CHAPTER 42 - 1

# CHAPTER 42 - 1 - Note e/4p =(1.60 10 0 2 19 C/4p(8.85 10 2...

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CHAPTER 42 – Nuclear Physics and Radioactivity Note: A factor that appears in the analysis of energies is e 2 /4p Å 0  = (1.60 × 10 –19  C) 2 /4p(8.85 × 10 –12  C 2 /N     m 2 ) = 2.30 × 10 –28  J     m = 1.44 MeV     fm. 1. To find the rest mass of an  α  particle, we subtract the rest mass of the two electrons from the rest mass of a helium atom: m   =  M He  – 2 m e   = (4.002603 u)(931.5 MeV/u c 2 ) – 2(0.511 MeV/ c 2 ) =       3727 MeV/ c 2 . 2. We convert the units: m  = (139 MeV/ c 2 )/(931.5 MeV/u c 2 ) =      0.149 u . 3. The   particle is a helium nucleus: r  = (1.2 × 10 –15  m) A 1/3  = (1.2 × 10 –15  m)(4) 1/3  =       1.9 × 10 –15  m       = 1.9 fm. 4. The radius of a nucleus is  r  = (1.2 × 10 –15  m) A 1/3 . If we form the ratio for the two isotopes, we get

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CHAPTER 42 - 1 - Note e/4p =(1.60 10 0 2 19 C/4p(8.85 10 2...

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