Which of the following decays cannot occur because the law of conservation of lepton number is violated? (a) \(\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}\) (b) \(\mu^{+} \rightarrow \mathrm{e}^{+}+v_{\mathrm{e}}\) (c) \(\pi^{+} \rightarrow \mathrm{e}^{+}+v_{\mathrm{e}}+\bar{v}_{\mu}\) (d) \(\mathrm{p} \rightarrow \mathrm{n}+\mathrm{e}^{+}+v_{\mathrm{e}}\) (e) \(\pi^{-} \rightarrow \mathrm{e}^{-}+\bar{v}_{\mathrm{e}}\) (f) \(\mu^{-} \rightarrow \mathrm{e}^{-}+\bar{v}_{\mathrm{e}}+v_{\mu}\) (g) \(\Lambda^{0} \rightarrow \pi^{-}+\mathrm{p}\) (h) \(\mathbf{K}^{+} \rightarrow \mu^{+}+v_{\mu}\)

Leptons can be any of the following particles: electron (\(\mathrm{e}^{-}\)), muon (\(\mu^{-}\)), and their neutrinos (\(\nu_{\mathrm{e}}\) and \(\nu_{\mu}\)). Antileptons are the corresponding antiparticles of leptons: positron (\(\mathrm{e}^{+}\)), antimuon (\(\mu^{+}\)), and their antineutrinos (\(\bar{\nu}_{\mathrm{e}}\) and \(\bar{\nu}_{\mu}\)).

Let's check each decay to see if the lepton number is conserved. (a) \(\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}\) Both sides have lepton numbers equal to zero, so lepton number is conserved. (b) \(\mu^{+} \rightarrow \mathrm{e}^{+}+v_{\mathrm{e}}\) Left side: \((-1)\) Right side: \((-1) + 1 = 0\) Lepton number is not conserved. (c) \(\pi^{+} \rightarrow \mathrm{e}^{+}+v_{\mathrm{e}}+\bar{v}_{\mu}\) Both sides have lepton numbers equal to zero, so lepton number is conserved. (d) \(\mathrm{p} \rightarrow \mathrm{n}+\mathrm{e}^{+}+v_{\mathrm{e}}\) Both sides have lepton numbers equal to zero, so lepton number is conserved. (e) \(\pi^{-} \rightarrow \mathrm{e}^{-}+\bar{v}_{\mathrm{e}}\) Both sides have lepton numbers equal to zero, so lepton number is conserved. (f) \(\mu^{-} \rightarrow \mathrm{e}^{-}+\bar{v}_{\mathrm{e}}+v_{\mu}\) Left side: \(1\) Right side: \(1 - 1 + 1 = 1\) Lepton number is conserved. (g) \(\Lambda^{0} \rightarrow \pi^{-}+\mathrm{p}\) Both sides have lepton numbers equal to zero, so lepton number is conserved. (h) \(\mathbf{K}^{+} \rightarrow \mu^{+}+v_{\mu}\) Both sides have lepton numbers equal to zero, so lepton number is conserved.

Out of all the given decays, decay (b) \(\mu^{+} \rightarrow \mathrm{e}^{+}+v_{\mathrm{e}}\) is the only one that violates the law of conservation of lepton number.