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.

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.