Explain the source of stellar energy. Explain the carbon – nitrogen cycle, proton – proton cycle occuring in stars.

In the fusion reaction, for the formation of Helium nucleus by fusing four hydrogen nuclei, they released (26.7 MeV) energy is called stellar energy.The reaction is as follows
$4_1{\mathrm{H}}^1{\to }_2{\mathrm{He}}^4+2_{+1}{\mathrm{e}}^{\mathrm{o}}+26.7\mathrm{Mev}$
In this reaction the mass of the helium nucleus is smaller than the sum of the masses of four protons. This difference in mass appears as liberation of energy in this process.
Carbon – Nitrogen cycle:-
Bothe suggested the following carbon – nitrogen cycle as one of the most important nuclear reactions for the production of solar energy by fusion.
This cycle consists of nuclear reactions in which hydrogen in converted into Helium, with the help of carbon and nitrogen as catalysts. These are as given below.
$\begin{array}{l}{ }_1{\mathrm{H}}^1{+}_6{\mathrm{C}}^{12}{\to }_7{\mathrm{N}}^{13}+\mathrm{\gamma }-\text{ ray }\\ { }_7{\mathrm{N}}^{13}{\to }_6{\mathrm{C}}^{13}{+}_1{\mathrm{e}}^0+\mathrm{v}(\text{ neutrino) }\\ { }_1{\mathrm{H}}^1{+}_6{\mathrm{C}}^{13}{\to }_7{\mathrm{N}}^{14}+\mathrm{\gamma }-\text{ ray }\\ { }_1{\mathrm{H}}^1{+}_7{\mathrm{N}}^{14}{\to }_8{\mathrm{O}}^{15}+\mathrm{\gamma }-\text{ ray }\\ { }_8{\mathrm{O}}^{15}{\to }_7{\mathrm{N}}^{15}{+}_1{\mathrm{e}}^0+\mathrm{v}\left(\text{ neutrino }\right)\\ { }_1{\mathrm{H}}^1{+}_7{\mathrm{N}}^{15}{\to }_6{\mathrm{C}}^{12}{+}_2{\mathrm{He}}^4\end{array}$
The net result of the above reactions is,
$4_1{\mathrm{H}}^1{⟶}_2{\mathrm{He}}^4+2_1{\mathrm{e}}^0+3\mathrm{\gamma }+2\mathrm{v}$
The energy released during this process is 26.7 MeV
Proton - proton cycle: The cycle consist of nuclear reaction in which four protons convert into helium with release of total energy 26.7 MeV.
$\begin{array}{l}{ }_1^1\mathrm{H}{+}_1^1\mathrm{H}{\to }_1^2\mathrm{H}+{\mathrm{e}}^{+}+\mathrm{\vartheta }+0.42\mathrm{MeV} \left(\mathrm{i}\right)\\ {\mathrm{e}}^{+}+{\mathrm{e}}^{-}\to \mathrm{\gamma }+\mathrm{\gamma }+1.02\mathrm{MeV} \left(\mathrm{ii}\right)\\ { }_1^2\mathrm{H}{+}_1^1\mathrm{H}{\to }_2^3\mathrm{He}+\mathrm{\gamma }+5.49\mathrm{MeV} \left(\mathrm{iii}\right)\\ { }_2^3\mathrm{He}{+}_2^3\mathrm{He}{\to }_2^4\mathrm{He}{+}_1^1\mathrm{H}{+}_1^1\mathrm{H}+12.86\mathrm{MeV} \left(\mathrm{iv}\right)\end{array}$
For the fourth reaction to occur, the first three reactions must occur twice, to form ordinary helium nucleus. If we consider the combination
$2\left(\mathrm{i}\right)+2\text{ (ii) }+2\text{ (iii) }+\text{ (iv), the net effect is }$
$\begin{array}{l}{4}_1^1\mathrm{H}+2{\mathrm{e}}^{-}{\to }_2^4\mathrm{He}+2\mathrm{\vartheta }+2\mathrm{\gamma }+26.7\mathrm{MeV}\\ \left(4_1^1\mathrm{H}+4{\mathrm{e}}^{-}\right)\to \left({ }_2^4\mathrm{He}+2{\mathrm{e}}^{-}\right)+2\mathrm{\vartheta }+6\mathrm{\gamma }+26.7\mathrm{MeV}\end{array}$
Thus, four hydrogen atoms combine to form an${ }_2^4\mathrm{He}$ atom with a release of 26.7 MeV of energy.