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CHAPTER 42 - 2

# CHAPTER 42 - 2 - 8 r=r A 0 1/3 r =(1.2fm(4 =1.9 fm r 1/3...

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8. We find the radii of the two nuclei from r  =  r 0 A 1/3 ; r α  = (1.2 fm)(4) 1/3  = 1.9 fm; r Am  = (1.2 fm)(243) 1/3  = 7.5 fm. We assume that the nucleus is so much heavier than the   particle that we can ignore the recoil of the nucleus.  We find the  kinetic energy of the   particle from the conservation of energy: K i  +  U i  =  K f  +  U f     ; 0 +  Z Z Am e 2 /4p Å 0 ( r  +  r Am ) =  K f   + 0; (2)(95)(1.44 MeV     fm)/(1.9 fm + 7.5 fm) =  K f   , which gives  K f  =       29 MeV . 9. The radius of a nucleus is  r  = (1.2 × 10 –15  m) A 1/3 . If we form the ratio for the two nuclei, we get r X / r U  = ( A X / A U ) 1/3 ; !  = ( A X /238) 1/3 , which gives  A X  = 30. From the Appendix, we see that the stable nucleus could be     . 10. From Figure 42–1, we see that the average binding energy per nucleon at

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CHAPTER 42 - 2 - 8 r=r A 0 1/3 r =(1.2fm(4 =1.9 fm r 1/3...

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