The law of conservation of lepton number states that the total number of leptons minus the total number of antileptons is conserved in any particle interaction. Therefore, any decay that violates this conservation law cannot occur. Let's check each decay:
(a) n→p+e−: This decay is allowed since the neutron (n) is a hadron and does not involve any leptons.
(b) μ+→e++ve: This decay violates the conservation of lepton number, since the muon (μ+) is a lepton and its decay products include an electron (e+) and a neutrino (νe), which are also leptons. Therefore, this decay cannot occur.
(c) π+→e++ve+vˉμ: This decay violates the conservation of lepton number, since the pion (π+) is a hadron and its decay products include an electron (e+), a neutrino (νe), and an antineutrino (νμ), which are all leptons. Therefore, this decay cannot occur.
(d) p→n+e++ve: This decay is allowed since the proton (p) is a hadron and does not involve any leptons.
(e) π−→e−+vˉe: This decay is allowed since the pion (π-) is a hadron and involves a lepton (e-) and an antineutrino (νe).
(f) μ−→e−+vˉe+vμ: This decay violates the conservation of lepton number, since the muon (μ-) is a lepton and its decay products include an electron (e-), an antineutrino (νe), and a neutrino (νμ), which are all leptons. Therefore, this decay cannot occur.
(g) Λ0→π−+p: This decay is allowed since the lambda (Λ0) is a hadron and does not involve any leptons.
(h) K+→μ++vμ: This decay is allowed since the kaon (K+) is a hadron and involves a lepton (μ+) and a neutrino (νμ).
Therefore, the decays that violate the conservation of lepton number and cannot occur are (b) μ+→e++ve and (c) π+→e++ve+vˉμ.