Chapter 31, Problem 35P

(a)

To determine

To show:In the cycle, no carbon is consumed

Explanation of Solution

Carbon cycle of massive star

All the reactants and all the products are

  $\begin{array}{c}{}_6^{12}\text{C}{+}_1^1\text{H}{\to }_7^{13}\text{N}+\mathrm{\gamma }\\ {}_7^{13}\text{N}{\to }_6^{13}\text{C}+{\text{e}}^{+}+\mathrm{v}\\ {}_6^{13}\text{C}{+}_1^1\text{H}{\to }_7^{14}\text{N}+\mathrm{\gamma }\\ {}_7^{14}\text{N}{+}_1^1\text{H}{\to }_8^{15}\text{O}+\mathrm{\gamma }\\ {}_8^{15}\text{O}{\to }_7^{15}\text{N}+{\text{e}}^{+}+\mathrm{v}\\ {}_7^{15}\text{N}{+}_1^1\text{H}{\to }_6^{12}\text{C}{+}_2^4\text{He}\end{array}$

Adding all the reactants and all the products

  $\begin{array}{l}{}_6^{12}\text{C}{+}_1^1\text{H}{+}_7^{13}\text{N}{+}_6^{13}\text{C}{+}_1^1\text{H}{+}_7^{14}\text{N}{+}_1^1\text{H}{+}_8^{15}\text{O}{+}_7^{15}\text{N}{+}_1^1\text{H}\\ \to {}_7^{13}\text{N}+\mathrm{\gamma }{+}_6^{13}\text{C}+{\text{e}}^{+}+\mathrm{v}{+}_7^{14}\text{N}+\mathrm{\gamma }{+}_8^{15}\text{O}+\mathrm{\gamma }{+}_7^{15}\text{N}+{\text{e}}^{+}+\mathrm{v}{+}_6^{12}\text{C}{+}_2^4\text{He}\end{array}$

Cancelling those that appear on both sides of the reaction.

  $4_1^1\text{H}{\to }_2^4\text{He}+2{\text{e}}^{+}+2\mathrm{v}+3\mathrm{\gamma }$

No carbon is consumed in this cycle because one ${}_6^{12}\text{C}$ nucleus is required in the first step of the cycle, and one ${}_6^{12}\text{C}$ nucleus is produced in the last step of the cycle.

(b)

To determine

To find:The total energy released

Answer to Problem 35P

  $26.73\text{MeV}$

Explanation of Solution

Reaction in star

  $4_1^1\text{H}{\to }_2^4\text{He}+2{\text{e}}^{+}+2\mathrm{v}+3\mathrm{\gamma }$

The number of electrons is balanced well as the number of protons and neutrons. The above "net" equation does not consider the electrons that neutral nuclei would have, because it does not conserve charge.

What the above reaction really represents (ignoring the gammas and neutrinos) is the following:

  $4_1^1\text{H}{\to }_2^4\text{He}+2{\text{e}}^{+}\Rightarrow 4_1^1\text{p}\to 2_1^1\text{p}+2_0^1\text{n}+2{\text{e}}^{+}$

Each positron-electron annihilation produces another $1.02\text{MeV},$

To use the values from Appendix B, we must add 4 electrons to each side of the reaction.

  $\begin{array}{l}\left(4_1^1\mathrm{p}+4{\mathrm{e}}^{-}\right)\to \left(2_1^1\mathrm{p}+2{\mathrm{e}}^{-}+2_0^1\mathrm{n}\right)+2{\mathrm{e}}^{+}+2{\mathrm{e}}^{-}\\ \Rightarrow 4_1^1\mathrm{H}{\to }_2^4\mathrm{H}\mathrm{e}+2{\mathrm{e}}^{+}+2{\mathrm{e}}^{-}\end{array}$

The energy produced in the reaction is the $\mathrm{Q}$ -value.

  $\begin{array}{l}\mathrm{Q}=4{\mathrm{m}}_{\text{H}}{\mathrm{c}}^2-{\mathrm{m}}_{\text{H}\text{e}}{\mathrm{c}}^2-4{\mathrm{m}}_{\text{e}}\\ =[4(1.007825\text{u})-4.002603\text{u}-4(0.000549\text{u})]\left(931.5\frac{\text{MeV}/{\mathrm{c}}^2}{\text{u}}\right){\mathrm{c}}^2=24.69\text{MeV}\end{array}$

The total energy released is $24.69\text{MeV}+2(1.02\text{MeV})=26.73\text{MeV}$

(c)

To determine

To find:Energy output for each reaction

Explanation of Solution

The first equation: ${}_6^{12}\text{C}{+}_1^1\text{H}{\to }_7^{13}\text{N}+\mathrm{\gamma }$

The first equation in the carbon cycle is electron-balanced,

The Q-value

  $\begin{array}{l}\mathrm{Q}={\mathrm{m}}_{}{{}_{\text{C}}}^{\mathrm{c}}+{\mathrm{m}}_{\text{H}}{\mathrm{c}}^2-{\mathrm{m}}_{\text{N}}{\mathrm{c}}^2\\ =[12.000000\text{u}+1.007825\text{u}-13.005739\text{u}]\left(931.5\frac{\text{MeV}/{\mathrm{c}}^2}{\text{u}}\right){\mathrm{c}}^2=1.943\text{MeV}\end{array}$

