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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

Expert verified

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

01

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.

02

Information Provided

  • The proton-proton cycle reaction is: \({\rm{2}}{{\rm{e}}^{\rm{ - }}}{\rm{ + }}{{\rm{4}}^{\rm{1}}}{\rm{H}}{ \to ^{\rm{4}}}{\rm{He + 2}}{{\rm{\nu }}_{\rm{e}}}{\rm{ + 6}}\gamma \).
03

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.

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