TAILIEUCHUNG - Physics exercises_solution: Chapter 44

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 44 | a K mc2 k Ti 1 2 2 - v I c I -1 J m x 10-31 kg so K x 10-14 J b The total energy of each electron or positron is E K mc2 x 10-14 J. 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 X means equal energy so each photon has energy x 10-14 J. c E hc 1 so 1 hc E hc x 10 14 J pm The wavelength calculated in Example is pm. When the particles also have kinetic energy the energy of each photon is greater so its wavelength is less. The total energy of the positron is E K mc2 MeV MeV . We can calculate the speed of the positron from Eq. mc2 X 2 V v mc I - 1 -I-I c V I E J E JX 2 MeV I . k MeV J Each photon gets half of the energy of the pion 1 1 Ey -m c2 270 me c2 270 MeV 69 MeV E _ x 107 eV x 10-1 J eV 22 T ------------------ ----------- X 10 HZ h x 10J s c 108 m s 14 1 --------- x 10 m gamma ray. f x 1022 Hz x hc hc h x 10-34 J s a 1 -----v ---- --------------- i E m c1 mf 207 10-31 kg 108 m s x 10-14 m pm. 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. a Am mn - m 270 me - 207 me 63 me E 63 MeV 32 MeV. 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. a The energy will be the proton rest energy MeV corresponding to a frequency of x 1023 Hz and a wavelength of x 10 m. b The energy of .

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