Question 22M.2.SL.TZ1.c
| Date | May 2022 | Marks available | [Maximum mark: 2] | Reference code | 22M.2.SL.TZ1.c |
| Level | SL | Paper | 2 | Time zone | TZ1 |
| Command term | Explain, State | Question number | c | Adapted from | N/A |
The K+ meson can decay as
K+ → μ+ + vμ.
State and explain the interaction that is responsible for this decay.
[2]
weak «interaction» ✓
strangeness is not conserved and this is possible only in weak interactions
OR
the weak interaction allows change of quark flavour
OR
only the weak interaction has a boson / an exchange particle / a W+ to conserve the charge
OR
neutrinos are only produced via the weak interaction ✓
This was another item where some candidates simply described the particles without specifying the weak interaction. The second marking point was for a justification based on an aspect of this decay that could only be true of the weak nuclear force. A commonly incorrect answer was that this was the only force that acted on quarks and leptons, which was not accepted due to the fact that the gravitational force also acts on these particles as well. Another common incorrect answer among SL candidates was to assume that this was an example of beta negative decay due to the presence of a neutrino.
