Question

The ${\mathit{A}}_{\mathbf{2}}^{\mathbf{+}}$ particle and its products decay according to the scheme

$$\begin{gathered} {\mathit{A}}_{\mathbf{2}}^{\mathbf{+}}\mathbf{\to }{\mathit{\rho }}^{\mathbf{0}}\mathbf{+}{\mathit{\pi }}^{\mathbf{+}} \\ {\mathit{\rho }}^{\mathbf{0}}\mathbf{\to }{\mathit{\pi }}^{\mathbf{+}}\mathbf{+}{\mathit{\pi }}^{\mathbf{-}} \\ {\mathit{\pi }}^{\mathbf{+}}\mathbf{\to }{\mathit{\mu }}^{\mathbf{+}}\mathbf{+}\mathit{\nu } \\ {\mathit{\mu }}^{\mathbf{+}}\mathbf{\to }{\mathit{e}}^{\mathbf{+}}\mathbf{+}\mathit{\nu }\mathbf{+}\overline{\mathbf{\nu }} \\ {\mathit{\pi }}^{\mathbf{-}}\mathbf{\to }{\mathit{\mu }}^{\mathbf{-}}\mathbf{+}\overline{\mathbf{\nu }} \\ {\mathit{\mu }}^{\mathbf{-}}\mathbf{\to }{\mathit{e}}^{\mathbf{-}}\mathbf{+}\mathit{\nu }\mathbf{+}\overline{\mathbf{\nu }} \end{gathered}$$

(a) What are the final stable decay products? From the evidence,(b) is the ${\mathit{A}}_{\mathbf{2}}^{\mathbf{+}}$particle, a fermion or a boson and (c) is it a meson or a baryon?(d) What is its baryon number?

Short answer

(a) The final stable decay products are two electrons, one positron, five neutrinos and four antineutrinos.

(b) The particle is a boson.

(c) The particle is a meson.

(d) The baryon number of the particle is 0.

Step-by-step solution

Given data

The decay scheme of the particle is as below.

$\begin{array}{rcl}{A}_2^{+}& \to & {\rho }^0+{\pi }^{+}\\ {\rho }^0& \to & {\pi }^{+}+{\pi }^{-}\\ {\pi }^{+}& \to & {\mu }^{+}+\nu \\ {\mu }^{+}& \to & e^{+}+\nu +\overline{\mathrm{\nu }}\\ {\pi }^{-}& \to & {\mu }^{-}+\overline{\mathrm{\nu }}\\ {\mu }^{-}& \to & e^{-}+\nu +\overline{\mathrm{\nu }}\end{array}$

Spins of pion and rho-meson 

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 ${\pi }^{+},{\pi }^{-}$ have spin 0 and the rho-meson ${\rho }^0$ has spin 1.

(a) Determining the final decay products of A2+

The $A_2^{+}$ decays to a ${\pi }^{+}$ and a ${\rho }^0$. The${\rho }^0$decays to a${\pi }^{+}$ and a ${\pi }^{-}$. Thus $A_2^{+}$ decays to two ${\pi }^{+}$ and one ${\pi }^{-}$.

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

Thus the final decay products are,

$\begin{array}{rcl}{A}_2^{+}& \to & {\rho }^0+{\pi }^{+}\\ & \to & {\pi }^{+}+{\pi }^{-}+{\pi }^{+}\\ & \to & {\mu }^{+}+\nu +{\mu }^{-}+\overline{\nu }+{\mu }^{+}+\nu \\ & \to & e^{+}+\nu +\overline{\nu }+\nu +e^{-}+\nu +\overline{\nu }+\overline{\nu }+e^{+}+\nu +\overline{\nu }+\nu \end{array}$

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

(b) Determining whether  A2+ is a fermion or a boson

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.

(c) Determining whether A2+  is a meson or a baryon:

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.

(d) Determining the baryon number of  A2+ 

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