The second equation: ${}_7^{13}\text{N}{\to }_6^{13}\text{C}+{\text{e}}^{+}+\mathrm{v}$

The second equation needs to have another electron, so that ${}_7^{13}\text{N}{\to }_6^{13}\text{C}+{\text{e}}^{-}+{\text{e}}^{+}+\mathrm{v}$

  $\begin{array}{l}\mathrm{Q}={\mathrm{m}}_{{}_7^{13}\text{N}}{\mathrm{c}}^2-{\mathrm{m}}_{\mathrm{c}}{}^2-2{\mathrm{m}}_{\text{e}}{\mathrm{c}}^2\\ =[13.005739\text{u}-13.003355\text{u}-2(0.000549\text{u})]\left(931.5\frac{\text{MeV}/{\mathrm{c}}^2}{\text{u}}\right){\mathrm{c}}^2\\ =1.198\text{MeV}\end{array}$

We must include an electron-positron annihilation in this reaction. $1.198\text{MeV}+1.02\text{MeV}=2.218\text{MeV}$

The third equation: ${}_6^{13}\text{C}{+}_1^1\text{H}{\to }_7^{14}\text{N}+\mathrm{\gamma }$

The third equation of the carbon cycle is electron-balanced. $\begin{array}{l}\mathrm{Q}={\mathrm{m}}_{\mathrm{c}}{\mathrm{c}}^2+{\mathrm{m}}_{\text{H}}{\mathrm{c}}^2-{\mathrm{m}}_{\text{N}}{\text{c}}^2\\ =[13.003355\text{u}+1.007825\text{u}-14.003074\text{u}]\left(931.5\frac{\text{MeV}/{\mathrm{c}}^2}{\text{u}}\right){\mathrm{c}}^2\\ =7.551\text{MeV}\end{array}$

The fourth equation: ${}_7^{14}\text{N}{+}_1^1\text{H}{\to }_8^{15}\text{O}+\mathrm{\gamma }$

The fourth equation of the carbon cycle is also electron-balanced.

  $\begin{array}{l}\mathrm{Q}={\mathrm{m}}_{\text{N}}{\text{c}}^2+{\mathrm{m}}_{\text{H}}{\mathrm{c}}^2-{\mathrm{m}}_{\text{O}}{\mathrm{c}}^2\\ =[14.003074\text{u}+1.007825\text{u}-15.003066\text{u}]\left(931.5\frac{\text{MeV}/{\text{c}}^2}{\text{u}}\right){\mathrm{c}}^2\\ =7.296\text{MeV}\end{array}$

The fifth equation: ${}_8^{15}\text{O}{\to }_7^{15}\text{N}+{\text{e}}^{+}+\mathrm{v}$

The fifth equation needs to have another electron, so ${}_8^{15}\text{O}{\to }_7^{15}\text{N}+{\text{e}}^{-}+{\text{e}}^{+}+\mathrm{v}$

  $\begin{array}{l}\mathrm{Q}={\mathrm{m}}_{15}{}_8{\text{O}}^2-{\mathrm{m}}_{15}{\text{N}}^2-2{\mathrm{m}}_{\text{e}}{\mathrm{c}}^2\\ =[15.003066\text{u}-15.000109\text{u}-2(0.000549\text{u})]\left(931.5\frac{\text{MeV}/{\mathrm{c}}^2}{\text{u}}\right){\mathrm{c}}^2\\ =1.732\text{MeV}\end{array}$

We must include an electron-positron annihilation in this reaction.

  $1.732\text{MeV}+1.02\text{MeV}=2.752\text{MeV}$ .

The sixth equation: ${}_7^{15}\text{N}{+}_1^1\text{H}{\to }_6^{12}\text{C}{+}_2^4\text{He}$

The sixth equation is electron-balanced.

  $\begin{array}{l}\mathrm{Q}={\mathrm{m}}_{\text{N}}{\mathrm{c}}^2+{\mathrm{m}}_{\text{H}}{\mathrm{c}}^2-{\mathrm{m}}_{\text{C}}{\mathrm{c}}^2-{\mathrm{m}}_{\text{H}\text{e}}{\mathrm{c}}^2\\ =[15.000109\text{u}+1.007825\text{u}-12.000000\text{u}-4.002603\text{u}]\left(931.5\frac{\text{MeV}/{\text{c}}^2}{\text{u}}\right){\mathrm{c}}^2\\ =4.966\text{MeV}\end{array}$

The total energy released is found by summing the energy released in each process.

  $\begin{array}{l}1.943\text{MeV}+2.218\text{MeV}+7.551\text{MeV}+7.296\text{MeV}+2.752\text{MeV}+4.966\text{MeV}\hfill \\ =26.73\text{MeV}\hfill \end{array}.$

(d)

To determine

To find:The reason carbon cycle require higher temperature than proton-proton cycle

Explanation of Solution

The carbon and nitrogen nuclei have higher Z values, leading to a greater Coulomb repulsion.To overcome the Coulomb repulsion between the nuclei, the carbon cycle need higher temperature so that the reactants have more initial kinetic energy.

Since, carbon has higher Coulomb repulsion than proton. So the temperature should be higher for carbon cycle reaction.