Question

(a) Show that the conjectured decay of the proton, \({\rm{p}} \to {\pi ^{\rm{0}}}{\rm{ + }}{{\rm{e}}^{\rm{ + }}}\), violates conservation of baryon number and conservation of lepton number.

(b) What is the analogous decay process for the antiproton?

Short answer

(a) The decay\({\rm{p}} \to {\pi ^{\rm{0}}}{\rm{ + }}{{\rm{e}}^{\rm{ + }}}\)violates conservation of baryon and lepton numbers\(\left( {\Delta B = - 1,\Delta {L_e} = 1} \right)\).

(b) The analogous decay process for the antiproton is \({\rm{\bar p}} \to {\pi ^{\rm{ + }}}{\rm{ + }}{{\rm{e}}^{\rm{ - }}}\).

Step-by-step solution

Concept Introduction

The sum of the baryon numbers of all entering particles equals the sum of the baryon numbers of all particles produced by the reaction, according to the rule of conservation of baryon number.

Conservation of Lepton says that when a lepton of a given generation is formed or destroyed in a reaction, a matching antilepton of the same generation must also be created or destroyed.

Decay of \({\rm{p}} \to {\pi ^{\rm{0}}}{\rm{ + }}{{\rm{e}}^{\rm{ + }}}\)

(a)

Prove that the reaction\({\rm{p}} \to {\pi ^{\rm{0}}}{\rm{ + }}{{\rm{e}}^{\rm{ + }}}\)violates conservation of baryon number and conservation of lepton numbers. Simply, to prove such thing check these numbers before and after the reaction, as follow –

	Before	After
\({\rm{B}}\)	\({\rm{1}}\)	\({\rm{0 + 0 = 0}}\)
\({\rm{Q}}\)	\({\rm{1}}\)	\({\rm{0 + 1 = 1}}\)
\({{\rm{L}}_{\rm{e}}}\)	\({\rm{0}}\)	\({\rm{0 + 1 = 1}}\)
\({{\rm{L}}_{\rm{\mu }}}\)	\({\rm{0}}\)	\({\rm{0}}\)
\({{\rm{L}}_{\rm{\tau }}}\)	\({\rm{0}}\)	\({\rm{0}}\)

Therefore, this decay violates the conservation of baryon and lepton numbers.

Analogous decay process

(b)

Write the same decay but for the anti-proton particle. The decay can be written simply by replacing each particle by its anti-particle, as follows –

\(\begin{aligned}{c}{\rm{p}} \to {\rm{\bar p}},\\{\pi ^{\rm{ - }}} \to {\pi ^{\rm{ + }}},\\{{\rm{e}}^{\rm{ + }}} \to {{\rm{e}}^{\rm{ - }}}\end{aligned}\)

Where,\({\rm{\bar p}}\) is the anti-proton,\({\pi ^{\rm{ + }}}\)is the anti-particle for\({\pi ^{\rm{ - }}}\), and the electron\({{\rm{e}}^{\rm{ - }}}\)is the anti-particle for positron\({{\rm{e}}^{\rm{ + }}}\).

Hence, the decay is –

\({\rm{\bar p}} \to {\pi ^{\rm{ + }}}{\rm{ + }}{{\rm{e}}^{\rm{ - }}}\)

Therefore, the decay reaction is obtained as\({\rm{\bar p}} \to {\pi ^{\rm{ + }}}{\rm{ + }}{{\rm{e}}^{\rm{ - }}}\).