total energy of each electron or positron is = = + = 2 2 1547 . 1 mc mc K E J. 10 46 . 9 14 - × The total energy of the electron and positron is converted into the total energy of the two photons. The initial momentum of the system in the lab frame is zero (since the equal-mass particles have equal speeds in opposite directions), so the final momentum must also be zero. The photons must have equal wavelengths and must be traveling in opposite directions. Equal λ means equal energy, so each photon has energy J. 10 46 . 9 14 - × c) pm 10 . 2 ) J 10 46 . 9 ( λ so λ 14 = × = = = - hc E hc hc E The wavelength calculated in Example 44.1 is 2.43 pm. When the particles also have kinetic energy, the energy of each photon is greater, so its wavelength is less. 44.2: The total energy of the positron is MeV. 5.51 MeV 0.511 MeV 00 . 5 2 = + = + = mc K E We can calculate the speed of the positron from Eq. 37.38 . 996 . 0 MeV 5.51 MeV 511 . 0 1 1 1 2 2 2 2 2 2 = - = - = - = E mc c v mc E c v 44.3: Each photon gets half of the energy of the pion ray. gamma m 10 8 . 1 Hz 10 7 . 1 s m 10 00 . 3 λ Hz 10 7 . 1 s) J 10 63 . 6 ( ) eV J 10 6 . 1 ( ) eV 10 9 . 6 ( MeV 69 MeV) 511 . 0 ( ) 270 ( 2 1 ) 270 ( 2 1 2 1 14 22 8 22 34 19 7 2 e 2 γ - - - × = × × = = × = × × × = = = = = = f c h E f c m c m E π 44.4: a) ) s m 10 (3.00 kg) 10 (9.11 (207) s) J 10 626 . 6 ( λ 8 31 34 2 × × × = = = = - - c m h c m hc E hc μ μ pm. 0.0117 m 10 17 . 1 14 = × = - In this case, the muons are created at rest (no kinetic energy). b) Shorter wavelengths would mean higher photon energy, and the muons would be created with non-zero kinetic energy. 44.5: a) e e e 63 207 270 m m m m m m = - = - = + + μ π MeV. 32 MeV) 511 . 0 ( 63 = = E b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you could always find a frame where energy was not conserved. This cannot occur.
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44.6: a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of Hz 10 27 . 2 23 × and a wavelength of m. 10 32 . 1 15 - × b) The energy of each photon will be MeV, 1768 MeV 830 MeV 3 . 938 = + with frequency Hz 10 8 . 42 22 × and wavelength m. 10 02 . 7 16 - × 44.7: J. 10 20 . 7 ) s m 10 00 . 3 ( ) kg 400 kg 400 ( ) ( 19 2 8 2 × = × + = = c m E 44.8: n C Be He 1 0 12 6 9 4 4 2 + + We take the masses for these reactants from Table 43.2, and use Eq. 43.23 reaction. exoergic an is This MeV. 701 . 5 ) u MeV (931.5 u) 1.008665 u 12.000000 u 9.012182 u 002603 . 4 ( = - - + = Q 44.9: He Li B n 4 2 7 3 10 5 1 0 + + MeV 79 . 2 ) u MeV (931.5 u) (0.002995 u; 0.002995 u 11.018607 u 4.002603 u 7.016004 He) Li ( u 11.021602 u 10.012937 u 1.008665 B) n ( 4 2 7 3 10 5 1 0 = = = + = + = + = + m m m The mass decreases so energy is released and the reaction is exoergic. 44.10: a) The energy is so high that the total energy of each particle is half of the available energy, 50 GeV. b) Equation (44.11) is applicable, and MeV. 226 a = E 44.11: a) q mf q m B m B q π ϖ ϖ 2 = = = T 18 . 1 C 10 60 . 1 Hz) 10 (9.00 ) u kg 10 (1.66 u) 01 . 2 ( 2 19 6 27 = × × × = - - B B π b) J 10 47 . 5 ) u kg 10 66 . 1 ( ) u 01 . 2 ( 2 m) (0.32 T) (1.18 C) 10 60 . 1 ( 2 13 27 2 2 2 19 2 2 2 - - - × = × × = = m R B q K . s m 10 81 . 1 ) u kg 10 (1.66 u) (2.01 J) 10 47 . 5 ( 2 2 and MeV 3.42 eV 10 3.42 7 27 13 6 × = × × = = = × = - - m K v 44.12: a) ) c s. m 10 12 . 3 ) b s. 10 97 . 3 2 7 7 × = = × = = = m eBR R m eB f ϖ π π ϖ For three- figure precision, the relativistic form of the kinetic energy must be used, , )mc eV 2 1 - = V.
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