WHomework set 21_solns_2011

WHomework set 21_solns_2011 - Homework set # 21 Problems:...

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Homework set # 21 Problems: 26.16, 22; 27.9, 10 Solutions: 26.16 (a)  2 22 7 8 -13 1 MeV 1.67 10 kg 3.00 10 m s 939 MeV 1.60 10 J R Em c     (b) 2 2 3 2 1 939 MeV 3.01 10 MeV 3.01 GeV 1 0.950 R R E c E vc  (c) 33 939 MeV 2.07 10 MeV 2.07 GeV R KE E E   26.22 Let 1 m be the mass of the fragment moving at 1 0.868 , and 2 m be the mass moving at 2 0.987 . From conservation of mass-energy, 2 12 1 0.868 1 0.987 mc E mc   giving 27 2.01 6.22 3.34 10 kg mm  (1) The momenta of the two fragments must add to zero, so the magnitudes must be equal, giving pp or 111 2 22 mv . This yields     2.01 0.868 6.22 0.987 , or 3.52 (2) Substituting equation (2) into (1) gives   27 2 7.07 6.22 3.34 10 kg m , or 28 2 2.51 10 kg m  Equation (2) then yields   28 28 1 3.52 2.51 10 kg 8.84 10 kg m
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27.9 (a) From the photoelectric effect equation, the work function is max hc KE  , or  
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This note was uploaded on 05/12/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at University of Georgia Athens.

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WHomework set 21_solns_2011 - Homework set # 21 Problems:...

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