Feynman diagram for semileptonic decay of neutral kaon I am unsure how to draw a feynman diagram for a reaction that occurs as follows
$$ K^0  --> l^+\nu_l\pi^- $$
Any tips would be helpful.
 A: Drawing the most appropriate Feynman diagram can be a little tricky sometimes. I find it best to work backwards with the general rule of thumb being to try to minimize the number of vertices. We know that the lepton pair $l^{+}, \nu_{l}$ must come from a $W^{+}$ boson as that's the only mediator that conserves charge and can violate flavor. Since we are dealing with the $K^{0}$ meson whose quark configuration is $d \bar{s}$ and a daughter meson $\pi^{-}$ whose configuration is $d \bar{u}$, we notice that both the $K^{0}$ and $\pi^{-}$ contain a $d$-quark. This $d$-quark will act as a spectator quark (will remain unchanged in the scattering process). Now all that's left is recognizing that $W$ bosons connect the positively charged quarks to the negative and vice verse with the exception of not connecting particles to antiparticles. So for an example a $q^{+} \longrightarrow q^{-} + W^{+}$ where the positive superscript is merely to denote a positive charge and the negative denotes negative charge. 
Hopefully that's sufficient information to guide you to the proper Feynman Diagram.
A: Because neutral currents have not been observed and flavor changing can only be done with the emission of W bosons the strange antiquark will decay to an up antiquark and W- boson . The W- boson will decay to an electron and an antielectron neutrino. The pion formed by the up antiquark and the down quark will decay to a neutral pion and a W- boson.The W- boson will decay to an electron and an electron antineutrino.The neutral pion will decay to 2 gamma rays and all is left is 2 electrons and 2 electron antineutrinos and some light.
