An iodine nucleus decays into a nucleus of X e-^131, a beta-minus particle and particle Y - AQA - A-Level Physics - Question 12 - 2018 - Paper 1

To determine a property of particle Y in the decay process of iodine into xenon and a beta-minus particle, we first note the conservation of quantum numbers in particle decays.

Given the decay:

$^{131}_{53}I \rightarrow ^{131}_{54}Xe + ^{0}_{-1}e + Y$

1. The lepton number must balance. The beta-minus particle (electron) has a lepton number of +1, suggesting that particle Y must have a lepton number of -1 to conserve the total lepton number.

2. Particle Y, therefore, must be an antineutrino to maintain lepton number conservation in the decay process.

3. Since it's an antineutrino, it is not negatively charged (neutral instead) and is not an antiparticle in the traditional sense.

4. Additionally, the strong interaction does not exert force on leptons, hence particle Y does not experience the strong interaction.

Based on this analysis, the correct answer is:

A: It has a lepton number of +1, identifying Y as an antineutrino, thus confirming the decay process.