Chapter 30, Problem 41P

(a)

To determine

The temperature required for the carbon cycle.

Answer to Problem 41P

The temperature required for the carbon cycle is $\underset{\_}{5\times {10}^7\text{K}}$.

Explanation of Solution

Write the expression for the kinetic energy to overcome the columbic repulsion barrier in the proton-proton cycle reaction

    $K=\frac{k_e\left(e\right)\left(2e\right)}{r}$        (I)

Here, $K$ is the kinetic energy, $k_e$ is the columbic constant, $e$ is the electronic charge, $r$ is the separation between the charges.

Write the expression for the columbic barrier to Bethe’s fifth and eight reaction.

    $K_{\text{B}}=\frac{k_e\left(e\right)\left(7e\right)}{r}$        (I)

Here, $K_{\text{B}}$ is the columbic energy barrier.

Write the expression for the temperature required for the carbon cycle reaction using equation (I) and (II).

    $T_{\text{C}}=\frac{7}{2}\left(T_{\text{p}-\text{p}}\right)$        (III)

Here, $T_{\text{C}}$ is the temperature of carbon cycle, $T_{\text{p}-\text{p}}$ is the temperature of proton-proton cycle.

Conclusion:

Substitute $1.5\times {10}^7\text{K}$ for $T_{\text{p}-\text{p}}$ in equation (III) to find $T_{\text{C}}$.

    $\begin{array}{c}{T}_{\text{C}}=\frac{7}{2}\left(1.5\times {10}^7\text{K}\right)\\ =5.25\times {10}^7\text{K}\\ \approx \text{5}\times {\text{10}}^7\text{K}\end{array}$

Therefore, the temperature required for the carbon cycle is $\underset{\_}{5\times {10}^7\text{K}}$.

(b)

To determine

The energy released in each reaction in the carbon cycle and the overall energy released.

Answer to Problem 41P

The energy released $\underset{\_}{Q_1=1.94\text{MeV}}$, $\underset{\_}{Q_2=1.20\text{MeV}}$, $\underset{\_}{Q_3=1.02\text{MeV}}$, $\underset{\_}{Q_4=7.55\text{MeV}}$, $\underset{\_}{Q_5=7.30\text{MeV}}$, $\underset{\_}{Q_6=1.73\text{MeV}}$, $\underset{\_}{Q_7=1.02\text{MeV}}$, $\underset{\_}{Q_8=4.97\text{MeV}}$ and the overall energy $\underset{\_}{Q=26.7\text{MeV}}$.

Explanation of Solution

Let the general reaction in the carbon cycle is as follows.

    $\text{a}+\text{X}\to \text{Y}+\text{b}$        (IV)

Here, $\text{a}$ is the bombarding particle, $\text{X}$ is the target nucleus, $\text{b}$ is the outgoing particle, $\text{Y}$ is the product nucleus.

Write the expression for the energy released in the reaction given in equation (I).

    $Q=\left[M\left(\text{a}\right)+M\left(\text{X}\right)-M\left(\text{Y}\right)-M\left(\text{b}\right)\right]\left(931.5\text{MeV}/\text{u}\right)$        (V)

Here, $M\left(\text{a}\right)$ is the bombarding particle, $M\left(\text{X}\right)$ is the mass of target nucleus, $M\left(\text{Y}\right)$ is the mass of the product nucleus, $M\left(\text{b}\right)$ is the mass of the outgoing particle.

Write the expression for the total energy released in the carbon-cycle reaction.

    $\begin{array}{c}Q=Q_1+Q_2+Q_3+Q_4+Q_5+Q_6+Q_7+Q_8\\ ={\displaystyle \sum _{i=1}^8{Q}_i}\end{array}$        (VI)

Here, $Q$ is the total energy released in the carbon cycle reaction, $Q_i$ is the energy released in each reaction.

Conclusion:

Substitute $12.000000\text{u}$ for $M\left({}^{12}\text{C}\right)$, $1.007825\text{u}$ for $M\left({}^1\text{H}\right)$, $13.005739\text{u}$ for $M\left({}^{13}\text{N}\right)$, $0\text{u}$ for $M\left(\gamma \right)$ in equation (V) to solve for $Q_1$.

    $\begin{array}{c}{Q}_1=\left[12.000000\text{u}+1.007825\text{u}-13.005739\text{u}\right]\left(931.5\text{MeV}/\text{u}\right)\\ =1.94\text{MeV}\end{array}$

Substitute $13.005739\text{u}$ for $M\left({}^{13}\text{N}\right)$, $13.003355\text{u}$ for $M\left({}^{13}\text{C}\right)$, $0.000549\text{u}$ for $M\left(e^{-}\right)$ in equation (V) to find $Q_2$.

    $\begin{array}{c}{Q}_2=\left[13.005739\text{u}-13.003355\text{u}-2\left(0.000549\text{u}\right)\right]\left(931.5\text{MeV}/\text{u}\right)\\ =1.20\text{MeV}\end{array}$

Substitute $0.000549\text{u}$ for $M\left(e^{+}\right)$, $0.000549\text{u}$ for $M\left(e^{-}\right)$, $0\text{u}$ for $M\left(\gamma \right)$ in equation (V) to find $Q_3$.

