Question

There are particles called bottom mesons or \({\rm{B}}\)-mesons. One of them is the \({{\rm{B}}^{\rm{ - }}}\)meson, which has a single negative charge; its baryon number is zero, as are its strangeness, charm, and topness. It has a bottomness of \({\rm{ - 1}}\). What is its quark configuration?

Short answer

The quark configuration of the \({\rm{B - }}\)meson particle is \({B^ - } = \left( {b\bar u} \right)\).

Step-by-step solution

Concept Introduction

Mesons are hadronic subatomic particles made up of an equal number of quarks and antiquarks, generally one of each, and linked together by strong interactions in particle physics.

A quark is a basic ingredient of matter and a sort of elementary particle.

Antiquarks are the antiparticles that correspond to each flavour of quark.

Quantum numbers for \({{\rm{B}}^{\rm{ - }}}\)

Find the quark compositions for\({{\rm{B}}^{\rm{ - }}}\)given that all its quantum numbers as follows –

	\({\rm{B}}\)	\({\rm{Q}}\)	\({\rm{S}}\)	\({\rm{t}}\)	\({\rm{b}}\)	\({\rm{c}}\)
\({{\rm{B}}^{\rm{ - }}}\)	\({\rm{0}}\)	\({\rm{ - 1}}\)	\({\rm{0}}\)	\({\rm{0}}\)	\({\rm{ - 1}}\)	\({\rm{0}}\)

Quark configuration for \({{\rm{B}}^{\rm{ - }}}\)

Now, find its quark configuration –

Since \({{\rm{B}}^{\rm{ - }}}\) is a meson. Thus, it consists of quark and anti-quark.

\({B^ - } = \left( {q\bar q} \right)\)

Since it has a bottomness of \(b = - 1\) Therefore, the quark is the bottom \({\rm{b}}\).

\({B^ - } = \left( {b\bar q} \right)\)

Since its charge is \({\rm{ - 1}}\) and the charge of the bottom quark is \({\rm{ - 1/3}}\). Thus, an anti-quark with charge \({\rm{ - 2/3}}\) is needed. The options available are – \({\rm{\bar u,\bar c,}}\) and \({\rm{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftarrow$}} \over t} }}\).

Since its topness and charm are zero. Therefore,\({\rm{\bar c,}}\)and \({\rm{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftarrow$}} \over t} }}\) cannot be considered.

Hence, the quark configuration for the \({\rm{B}}\)-meson \({{\rm{B}}^{\rm{ - }}}\) is –

\({B^ - } = \left( {b\bar u} \right)\)

Therefore, the quark configuration is obtained as \({B^ - } = \left( {b\bar u} \right)\).