Date | November 2012 | Marks available | 1 | Reference code | 12N.3.HL.TZ0.22 |
Level | Higher level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | State | Question number | 22 | Adapted from | N/A |
Question
This question is about quarks.
State the name of a particle that is its own antiparticle.
The meson K0 consists of a d quark and an anti s quark. The K0 decays into two pions as shown in the Feynman diagram.
(i) State a reason why the kaon K0 cannot be its own antiparticle.
(ii) Explain how it may be deduced that this decay is a weak interaction process.
(iii) State the name of the particle denoted by the dotted line in the diagram.
(iv) The mass of the particle in (b)(iii) is approximately 1.6×10–25kg. Determine the range of the weak interaction.
Markscheme
photon / graviton / Z / Higgs;
(i) K0 has a strangeness of +1, its antiparticle has strangeness –1 and so are different;
the antiparticle is s, \(\overline d \) and so is different;
(ii) strangeness is violated in this decay;
this can only happen with the weak interaction;
(iii) Z0 / Z;
(iv) \(R = \left( {\frac{h}{{4\pi mc}} = } \right)\frac{{6.6 \times {{10}^{ - 34}}}}{{4 \times \pi \times 1.6 \times {{10}^{ - 25}} \times 3.0 \times {{10}^8}}}\);
R≈10−18m;
Award [2] for a bald correct answer.