Chapter 31, Problem 36P

(a)

To determine

To find:The ratio of energy of carbon cycle and deuterium-tritium reaction

Explanation of Solution

The radius of atom that increases with atomic number $\mathrm{A}$

  $\mathrm{r}\approx \left(1.2\times {10}^{-15}\text{m}\right)\left({\mathrm{A}}^{\frac{1}{3}}\right)$

The carbon cycle reaction is ${}_6^{12}\text{C}{+}_1^2\text{H}{\to }_7^{13}\text{N}+\mathrm{\gamma }.$

We assume the nuclei approach head-on, each with kinetic energy,and that the nuclear force comes into play when the distance between their centers equals the sum of their nuclear radii.

The electrostatic potential energyof the two particles at this distance equals the minimum totalkinetic energy of the two particles when far apart.

  $\begin{array}{c}{\text{KE}}_{\text{C}}=\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_0}\frac{{\mathrm{Q}}_{\text{C}}{\mathrm{Q}}_{\text{H}}}{\left({\mathrm{r}}_{\text{C}}+{\mathrm{r}}_{\text{H}}\right)}\\ =\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_0}\frac{{\mathrm{Q}}_{\text{C}}{\mathrm{Q}}_{\text{H}}}{\left(1.2\times {10}^{-15}\text{m}\right)\left({\mathrm{A}}_{\mathrm{C}}^{1/3}+{\mathrm{A}}_{\mathrm{H}}^{1/3}\right)}\\ =\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{(6)(1){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{\left(1.2\times {10}^{-15}\text{m}\right)\left(1^{1/3}+{12}^{1/3}\right)}\left(\frac{1\text{MeV}}{1.60\times {10}^{-13}\text{J}}\right)\\ =2.19\text{MeV}\end{array}$

Similarly, for deuterium-tritium reaction

  $\begin{array}{c}{\text{KE}}_{\text{d-t}}\approx \frac{1}{4\mathrm{\pi }{\mathrm{ϵ}}_0}\frac{{\mathrm{Q}}_{\mathrm{d}}{\mathrm{Q}}_{\mathrm{t}}}{\left({\mathrm{r}}_{\text{d}}+{\mathrm{r}}_{\text{t}}\right)}\\ =\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_0}\frac{{\mathrm{e}}^2}{\left(1.2\times {10}^{-15}\text{m}\right)\left({\mathrm{A}}_{\mathrm{d}}^{1/3}+{\mathrm{A}}_{\mathrm{t}}^{1/3}\right)}\\ \approx \left(9.0\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2}{\left(1.2\times {10}^{-15}\text{m}\right)\left(2^{1/3}+3^{1/3}\right)}\left(\frac{1\text{MeV}}{1.60\times {10}^{-13}\text{J}}\right)\\ \approx 0.45\text{MeV}\end{array}$

The ratio of energy of carbon cycle and deuterium-tritium reaction

  $\frac{{\text{KE}}_{\text{C}}}{{\text{KE}}_{\text{d}-\text{t}}}=\frac{2.19\text{MeV}}{0.45\text{MeV}}=4.9$

Conclusion:

The ratio of energy of carbon cycle and deuterium-tritium reaction is 5

(b)

To determine

To find:Temperature needed for the first carbon-cycle reaction.

Answer to Problem 36P

  $\mathrm{T}\approx 1.5\times {10}^9\text{K}$

Explanation of Solution

If a deuterium-tritium reaction actually requires a temperature ${\mathrm{T}}_{\text{deuterium-tritium}}\approx 3\times {10}^8\text{K}$

The ratio of energy of carbon cycle and deuterium-tritium reaction is 5

The kinetic energy is proportional to the temperature by $\overline{\text{KE}}=\frac{3}{2}\mathrm{k}\mathrm{T}.$

Since, the kinetic energy has to increase by a factor of 5, so does the temperature.

Temperature needed for the first carbon-cycle reaction,

  $\begin{array}{c}{\mathrm{T}}_{\text{carbon-cycle}}=5\times {\mathrm{T}}_{\text{deuterium-tritium}}\\ =5\left(3\times {10}^8\text{K}\right)\\ =1.5\times {10}^9\text{K}\end{array}$

Therefore, temperature needed for the first carbon-cycle reaction, $1.5\times {10}^9\text{K}$