In BETA-MINUS ($\beta^-$) decay, the parent nucleus emits an electron AND another particle. The other particle is
Aa proton
Ba neutrino
Ca photon
Dan antineutrino
Answer & Solution
Correct answer: D. an antineutrino
1. The full $\beta^-$ decay equation is: $\,^{A}_{Z}\mathrm{X} \to \,^{A}_{Z+1}\mathrm{Y} + e^- + \bar{\nu}_e$.
2. The third product is an ANTI-neutrino (specifically the electron antineutrino $\bar{\nu}_e$), not a neutrino.
3. Why antineutrino? Lepton-number conservation. The electron $e^-$ has lepton number $+1$; to keep total lepton number zero on the right (as on the left), an antiparticle with lepton number $-1$ is required — that is $\bar{\nu}_e$.
4. At the quark level, $\beta^-$ decay is $n \to p + e^- + \bar{\nu}_e$ — a neutron in the nucleus converts to a proton with the emission of $e^-$ and $\bar{\nu}_e$.
5. The antineutrino is what carries away the missing energy and momentum that, without it, would have made the beta spectrum a sharp line (it is actually a continuum) — historically this is HOW the neutrino was predicted by Pauli in 1930.
6. Option A is wrong — a proton stays in the nucleus. Option B (neutrino) is reserved for $\beta^+$ decay where the daughter emits a positron and a neutrino. Option D (photon) is gamma decay, a distinct process.
_Source: NCERT Class 12 Physics Part 2, Ch 13, §13.6 (Radioactivity — beta decay), p. 8–9._
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