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I just recently learned how to draw feynman diagrams by looking at an equation such as one for $\beta-$ Decay: $$n = p + e^- + \bar{v}_e$$

I was wondering in $\beta-$ why an electron-antineutrino was produced and not just an electron-neutrino? Similarly, why does $\beta+$ produce an electron-neutrino and not an electron-antineutrino.

$\beta+$ Decay: $$p = n + e^+ + v_e$$

Feynman Diagram for $\beta-$ Decay: Feynman diagram of negative beta decay

Feynman Diagram for $\beta+$ Decay:

Feynman diagram of beta decay

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I was wondering in $\beta^{-}$ why an electron-antineutrino was produced and not just an electron-neutrino?

Because lepton number has to be conserved. In the LHS the lepton number is zero so on the RHS the lepton number must also be zero. All leptons have assigned a value of $+1$ and antileptons have $-1$. The conservation of lepton number is satisfied by the fact that $\beta^{-}$ decay produces an antineutrino instead of a neutrino.

Similarly, why does $\beta^{+}$ produce an electron-neutrino and not an electron-antineutrino.

$\beta^{+}$ produces a positron and a neutrino. The explanation is the same as the one above. Positron has a lepton number equal to $-1$ (because its an antilepton) and neutrino has $+1$, so that the total lepton number on the RHS is equal to the total lepton number on the LHS.

The second Feynman diagram is wrong. $W^{+}$ should decay into a positron-neutrino pair: $$W^{+}\rightarrow e^{+}\nu_{e}$$

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