(a) Find the total energy released in \(MeV\) in each carbon cycle (elaborated in the above problem) including the annihilation energy.

(b) How does this compare with the proton-proton cycle output?

The proton-proton chain, also referred to as the\({\rm{p - p}}\)chain, is one of two known sets of nuclear fusion events that stars use to convert hydrogen to helium. The second known reaction, the\({\rm{CNO}}\)cycle, is dominant in stars with masses greater than or equal to\({\rm{1}}{\rm{.3}}\)times that of the Sun, according to theoretical models, but it is dominating in stars with masses less than or equal to that of the Sun.

• Reactions from the carbon cycle are:\(\begin{array}{*{20}{l}}{^{12}C{ + ^1}H}&{{ \to ^{13}}N + \gamma ,}\\{^{13}N}&{{ \to ^{13}}C + {e^ + } + {v_e},}\\{^{13}C{ + ^1}H}&{{ \to ^{14}}N + \gamma ,}\\{^{14}N{ + ^1}H}&{{ \to ^{15}}O + \gamma ,}\\{^{15}O}&{{ \to ^{15}}N + {e^ + } + {v_e},}\\{^{15}N{ + ^1}H}&{{ \to ^{12}}C{ + ^4}He.}\end{array}\).
• The proton-proton cycle reaction is: \(2{e^ - } + {4^1}H{ \to ^4}He + 2{v_e} + 6\gamma \).

(a)

The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. Energydifference can be calculated by the free mass-energy formula –

\(E = \Delta m{c^2}\)

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all otherby-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to \(eV\) –

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now analyse the reaction –

\(^{12}C{ + ^1}H{ \to ^{13}}N + \gamma \)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}^1H = 1.007825{\rm{ }}u\\^{12}C = 12.000{\rm{ }}u{\rm{ }}\\^{13}N = 13.005739{\rm{ }}u\\{\nu _e} = 0\end{array}\)

Subtracting the masses of parent and daughter nuclei, it is obtained –

\(\Delta m = \left( {{m_{{C^{14}}}} + {m_{{H^1}}} - {m_{{N^{13}}}} - {m_\gamma }} \right)\)

Plugging in the values –

\(\begin{array}{c}\Delta m = \left( {1.007825{\rm{ }}u + 12.000000{\rm{ }}u - 13.005739{\rm{ }}u - 0} \right)\\\Delta m = 0.002086{\rm{ }}u\end{array}\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.002086 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E = 1.9431{\rm{ }}MeV\end{array}\)

Therefore, the energy value for the reaction is obtained as \(E = 1.9431{\rm{ }}MeV\).

The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. Energydifference can be calculated by the free mass-energy formula –

\(E = \Delta m{c^2}\)

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all otherby-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to \(eV\) –

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now analyse the reaction –

\(^{13}N{ \to ^{13}}C + {e^ + } + {v_e}\)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}^1H = 1.007825{\rm{ }}u\\^{13}C = 13.003355{\rm{ }}u{\rm{ }}\\{e^ + } = 0.000549{\rm{ }}u\\{\nu _e} = 0\end{array}\)

Subtracting the masses of parent and daughter nuclei, it is obtained –

\(\Delta m = \left( {{m_{{N^{13}}}} - {m_{{e^{13}}}} - {m_{{e^ + }}} - {m_{{v_e}}}} \right)\)

Plugging in the values –

\(\begin{array}{c}\Delta m = \left( {13.005739{\rm{ }}u - 13.003355{\rm{ }}u - 0.000549{\rm{ }}u - 0} \right)\\\Delta m = 0.001835{\rm{ }}u\end{array}\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.001835 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E = 1.7093{\rm{ }}MeV\end{array}\)

Therefore, the energy value for the reaction is obtained as \(E = 1.7093{\rm{ }}MeV\).

The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. Energydifference can be calculated by the free mass-energy formula –

\(E = \Delta m{c^2}\)

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all otherby-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to \(eV\) –

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now analyse the reaction –

\(^{13}C{ + ^1}H{ \to ^{14}}N + \gamma \)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}^1H = 1.007825{\rm{ }}u\\^{13}C = 13.003355{\rm{ }}u{\rm{ }}\\^{14}N = 14.003074{\rm{ }}u\end{array}\)

Subtracting the masses of parent and daughter nuclei, it is obtained –

\(\Delta m = \left( {{m_{{C^{13}}}} + {m_{{H^1}}} - {m_{{N^{14}}}}} \right)\)

Plugging in the values –

\(\begin{array}{c}\Delta m = \left( {13.003355{\rm{ }}u + 1.007825{\rm{ }}u - 14.003074{\rm{ }}u} \right)\\\Delta m = 0.008106{\rm{ }}u\end{array}\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.008106 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E = 7.5507{\rm{ }}MeV\end{array}\)

Therefore, the energy value for the reaction is obtained as \(E = 7.5507{\rm{ }}MeV\).

