Chapter 30, Problem 37P

(a)

To determine

The repulsive energy barrier in the carbon fusion reaction.

Answer to Problem 37P

The repulsive energy barrier in the carbon fusion reaction is $\underset{\_}{8\times {10}^4\text{eV}}$.

Explanation of Solution

Write the expression for the average kinetic energy of the carbon atom.

    $K_{\text{avg}}=\frac{3}{2}{k}_{\text{B}}T$        (I)

Here, $K_{\text{avg}}$ is the average kinetic energy, $k_{\text{B}}$ is the Boltzmann constant, $T$ is the temperature.

Conclusion:

Substitute $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $k_{\text{B}}$, $6\times {10}^8\text{K}$ for $T$ in equation (I) to find $K_{\text{avg}}$.

    $\begin{array}{c}{K}_{\text{avg}}=\frac{3}{2}\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(6\times {10}^8\text{K}\right)\\ =8\times {10}^4\text{eV}\end{array}$

Therefore, the repulsive energy barrier in the carbon fusion reaction is $\underset{\_}{8\times {10}^4\text{eV}}$.

(b)

To determine

The energy released in each of the carbon burning reaction.

Answer to Problem 37P

The energy released in first reaction is $\underset{\_}{4.62\text{MeV}}$ and the second reaction is $\underset{\_}{13.9\text{MeV}}$.

Explanation of Solution

Write the expression for the energy released in the first reaction.

    $Q_1=\left[2M_{{\text{C}}^{12}}-M_{{\text{Ne}}^{20}}-M_{{\text{He}}^4}\right]\left(931.5\text{MeV}/\text{u}\right)$        (II)

Here, $Q_1$ is the energy released, $M_{{\text{C}}^{12}}$ is the mass of ${\text{C}}^{12}$, $M_{{\text{Ne}}^{20}}$ is the mass of ${\text{Ne}}^{20}$, $M_{{\text{He}}^4}$ is the mass of ${\text{He}}_4^{}$.

Write the expression for the energy released in the second reaction.

    $Q_2=\left[2M_{{\text{C}}^{12}}-M_{{\text{Mg}}^{24}}\right]\left(931.5\text{MeV}/\text{u}\right)$        (III)

Here, $Q_2$ is the energy released, $M_{{\text{Mg}}^{24}}$ is the mass of ${\text{Ne}}^{20}$.

Conclusion:

Substitute $12.000000\text{u}$ for $M_{{\text{C}}^{12}}$, $19.992440\text{u}$ for $M_{{\text{Ne}}^{20}}$, $4.002603\text{u}$ for $M_{{\text{He}}^4}$ in equation (II) to find $Q_1$.

    $\begin{array}{c}{Q}_1=\left[2\left(12.000000\text{u}\right)-19.992440\text{u}-4.002603\text{u}\right]\left(931.5\text{MeV}/\text{u}\right)\\ =4.62\text{MeV}\end{array}$

Substitute $12.000000\text{u}$ for $M_{{\text{C}}^{12}}$, $23.985042\text{u}$ for $M_{{\text{Mg}}^{24}}$ in equation (III) to find $Q_2$.

    $\begin{array}{c}{Q}_2=\left[2\left(12.000000\text{u}\right)-23.985042\text{u}\right]\left(931.5\text{MeV}/\text{u}\right)\\ =13.9\text{MeV}\end{array}$

Therefore, the energy released in fist reaction is $\underset{\_}{4.62\text{MeV}}$ and the second reaction is $\underset{\_}{13.9\text{MeV}}$.

(c)

To determine

The energy released when $2.00\text{kg}$ of carbon completely fuse according to the first equation.

Answer to Problem 37P

The energy released when $2.00\text{kg}$ of carbon completely fuse according to the first equation is $\underset{\_}{1.03\times {10}^7\text{kWh}}$.

Explanation of Solution

Two carbon nuclei are used per reaction. The number of events take place in the fusion process is $\frac{N}{2}$.

Write the expression for the energy released in the first reaction.

    $E_1=\frac{N}{2}{Q}_1$        (IV)

Here, $N$ is the number of carbon nuclei take place

Write the expression for $N$.

    $N=\frac{m{N}_{\text{A}}}{M}$        (V)

Here, $N_{\text{A}}$ is the Avogadro number, $M$ is the molar mass, $m$ is the mass of the fuel.

Use equation (II) in (I) to solve for $E_1$.

    $E_1=\frac{1}{2}\left(\frac{m{N}_{\text{A}}}{M}\right)Q_1$        (VI)

Conclusion:

Substitute $2.00\times {10}^3\text{kg}$ for $m$, $6.02\times {10}^{23}\text{atoms}/\text{mol}$ for $N_{\text{A}}$, $12.0\text{g}/\text{mol}$ for $M$, $4.62\text{MeV}$ for $Q_1$ in equation (VI) to find $E_1$.

    $\begin{array}{c}{E}_1=\frac{1}{2}\left[\frac{\left(2.00\times {10}^3\text{kg}\right)\left(6.02\times {10}^{23}\text{atoms}/\text{mol}\right)}{\left(12.0\text{g}/\text{mol}\right)}\right]\left(4.62\text{MeV}\times \frac{1\text{kWh}}{2.25\times {10}^{19}\text{MeV}}\right)\\ =\frac{\left(1.00\times {10}^{26}\right)}{2\left(2.25\times {10}^{19}\right)}\text{kWh}\\ \text{=1}\text{.03}\times {\text{10}}^7\text{kWh}\end{array}$

Therefore, the energy released when $2.00\text{kg}$ of carbon completely fuse according to the first equation is $\underset{\_}{1.03\times {10}^7\text{kWh}}$.