Question

Verify by listing the number of nucleons, total charge, and electron family number before and after the cycle that these quantities are conserved in the overall proton-proton cycle in \(2{e^ - } + {4^1}H{ \to ^4}He + 2{\nu _e} + 6\gamma \).

Short answer

The listing is done as follows –

	Nucleons	Charges	Family
\({{\rm{e}}^{\rm{ - }}}\)	\({\rm{0}}\)	\({\rm{ - 2}}\)	\({\rm{ + 1}}\)
\(^1H\)	\({\rm{1}}\)	\({\rm{1}}\)	\({\rm{0}}\)
\(^{\rm{4}}{\rm{He}}\)	\({\rm{4}}\)	\({\rm{2}}\)	\({\rm{0}}\)
\(\gamma \)	\({\rm{0}}\)	\({\rm{0}}\)	\({\rm{0}}\)

The listing suggests that all the quantities are conserved in the overall proton-proton cycle in \(2{e^ - } + {4^1}H{ \to ^4}He + 2{\nu _e} + 6\gamma \).

Step-by-step solution

Concept Introduction

One of two known sets of nuclear fusion events by which stars convert hydrogen to helium is the proton–proton chain, often known as the\({\rm{p - p}}\)chain. It is dominant in stars with masses less than or equal to that of the Sun, but theoretical models imply that the\({\rm{CNO}}\)cycle, the second known reaction, is dominant in stars with masses more than around\({\rm{1}}{\rm{.3}}\)times that of the Sun.

Information Provided

• The proton-proton cycle reaction is: \({\rm{2}}{{\rm{e}}^{\rm{ - }}}{\rm{ + }}{{\rm{4}}^{\rm{1}}}{\rm{H}}{ \to ^{\rm{4}}}{\rm{He + 2}}{{\rm{\nu }}_{\rm{e}}}{\rm{ + 6}}\gamma \).

List of nucleons, total charge and electron family number

The following is the reaction –

\({\rm{2}}{{\rm{e}}^{\rm{ - }}}{\rm{ + }}{{\rm{4}}^{\rm{1}}}{\rm{H}}{ \to ^{\rm{4}}}{\rm{He + 2}}{{\rm{\nu }}_{\rm{e}}}{\rm{ + 6}}\gamma \)

Analyse the number of charges, nucleons, and family for each constituent –

	Nucleons	Charges	Family
\({{\rm{e}}^{\rm{ - }}}\)	\({\rm{0}}\)	\({\rm{ - 2}}\)	\({\rm{ + 1}}\)
\(^{\rm{1}}{\rm{H}}\)	\({\rm{1}}\)	\({\rm{1}}\)	\({\rm{0}}\)
\(^{\rm{4}}{\rm{He}}\)	\({\rm{4}}\)	\({\rm{2}}\)	\({\rm{0}}\)
\(\gamma \)	\({\rm{0}}\)	\({\rm{0}}\)	\({\rm{0}}\)

Conservations:

• Charge - Before \( - 2 + 4 = 2\), after only helium (two protons, two neutrons) is obtained, so now it is \(2 - 2\) which is conserved.
• Nucleons - Before \({\rm{41}}\) (hydrogen), helium has four nucleons (two protons, two neutrons), \(4 = 4\), which is conserved.
• Lepton family - Electron and their neutrinos are having positive lepton family number. With two electrons on the left and two-electron neutrinos on the right, it is \(2 - 2\), which means the lepton family number is conserved.

Therefore, all the quantities are conserved.