91Ó°ÊÓ

The decay scheme of the particle is as below.

$\begin{array}{rcl}{A}_2^{+}& →& {\mathrm{ÒÏ}}^0+{\mathrm{Ï€}}^{+}\\ {\mathrm{ÒÏ}}^0& →& {\mathrm{Ï€}}^{+}+{\mathrm{Ï€}}^{-}\\ {\mathrm{Ï€}}^{+}& →& {\mathrm{μ}}^{+}+\mathrm{ν}\\ {\mathrm{μ}}^{+}& →& e^{+}+\mathrm{ν}+\stackrel{¯}{\mathrm{ν}}\\ {\mathrm{Ï€}}^{-}& →& {\mathrm{μ}}^{-}+\stackrel{¯}{\mathrm{ν}}\\ {\mathrm{μ}}^{-}& →& e^{-}+\mathrm{ν}+\stackrel{¯}{\mathrm{ν}}\end{array}$

Quark modeling of the pion and rho classifies the pion as a pseudoscalar meson with zero angular momentum. Quarks have spin 1/2 and the spins are "paired" or anti-aligned. In the rho meson, a vector meson, the angular momentum is j = 2 , indicating parallel rotations.

The pions ${\mathrm{Ï€}}^{+},{\mathrm{Ï€}}^{-}$ have spin 0 and the rho-meson ${\mathrm{ÒÏ}}^0$ has spin 1.

The $A_2^{+}$ decays to a ${\mathrm{Ï€}}^{+}$ and a ${\mathrm{ÒÏ}}^0$. The${\mathrm{ÒÏ}}^0$decays to a${\mathrm{Ï€}}^{+}$ and a ${\mathrm{Ï€}}^{-}$. Thus $A_2^{+}$ decays to two ${\mathrm{Ï€}}^{+}$ and one ${\mathrm{Ï€}}^{-}$.

A ${\mathrm{π}}^{+}$ decays to${\mathrm{μ}}^{+}$ and a neutrino and the${\mathrm{μ}}^{+}$ decays to an electron and a neutrino-antineutrino pair. The ${\mathrm{π}}^{-}$decays to a${\mathrm{μ}}^{-}$ and an antineutrino and the ${\mathrm{μ}}^{-}$ decays to a positron and a neutrino-antineutrino pair.

Thus the final decay products are,

$\begin{array}{rcl}{A}_2^{+}& →& {\mathrm{ÒÏ}}^0+{\mathrm{Ï€}}^{+}\\ & →& {\mathrm{Ï€}}^{+}+{\mathrm{Ï€}}^{-}+{\mathrm{Ï€}}^{+}\\ & →& {\mathrm{μ}}^{+}+\mathrm{ν}+{\mathrm{μ}}^{-}+\stackrel{¯}{\mathrm{ν}}+{\mathrm{μ}}^{+}+\mathrm{ν}\\ & →& e^{+}+\mathrm{ν}+\stackrel{¯}{\mathrm{ν}}+\mathrm{ν}+e^{-}+\mathrm{ν}+\stackrel{¯}{\mathrm{ν}}+\stackrel{¯}{\mathrm{ν}}+e^{+}+\mathrm{ν}+\stackrel{¯}{\mathrm{ν}}+\mathrm{ν}\end{array}$

Thus, the final products are two electrons, one positron, five neutrinos and four antineutrinos.

The spin of the$A_2^{+}$ is the sum of spins of the pion and the rho-meson. Hence its spin is 0 + 1 = 1 .

Hence, the $A_2^{+}$ is a boson.

As stated in the previous step, the $A_2^{+}$ is a boson. Baryons are fermions because they have half-integer spins. Since the $A_2^{+}$ is a boson, it is also a meson.

As obtained in the previous step, the$A_2^{+}$ is not a baryon, hence its baryon number is 0.