The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. Energydifference can be calculated by the free mass-energy formula –

\(E = \Delta m{c^2}\)

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all otherby-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to \(eV\) –

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now analyse the reaction –

\(^{14}N{ + ^1}H{ \to ^{15}}O + \gamma \)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}^1H = 1.007825{\rm{ }}u\\^{15}O = 13.003066{\rm{ }}u{\rm{ }}\\^{14}N = 14.003074{\rm{ }}u\end{array}\)

Subtracting the masses of parent and daughter nuclei, it is obtained –

\(\Delta m = \left( { - {m_{{O^{15}}}} + {m_{{H^1}}} + {m_{{N^{14}}}}} \right)\)

Plugging in the values –

\(\begin{array}{c}\Delta m = \left( { - 15.003066{\rm{ }}u + 1.007825{\rm{ }}u + 14.003074{\rm{ }}u} \right)\\\Delta m = 0.007833{\rm{ }}u\end{array}\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.007833 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E = 7.2964{\rm{ }}MeV\end{array}\)

Therefore, the energy value for the reaction is obtained as \(E = 7.2964{\rm{ }}MeV\).

The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. Energydifference can be calculated by the free mass-energy formula –

\(E = \Delta m{c^2}\)

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all otherby-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to \(eV\) –

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now analyse the reaction –

\(^{15}O{ \to ^{15}}N + {e^ + } + {v_e}\)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}{e^ + } = 0.000549{\rm{ }}u\\^{15}O = 13.003066{\rm{ }}u{\rm{ }}\\^{15}N = 15.000109{\rm{ }}u\end{array}\)

Subtracting the masses of parent and daughter nuclei, it is obtained –

\(\Delta m = \left( {{m_{{O^{15}}}} - {m_{{e^ + }}} - {m_{{N^{15}}}}} \right)\)

Plugging in the values –

\(\begin{array}{c}\Delta m = \left( {15.003066{\rm{ }}u - 15.000109{\rm{ }}u - 0.000549{\rm{ }}u} \right)\\\Delta m = 0.002408{\rm{ }}u\end{array}\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.002408 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E = 2.2431{\rm{ }}MeV\end{array}\)

Therefore, the energy value for the reaction is obtained as \(E = 2.2431{\rm{ }}MeV\).

The energy release is happening due to the different total mass of nuclei before and after the nuclear reaction. Energydifference can be calculated by the free mass-energy formula –

\(E = \Delta m{c^2}\)

To get these mass differences take parent nuclei and subtract from them daughter nuclei and masses of all otherby-products of the nuclear reaction. Since all atomic masses will be expressed with the unified atomic unit, remember how to convert it to \(eV\) –

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now analyse the reaction –

\(^{15}N{ + ^1}H{ \to ^{12}}C{ + ^4}He\)

Looking in the Appendix for the atomic masses the following is obtained–

\(\begin{array}{c}^1H = 1.007825{\rm{ }}u\\^4He = 4.002603{\rm{ }}u\\^{12}C = 12.000000{\rm{ }}u{\rm{ }}\\^{15}N = 15.000109{\rm{ }}u\end{array}\)

Subtracting the masses of parent and daughter nuclei, it is obtained –

\(\Delta m = \left( {{m_{{H^1}}} - {m_{{C^{12}}}} + {m_{{N^{15}}}} - {m_{H{e^4}}}} \right)\)

Plugging in the values –

\(\begin{array}{c}\Delta m = \left( {1.007825{\rm{ }}u + 15.000109{\rm{ }}u - 4.002603{\rm{ }}u - 12.000000{\rm{ }}u} \right)\\\Delta m = 0.005331{\rm{ }}u\end{array}\)

So, from the energy relation it is obtained that –

\(\begin{array}{c}E = \Delta m{c^2}\\ = 0.005331 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E = 4.9658{\rm{ }}MeV\end{array}\)

Therefore, the energy value for the reaction is obtained as \(E = 4.9658{\rm{ }}MeV\).

Now to calculate the overall energy released in the carbon cycle, just add up the energy from each reaction which is –

\(\begin{array}{c}{E_{tot}} = 1.9431{\rm{ }}MeV + 1.7093{\rm{ }}MeV + 7.5507{\rm{ }}MeV + 7.2964{\rm{ }}MeV + 2.2431{\rm{ }}MeV + 4.9658{\rm{ }}MeV\\{E_{tot}} = 25.7084{\rm{ }}MeV\end{array}\)

Which gives the result of –

\({E_{tot}} = 25.7084{\rm{ }}MeV\)

Therefore, comparing this with energy released in proton-proton cycle, it can be seen that the proton-proton cycle is releasing more energy than the carbon cycle.