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

In the decay process represented, particle Y is produced alongside a beta-minus particle (the

electron). Since a beta-minus decay involves the emission of an electron, it indicates that particle Y must have characteristics consistent with a neutrino. The relevant properties of particle Y can therefore be assessed.

Option A: It has a lepton number of +1 - This is incorrect as neutrinos (if we consider Y to be a neutrino) have a lepton number of 0.

Option B: It is an antiparticle - This is incorrect as Y is likely a neutrino, not an antiparticle.

Option C: It is negatively charged - This is incorrect; neutrinos are neutral.

Option D: It experiences the strong interaction - This is correct as the weak force governs beta decay, not the strong interaction.

Thus, the correct property of particle Y is that it experiences the strong interaction.