# Why is $\Delta^0$ not decaying weakly?

Given the lower decay I wonder why it happens this way. Wouldn't it be possible to decay via a weak process as well?

• Isn’t this just the fastest decay channel, not the only one? Strong implies fast; weak implies slow. – G. Smith Aug 21 at 17:41
• Hint: what's the typical lifetime for a weak decay? – probably_someone Aug 21 at 17:41
• – anna v Aug 21 at 18:11
• I take back my earlier comment. The experimental signature would be the charged lepton (we'd be looking at $\Delta^0 \to p^+ + W^- \to p^+ + l^- + \nu_l$). The rate will be small because of the difference in coupling constants and the mass difference being much smaller than the W mass, but at least you'll have something to look for. – dmckee Aug 21 at 19:19
• Yes the couplings, that is why the interaction is called weak. hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html – anna v Aug 21 at 20:38

\begin{align} \text{weak interactions:}\quad & \tau \approx 10^{-10}\text{ s}\\ \text{electromagnetic interactions:}\quad & \tau \approx 10^{-18}\text{ s}\\ \text{strong interactions:}\quad & \tau \approx 10^{-23}\text{ s} \end{align}