1246_Physics ProblemsTechnical Physics

# 1246_Physics ProblemsTechnical Physics - Chapter 44 (b)...

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Chapter 44 587 (b) First, calculate the Q -value for the reaction: Q MMMM c =+ N-14 He-4 O-17 H-1 2 Q = 14 003 074 4 002 603 16 999 132 1 007 825 1 19 .. . u 931.5 MeV u MeV bg . Then, EQ MM M th Xa X =− + L N M O Q P =−− + L N M O Q P = 119 1 4002603 14 003 074 153 . . . . MeV MeV a f . P44.57 1 1 3 7 0 1 HL i B en 4 7 +→ + QMM M M Q Q −+ + × i B MeV u u u u M e V u u MeV u MeV b g b g ej 931 5 1 007 825 7 016 004 7 016 929 1 008 665 931 5 1765 10 9315 1644 3 . . . Thus, KE m m Q min . . F H G I K J F H G I K J = 11 1007825 7016004 188 incident projectile target nucleus MeV MeV a f . P44.58 (a) N 0 27 24 100 166 10 252 10 == × mass mass per atom kg 239.05 u kg u . . . af (b) λ ×× −− ln ln . 22 2 412 10 3 156 10 9106 10 12 47 13 1 T yr s yr s e j RN 00 13 1 24 12 9 106 10 2 52 10 2 29 10 × × = × . s B q e j (c) RR e t = 0 , so t R R R R = F H G I K J = F H G I K J 0 0 λλ ln ln t = × × F H G I K J × F H G I K J 1 229 10 3 38 10 1 07 10 13 1 12 13 6 . ln . s Bq 0.100 Bq s 1 yr 3.156 10 s yr 7 P44.59 (a) 27 57 1 0 0 0 Co Fe e 26 57 →+ + + ν The Q -value for this positron emission is QM M m c e 57 57 2 2 Co Fe . Q = 56 936 296 56 935 399 2 0 000 549
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