CHAPTER 42 - 1 - CHAPTER42NuclearPhysicsandRadioactivity...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at University of California, Berkeley.

Page1 / 2

CHAPTER 42 - 1 - CHAPTER42NuclearPhysicsandRadioactivity...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online