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

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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 'Need help understanding this figure on energy levels'

Thread 'Need help understanding this figure on energy levels'

This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
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Thread 'Understanding how to "tack on" the time wiggle factor'

The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
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