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Found in: Page 1363

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The ${{\mathbit{A}}}_{{\mathbf{2}}}^{{\mathbf{+}}}$ particle and its products decay according to the scheme ${{\mathbit{A}}}_{{\mathbf{2}}}^{{\mathbf{+}}}{\mathbf{\to }}{{\mathbit{\rho }}}^{{\mathbf{0}}}{\mathbf{+}}{{\mathbit{\pi }}}^{{\mathbf{+}}}\phantom{\rule{0ex}{0ex}}{{\mathbit{\rho }}}^{{\mathbf{0}}}{\mathbf{\to }}{{\mathbit{\pi }}}^{{\mathbf{+}}}{\mathbf{+}}{{\mathbit{\pi }}}^{{\mathbf{-}}}\phantom{\rule{0ex}{0ex}}{{\mathbit{\pi }}}^{{\mathbf{+}}}{\mathbf{\to }}{{\mathbit{\mu }}}^{{\mathbf{+}}}{\mathbf{+}}{\mathbit{\nu }}\phantom{\rule{0ex}{0ex}}{{\mathbit{\mu }}}^{{\mathbf{+}}}{\mathbf{\to }}{{\mathbit{e}}}^{{\mathbf{+}}}{\mathbf{+}}{\mathbit{\nu }}{\mathbf{+}}\overline{\mathbf{\nu }}\phantom{\rule{0ex}{0ex}}{{\mathbit{\pi }}}^{{\mathbf{-}}}{\mathbf{\to }}{{\mathbit{\mu }}}^{{\mathbf{-}}}{\mathbf{+}}\overline{\mathbf{\nu }}\phantom{\rule{0ex}{0ex}}{{\mathbit{\mu }}}^{{\mathbf{-}}}{\mathbf{\to }}{{\mathbit{e}}}^{{\mathbf{-}}}{\mathbf{+}}{\mathbit{\nu }}{\mathbf{+}}\overline{\mathbf{\nu }}$(a) What are the final stable decay products? From the evidence, (b) is the ${{\mathbit{A}}}_{{\mathbf{2}}}^{{\mathbf{+}}}$ particle, a fermion or a boson and (c) is it a meson or a baryon? (d) What is its baryon number?

(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.

See the step by step solution

## Step 1: Given data

The decay scheme of the particle is as below.

$\begin{array}{rcl}{A}_{2}^{+}& \to & {\rho }^{0}+{\pi }^{+}\\ {\rho }^{0}& \to & {\pi }^{+}+{\pi }^{-}\\ {\pi }^{+}& \to & {\mu }^{+}+\nu \\ {\mu }^{+}& \to & {e}^{+}+\nu +\overline{\mathrm{\nu }}\\ {\pi }^{-}& \to & {\mu }^{-}+\overline{\mathrm{\nu }}\\ {\mu }^{-}& \to & {e}^{-}+\nu +\overline{\mathrm{\nu }}\end{array}$

## Step 2: 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.

## Step 3: (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.

## Step 4: (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.

## Step 5: (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.

## Step 6: (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.