What are the momenta of particles in beta emission of C14?

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SUMMARY

The discussion centers on the beta emission of Carbon-14 (C14), which disintegrates with a reaction energy (Q value) of 0.155 MeV. During this process, a beta particle with an energy of 0.025 MeV is emitted at an angle of 135° relative to the recoil nucleus's motion. The momenta of the beta particle, nitrogen-14 (14N), and the antineutrino must be calculated in MeV/c units, considering the mass of the beta particle (M0 = 0.511 MeV/c²) and the contributions of all particles involved in the reaction.

PREREQUISITES
  • Understanding of beta decay and particle physics
  • Familiarity with energy conservation principles in nuclear reactions
  • Knowledge of momentum calculations in relativistic physics
  • Basic understanding of MeV/c units and their significance in particle physics
NEXT STEPS
  • Study the conservation of momentum in beta decay processes
  • Learn about the role of antineutrinos in nuclear reactions
  • Explore the calculation of momenta using relativistic equations
  • Investigate the properties and behavior of beta particles and recoil nuclei
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This discussion is beneficial for physics students, particle physicists, and anyone studying nuclear decay processes, particularly those interested in the mechanics of beta emission and momentum calculations.

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Homework Statement



C14 disintegrates by b-emission with a reaction
energy (q value) of 0.155 MeV. A b-particles with an
energy of 0.025MeV is emitted in a direction at 135°
to the direction of motion of the recoil nucleus. Then
the momenta of the three particles (b = V, 14N)
involved in this disintegration in MeV/c units will
be. (where, c is speed of light in vacuum) (M0 = 0.511
MeV/c2 )

Homework Equations



Q=Total Energy. therefore Q= 0.025+ energy of nitrogen atom + antineutrino

The Attempt at a Solution


Q=P^2/2m(nitrogen) +p^2/2m(beta particle)
 
Physics news on Phys.org
I am confused about the effect of the antineutrino. Can you please explain that?
 

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