Question

Verify the quantum numbers given for the proton and neutron in Table \[33.2\] by adding the quantum numbers for their quark constituents as given in Table \[33.4\].

Short answer

The quantum numbers given for proton and neutron is\[B = 1,{\rm{ }}S = 0,{\rm{ }}{L_e} = {L_\mu } = {L_\tau } = 0\]which is verified with table\[{\rm{33}}{\rm{.2}}\].

Step-by-step solution

Concept Introduction

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 particles

Prove the quantum numbers for proton and neutron by using its quark compositions. Thus, the quantum numbers for proton and neutron from table\[{\rm{33}}{\rm{.2}}\]is –

	\[{\rm{B}}\]	\[{{\rm{L}}_{\rm{e}}}\]	\[{{\rm{L}}_{\rm{\mu }}}\]	\[{{\rm{L}}_{\rm{\tau }}}\]	\[{\rm{S}}\]
\[{\rm{p}}\]	\[{\rm{1}}\]	\[{\rm{0}}\]	\[{\rm{0}}\]	\[{\rm{0}}\]	\[0\]
\[{\rm{n}}\]	\[{\rm{1}}\]	\[{\rm{0}}\]	\[{\rm{0}}\]	\[{\rm{0}}\]	\[0\]

Now, from table\[{\rm{33}}{\rm{.4}}\], the quark compositions for\[{\rm{p}}\]and\[{\rm{n}}\]is –

\[\begin{array}{l}p = \left[ {uud} \right]\\n = \left[ {udd} \right]\end{array}\]

Quark Compositions

So, now for proton it is obtained –

\[\begin{array}{c}B = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\\S = 0 + 0 + 0 = 0\end{array}\]

Since, it is a baryon not a lepton. So, it is obtained –

\[{L_e} = {L_\mu } = {L_\tau } = 0\]

So, now for neutron it is obtained –

\[\begin{array}{c}B = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\\S = 0 + 0 + 0 = 0\end{array}\]

Since, it is a baryon not a lepton. So, it is obtained –

\[{L_e} = {L_\mu } = {L_\tau } = 0\]

Therefore, from the quark compositions for\[{\rm{p}}\]and\[{\rm{n}}\], it can be proved that its quantum numbers are given in table \[{\rm{33}}{\rm{.2}}\].