30_InstSolManual_PDF_Part16

# 30_InstSolManual_PDF_Part16 - 5 1 1 K 1 1 p 2 S 5 0 f S 5 2...

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The energy equivalent of the mass decrease is 30.65. Set Up: Table 30.5 gives and Solve: (a) Charge must be conserved, so is the only possible decay. (b) The mass decrease is The energy released is 219.1 MeV. 30.66. Set Up: Take the minimum uncertainty product, so Solve: The lifetime is 30.67. Set Up: Problem 30.66 gives Table 30.5 gives and Solve: (a) The mass decrease is A total of 32.6 MeV of energy is released, so the has kinetic energy (b) This is greater than The decay doesn’t occur because it isn’t energetically allowed. (c) Since the decay occurs and since has and has must have has strangeness and has strangeness These two decays don’t occur because they violate conservation of strangeness. Reflect: In an allowed decay, the mass of the parent nucleus must be greater than the total mass of the decay prod- ucts. Conservation laws must also be obeyed. 1 1 1 0 51 1. K 1 1m 2 1 1 1
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Unformatted text preview: 5 1 1 K 1 1 p 2 S 5 0. f S 5 2 1, K 2 S 5 1 1 K 1 f S K 1 1 K 2 m 1 f 2 . m 1 K 1 2 1 m 1 K 2 2 1 m 1 p 2 5 2 1 493.7 MeV / c 2 2 1 135.0 MeV / c 2 5 1122.4 MeV / c 2 . 1 2 1 32.6 MeV 2 5 16.3 MeV. K 1 32.6 MeV / c 2 . m 1 f 2 2 m 1 K 1 2 2 m 1 K 2 2 5 1020 MeV / c 2 2 2 1 493.7 MeV / c 2 2 5 m 1 p 2 5 135.0 MeV / c 2 . m 1 K 6 2 5 493.7 MeV / c 2 m 1 f 2 5 1020 MeV / c 2 . D t 5 h 2 p D E 5 1 4.136 3 10 2 15 eV # s 2 2 p 1 4.0 3 10 6 eV 2 5 1.6 3 10 2 22 s. D E D t 5 h 2 p m 1 K 1 2 2 m 1 p 2 2 m 1 p 1 2 5 493.7 MeV / c 2 2 135.0 MeV / c 2 2 139.6 MeV / c 2 5 219.1 MeV / c 2 . K 1 S p 1 p 1 m 1 p 6 2 5 139.6 MeV / c 2 . m 1 p 2 5 135.0 MeV / c 2 , m 1 K 1 2 5 493.7 MeV / c 2 , 1 3.81 3 10 2 3 u 21 931.5 Mev / u 2 5 355 MeV. D m 5 55.939847 u 2 55.934939 u 2 2 1 0.000549 u 2 5 3.81 3 10 2 3 u. 30-16 Chapter 30...
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## This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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