    $\begin{array}{c}{Q}_3=2\left(0.000549\text{u}\right)\left(931.5\text{MeV}/\text{u}\right)\\ =1.02\text{MeV}\end{array}$

Substitute $13.003355\text{u}$ for $M\left({}^{13}\text{C}\right)$, $1.007825\text{u}$ for $M\left({}^1\text{H}\right)$, $14.003074\text{u}$ for $M\left({}^{14}\text{N}\right)$, $0\text{u}$ for $M\left(\gamma \right)$ in equation (V) to find $Q_4$.

    $\begin{array}{c}{Q}_4=\left[13.003355\text{u}+1.007825\text{u}-14.003074\text{u}\right]\left(931.5\text{MeV}/\text{u}\right)\\ =7.55\text{MeV}\end{array}$

Substitute $15.003065\text{u}$ for $M\left({}^{15}\text{O}\right)$, $14.003074\text{u}$ for $M\left({}^{14}\text{N}\right)$, $1.007825\text{u}$ for $M\left({}^1\text{H}\right)$, $0\text{u}$ for $M\left(\gamma \right)$ in equation (V) to find $Q_5$.

    $\begin{array}{c}{Q}_5=\left[14.003074\text{u}+1.007825\text{u}-15.003065\text{u}\right]\left(931.5\text{MeV}/\text{u}\right)\\ =7.30\text{MeV}\end{array}$

Substitute $15.003065\text{u}$ for $M\left({}^{15}\text{O}\right)$, $0.000549\text{u}$ for $M\left(e^{+}\right)$, $15.000109\text{u}$ for $M\left({}^{15}\text{N}\right)$ in equation (V) to find $Q_6$.

    $\begin{array}{c}{Q}_6=\left[15.003065\text{u}-15.000109\text{u}-2\left(0.000549\text{u}\right)\right]\left(931.5\text{MeV}/\text{u}\right)\\ =1.73\text{MeV}\end{array}$

Substitute $0.000549\text{u}$ for $M\left(e^{+}\right)$, $0.000549\text{u}$ for $M\left(e^{-}\right)$, $0\text{u}$ for $M\left(\gamma \right)$ in equation (V) to find $Q_7$.

    $\begin{array}{c}{Q}_7=2\left(0.000549\text{u}\right)\left(931.5\text{MeV}/\text{u}\right)\\ =1.02\text{MeV}\end{array}$

Substitute $15.000109\text{u}$ for $M\left({}^{15}\text{N}\right)$, $1.007825\text{u}$ for $M\left({}^1\text{H}\right)$, $12.000000\text{u}$ for $M\left({}^{12}\text{C}\right)$, $4.002603\text{u}$ for $M\left({}^4\text{He}\right)$ in equation (V) to find $Q_8$.

    $\begin{array}{c}{Q}_8=\left[15.000109\text{u}+1.007825\text{u}-12.000000\text{u}-4.002603\text{u}\right]\left(931.5\text{MeV}/\text{u}\right)\\ =4.97\text{MeV}\end{array}$

Substitute $1.94\text{MeV}$ for $Q_1$, $1.20\text{MeV}$ for $Q_2$, $1.02\text{MeV}$ for $Q_3$, $7.55\text{MeV}$ for $Q_4$, $7.30\text{MeV}$ for $Q_5$, $1.73\text{MeV}$ for $Q_6$, $1.20\text{MeV}$ for $Q_7$, $4.97\text{MeV}$ for $Q_8$ in equation (VI) to find $Q$.

    $\begin{array}{c}Q=1.94\text{MeV}+1.20\text{MeV}+1.20\text{MeV}+7.55\text{MeV}+7.30\text{MeV}+1.73\text{MeV}+1.20\text{MeV}+4.97\text{MeV}\\ =26.7\text{MeV}\end{array}$

Therefore, the energy released $\underset{\_}{Q_1=1.94\text{MeV}}$, $\underset{\_}{Q_2=1.20\text{MeV}}$, $\underset{\_}{Q_3=1.02\text{MeV}}$, $\underset{\_}{Q_4=7.55\text{MeV}}$, $\underset{\_}{Q_5=7.30\text{MeV}}$, $\underset{\_}{Q_6=1.73\text{MeV}}$, $\underset{\_}{Q_7=1.02\text{MeV}}$, $\underset{\_}{Q_8=4.97\text{MeV}}$ and the overall energy $\underset{\_}{Q=26.7\text{MeV}}$.

(c)

To determine

Whether the energy carried off by the neutrinos is deposited in the star.

Answer to Problem 41P

Not all the energy appeared as the internal energy in the star. When the neutrino is created, it is likely to fly apart without interacting with any other particles.

Explanation of Solution

The amount of energy released in the reaction is not appeared as the internal energy in the star. When a neutrino is created, it is more likely to fly directly out of the star without any interaction with other particles.

Conclusion:

Therefore, all the energy is not appeared as the internal energy in the star. When the neutrino is created, it is likely to fly apart without interacting with any other particles.