CHAPTER 42 - 4

# CHAPTER 42 - 4 - H 1 3 → He 2 3 e – 1 ν n 1 → p 1 1...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: H 1 3 → He 2 3 + e – 1 + ν n 1 → p 1 1 + e –1 + ν C 6 11 → B 5 10 + p 1 1 Na 11 22 Mg 12 22 Na 11 22 → Ne 10 22 + β + + ν 20. The decay is . When we add an electron to both sides to use atomic masses, we see that the mass of the emitted β particle is included in the atomic mass of 3 He. Thus the energy released is Q = [ M ( 3 H) – M ( 3 He)] c 2 = [(3.016049 u) – (3.016029 u)] c 2 (931.5 MeV/u c 2 ) = 0.0186 MeV = 18.6 keV . 21. The decay is . We take the electron mass to use the atomic mass of 1 H. The kinetic energy of the electron will be maximum if no neutrino is emitted. If we ignore the recoil of the proton, the maximum kinetic energy is K = [ m ( 1 n) – M ( 1 H)] c 2 = [(1.008665 u) – (1.007825 u)] c 2 (931.5 MeV/u c 2 ) = 0.782 MeV ....
View Full Document

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

### Page1 / 2

CHAPTER 42 - 4 - H 1 3 → He 2 3 e – 1 ν n 1 → p 1 1...

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

View Full Document
Ask a homework question - tutors are online