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

decay

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    $\begingroup$ Isn’t this just the fastest decay channel, not the only one? Strong implies fast; weak implies slow. $\endgroup$ – G. Smith Aug 21 at 17:41
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    $\begingroup$ Hint: what's the typical lifetime for a weak decay? $\endgroup$ – probably_someone Aug 21 at 17:41
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    $\begingroup$ Have a look hyperphysics.phy-astr.gsu.edu/hbase/Particles/delta.html $\endgroup$ – anna v Aug 21 at 18:11
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    $\begingroup$ 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. $\endgroup$ – dmckee Aug 21 at 19:19
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    $\begingroup$ Yes the couplings, that is why the interaction is called weak. hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html $\endgroup$ – anna v Aug 21 at 20:38
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Particles can decay via three different kinds of processes, depending on which force is involved. These processes all have different time scales:

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

As you can see, the strong processes happen really fast. So if a particle can decay via a strong process it probably will.

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