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

AI Thread Summary
In the beta emission of C14, the reaction energy (q value) is 0.155 MeV, with a beta particle emitted at 0.025 MeV and at an angle of 135° relative to the recoil nucleus. The total energy equation includes the energies of the beta particle, nitrogen atom, and antineutrino. The attempt to solve for momenta involves kinetic energy equations, but there is confusion regarding the role of the antineutrino in the momentum calculations. Understanding the antineutrino's contribution is crucial for accurately determining the momenta of the emitted particles. Clarifying this aspect will lead to a complete solution for the momenta in MeV/c units.
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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)
 
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I am confused about the effect of the antineutrino. Can you please explain that?
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'

Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'

(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
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