Question

(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?

Short answer

(a) The total energy released in each carbon cycle is -

\(\begin{array}{*{20}{l}}{^{12}C{ + ^1}H}&{{ \to ^{13}}N + \gamma ,{\rm{ }}E = 1.9431{\rm{ }}MeV}\\{^{13}N}&{{ \to ^{13}}C + {e^ + } + {v_e},{\rm{ }}E = 1.7093{\rm{ }}MeV}\\{^{13}C{ + ^1}H}&{{ \to ^{14}}N + \gamma ,{\rm{ }}E = 7.5507{\rm{ }}MeV}\\{^{14}N{ + ^1}H}&{{ \to ^{15}}O + \gamma ,{\rm{ }}E = 7.2964{\rm{ }}MeV}\\{^{15}O}&{{ \to ^{15}}N + {e^ + } + {v_e},{\rm{ }}E = 2.2431{\rm{ }}MeV}\\{^{15}N{ + ^1}H}&{{ \to ^{12}}C{ + ^4}He.{\rm{ }}E = 4.9658{\rm{ }}MeV}\end{array}\)

(b) The proton-proton cycle releases more energy than the carbon cycle.

Step-by-step solution

Concept Introduction

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.

Information Provided

• 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: \(2{e^ - } + {4^1}H{ \to ^4}He + 2{v_e} + 6\gamma \).

Energy calculation for the first reaction

(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 \(E = 1.9431{\rm{ }}MeV\).

Energy calculation for the second reaction

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 \(E = 1.7093{\rm{ }}MeV\).

Energy calculation for the third reaction

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 \(E = 7.5507{\rm{ }}MeV\).

Energy calculation for the fourth reaction

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 \(E = 7.2964{\rm{ }}MeV\).

Energy calculation for the fifth reaction

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 \(E = 2.2431{\rm{ }}MeV\).

Energy calculation for the sixth reaction

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 \(E = 4.9658{\rm{ }}MeV\).

Comparison with p-p cycle

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.