Final answer:
The decay Σ^-→n+π^- conserves both baryon number and lepton number, making it a possible process. This conservation is evident when considering the quark constituents of each particle involved in the decay.
Step-by-step explanation:
The decay Σ^-→n+π^- is indeed possible considering the appropriate conservation laws. It conserves both baryon number and lepton number. The Σ^- baryon and the neutron (n) are both baryons, so their decay conserves the baryon number (baryon number for both is +1). The π^- (pion) does not affect baryon number as it is a meson. Lepton number is conserved as well since there are no leptons involved in this decay process.
To write this decay in terms of quark constituents, recall that Σ^- is made up of two down quarks (d) and a strange quark (s), while the neutron is made up of one up quark (u) and two down quarks (d), and π^- consists of a down quark and an anti-up quark (¯u). Thus, the decay in quark terms would be:
Σ^- (dds) → n (udd) + π^- (d¯u)