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Let P, Q and R be three different nuclei. Which one of the following nuclear processes is possible?

  • A.\(\nu_{e} + {}_{z}P^{A} \rightarrow {}_{z+1}Q^{A} + e^{-}\)
  • B.\(\nu_{e} + {}_{z}P^{A} \rightarrow {}_{z-1}R^{A} + e^{+}\)
  • C.\(\nu_{e} + {}_{z}P^{A} \rightarrow {}_{z}P^{A} + e^{+} + e^{-}\)
  • D.\(\nu_{e} + {}_{z}P^{A} \rightarrow {}_{z}P^{A} + \gamma\)

Answer: A

The possible process is \(\nu_{e} + {}_{z}P^{A} \rightarrow {}_{z+1}Q^{A} + e^{-}\).

This is neutrino capture, the inverse of beta decay: inside the nucleus a neutron absorbs the electron neutrino and turns into a proton, emitting an electron (νe + n → p + e-). Charge is conserved (0 + z = z + 1 - 1), the mass number is unchanged, and the lepton number (+1 on both sides) and electron flavour are conserved.

Producing a positron with a neutrino would violate lepton number: the left side has +1 and the right side -1. Creating an e+e- pair leaves the neutrino's lepton number unaccounted for, and the last option destroys the neutrino and creates only a photon, again violating lepton number.

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