Chapter 29

# Chapter 29 - CHAPTER 29 Quick Quizzes 1(c At the end of the...

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C HAPTER 29 Quick Quizzes 1. (c). At the end of the first half-life interval, half of the original sample has decayed and half remains. During the second half-life interval, half of the remaining portion of the sample decays. The total fraction of the sample that has decayed during the two half-lives is 1 1 1 3 2 2 2 4 + = . 2. (b). If the original activity is , the activity remaining after an elapsed time t is 0 R ( ) 12 0.693 0 0 T t t R R e R e λ = = . Solving for the half-life yields ( ) 12 0 0.693 ln T t R R = 0 0.96 = . If R R at 2.0 hr t = , the half-life is ( ) ( ) 12 0.693 2.0 h 34 h ln 0.96 T = = . 3. (a). Conservation of momentum requires the momenta of the two fragments be equal in magnitude and oppositely directed. Thus, from 2 2 KE p m = , the lighter alpha particle has more kinetic energy that the more massive daughter nucleus. 4. (a) and (b). Reactions (a) and (b) both conserve total charge and total mass number as required. Reaction (c) violates conservation of mass number with the sum of the mass numbers being 240 before reaction and being only 223 after reaction. 5. (b). In an endothermic reaction, the threshold energy exceeds the magnitude of the Q value by a factor of ( ) 1 m M + , where m is the mass of the incident particle and M is the mass of the target nucleus. 413

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C H A P T E R 2 9 Problem Solutions 29.1 The average nuclear radii are 13 0 r rA = , where and A is the mass number. F or 2 1 , 15 0 1.2 10 m 1.2 fm r = × = H ( )( ) 13 1.2 fm 2 r = = 1.5 fm F or 60 , 27 Co ( )( ) 13 1.2 fm 60 r = = 4.7 fm F or 197 79 , Au ( )( ) 13 1.2 fm 197 r = = 7.0 fm F or 239 94 , Pu ( )( ) 13 1.2 fm 239 r = = 7.4 fm 29.2 An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water molecule there are eight neutrons and ten protons. So protons and neutrons are nearly equally numerous in your body, each contributing (say) 35 kg out of a total body mass of 70 kg. -27 1 nucleon 35 kg 1.67 10 kg N = × 28 ~10 protons and 28 ~10 neutrons The electron number is precisely equal to the proton number, e N = 28 ~10 electrons 29.3 From 3 4 3 E n n M V r ρ ρ π = = , we find ( ) ( ) 13 13 24 17 3 35.98 10 kg 3 4 4 2.3 10 kg m E n M r π ρ π × = = = 2 1.8 10 m × × 29.4 The mass of the hydrogen atom is approximately equal to that of the proton, . If the radius of the atom is , then 27 1.67 10 kg × 10 0.53 10 m r = × ( ) ( ) ( ) 27 3 3 3 3 10 31.67 10 kg 2.7 10 kg m 43 4 0.53 10 m a m m V r ρ π π × = = = = × × 414
C H A P T E R 2 9 The ratio of the nuclear density to this atomic density is 17 3 3 3 2.3 10 kg m 2.7 10 kg m n a ρ ρ × = = × 13 8.6 10 × 29.5 (a) ( ) ( )( ) ( ) ( ) 2 9 2 2 19 1 2 2 2 14 8.99 10 N m C 2 6 1.60 10 C 1.00 10 m e m ax m in k q q F r × × = = = 27.6 N × (b) -27 27.6 N 6.64 10 kg m ax m ax F a m α = = = × 27 2 4.16 10 m s × (c) ( ) ( ) 14 1 2 27.6 N 1.00 10 m e m ax m ax m in m in k q q PE F r r = = = × 13 13 1 MeV 2.76 10 J 1.60 10 J = × = 1.73 MeV × 29.6 (a) From conservation of energy, KE PE = −∆ , or ( 2 1 2 m v q V ) = . Also, the centripetal acceleration is supplied by the magnetic force, so 2 m v qvB r = , or v qBr m = The energy equation then yields ( ) 2 2 r m V qB = For , And 12 C =12u m ( ) ( ) ( ) ( ) 27

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