Chapter 31, Problem 62GP

(a)

To determine

Energy released in the reaction.

Answer to Problem 62GP

The energy released is given as $\mathrm{Q}=13.93\text{ MeV}$

Explanation of Solution

Given:

Fusion of two $\text{C}$ nuclei into one $\text{M}\text{g}$ nucleus.

Calculation:

Energy released in the reaction can be calculated as follows:

  $\begin{array}{l}\mathrm{Q}=2{\mathrm{m}}_{\mathrm{C}}-{\mathrm{m}}_{\mathrm{M}\mathrm{g}}\\ \mathrm{Q}=\left(2\left(12\right)-23.985042\right)\left(931.5\right)\text{MeV}\\ \mathrm{Q}=13.93\text{ MeV}\end{array}$

Conclusion:

Thus, the energy released is given as $\mathrm{Q}=13.93\text{ MeV}$

(b)

To determine

Kinetic energy of the Carbon nucleus.

Answer to Problem 62GP

Kinetic energy of the Carbon nucleus is $\mathrm{E}=6.9\times {10}^{-13}\text{J}$

Explanation of Solution

Given:

Fusion of two $\text{C}$ nuclei into one $\text{M}\text{g}$ nucleus.

Distance, $\mathrm{r}=6.0\text{ fm}=6\times {10}^{-15}\text{m}$

Calculation:

The kinetic energy in such case can be calculated as follows:

  $\begin{array}{l}\mathrm{E}=\frac{1}{2}\left(\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_0}\frac{{\mathrm{e}}^2}{\mathrm{r}}\right)\\ \mathrm{E}=\frac{1}{2}\left(9\times {10}^9\text{N}{\text{.m}}^2{\text{/C}}^2\right)\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2}{\left(6\times {10}^{-15}\text{m}\right)}\\ \mathrm{E}=6.9\times {10}^{-13}\text{J}\end{array}$

Conclusion:

Thus, the kinetic energy of the Carbon nucleus is $\mathrm{E}=6.9\times {10}^{-13}\text{J}$

(c)

To determine

Approximate temperature required.

Answer to Problem 62GP

The approximate temperature is $\mathrm{T}=3.3\times {10}^{10}\text{K}$

Explanation of Solution

Given:

Fusion of two $\text{C}$ nuclei into one $\text{M}\text{g}$ nucleus.

Distance, $\mathrm{r}=6.0\text{ fm}=6\times {10}^{-15}\text{m}$

Calculation:

As we know that kinetic energy is given by:

  $\mathrm{E}=\frac{3}{2}\mathrm{k}\mathrm{T}$

So, rearranging the above expression, we get:

  $\begin{array}{l}\mathrm{T}=\frac{2\mathrm{E}}{3\mathrm{k}}\\ \mathrm{T}=\frac{2\left(6.9\times {10}^{-13}\text{J}\right)}{3\left(1.38\times {10}^{-23}\text{J/K}\right)}\\ \mathrm{T}=3.3\times {10}^{10}\text{K}\end{array}$

Conclusion:

Hence, the approximate temperature is $\mathrm{T}=3.3\times {10}^{10}\text{K}$