Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

The $A_{2}^{+}$ particle and its products decay according to the scheme
$A_{2}^{+} \to ρ^{0} + π^{+} ρ^{0} \to π^{+} + π^{-} π^{+} \to μ^{+} + ν μ^{+} \to e^{+} + ν + \bar{ν} π^{-} \to μ^{-} + \bar{ν} μ^{-} \to e^{-} + ν + \bar{ν}$

(a) What are the final stable decay products? From the evidence, (b) is the $A_{2}^{+}$ 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.

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The decay scheme of the particle is as below.

$\begin{array}{rcl} A_{2}^{+} & \to & ρ^{0} + π^{+} \\ ρ^{0} & \to & π^{+} + π^{-} \\ π^{+} & \to & μ^{+} + ν \\ μ^{+} & \to & e^{+} + ν + \bar{ν} \\ π^{-} & \to & μ^{-} + \bar{ν} \\ μ^{-} & \to & e^{-} + ν + \bar{ν} \end{array}$

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

Step 3: (a) Determining the final decay products of A2+

The $A_{2}^{+}$ decays to a $π^{+}$ and a $ρ^{0}$ . The $ρ^{0}$ decays to a $π^{+}$ and a $π^{-}$ . Thus $A_{2}^{+}$ decays to two $π^{+}$ and one $π^{-}$ .

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

Thus the final decay products are,

$\begin{array}{rcl} A_{2}^{+} & \to & ρ^{0} + π^{+} \\ \to & π^{+} + π^{-} + π^{+} \\ \to & μ^{+} + ν + μ^{-} + \bar{ν} + μ^{+} + ν \\ \to & e^{+} + ν + \bar{ν} + ν + e^{-} + ν + \bar{ν} + \bar{ν} + e^{+} + ν + \bar{ν} + ν \end{array}$

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.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Spins of pion and rho-meson

Step 3: (a) Determining the final decay products of A2+

Step 4: (b) Determining whether A2+ is a fermion or a boson

Step 5: (c) Determining whether A2+ is a meson or a baryon:

Step 6: (d) Determining the baryon number of A2+

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Spins of pion and rho-meson

Step 3: (a) Determining the final decay products of A2+

Step 4: (b) Determining whether A2+ is a fermion or a boson

Step 5: (c) Determining whether A2+ is a meson or a baryon:

Step 6: (d) Determining the baryon number of A2+

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