Chapter 33, Problem 47GP

(a)

To determine

Energy released in the reaction.

Answer to Problem 47GP

Energy released in the reaction is $\text{13}\text{.93 MeV}$ .

Explanation of Solution

Given:

Reaction

  ${}_6^{12}{\text{C + }}_6^{12}\text{C}{\to }_{12}^{24}\text{Mg}+\gamma$

Formula used:

Q-value or energy released is calculated as the difference between the mass of the reactant and the product.

Calculation:

The energy released or Q-value is calculated as

  $\begin{array}{l}{\text{Q = 2m}}_{\text{C}}{\text{c}}^2-{\text{m}}_{\text{Mg}}{\text{c}}^2\\ \text{Q = }[2(12.000000\text{ u})-23.985042\text{ u}]{\text{c}}^2(931.5{\text{ MeV/c}}^2)\\ \text{Q = 13}\text{.93 MeV}\end{array}$

Conclusion:

Thus, the energy released in the reaction is $\text{13}\text{.93 MeV}$ .

(b)

To determine

Kinetic energy of the reactants.

Answer to Problem 47GP

Kinetic energy of each carbon nucleus is $4.71\text{ MeV}$ .

Explanation of Solution

Given:

Radius of atomic nuclei is $1.2\times {10}^{-15}\text{ m}$

Formula Used:

Potential energy is given as

  $PE=\frac{1}{4\pi {\epsilon }_0}\frac{q_{nucleus}^2}{2r}$

Calculation:

Kinetic energy is equal to the electrical potential energy of the two nuclei when they come in contact. Thus, the kinetic energy of each nucleus is

  $\begin{array}{c}K{E}_{nucleus}=\frac{1}{2}PE\\ =\frac{1}{2}\frac{1}{4\pi {\epsilon }_0}\frac{q_{nucleus}^2}{2r}\\ K{E}_{nucleus}=\frac{1}{2}\frac{1}{4\pi {\epsilon }_0}\frac{q_{nucleus}^2}{2(1.2\times {10}^{-15}\text{ m})A^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\\ K{E}_{nucleus}=\frac{1}{2}(8.988\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2)\frac{{(6)}^2{(1.60\times {10}^{-19}\text{ C})}^2}{2(1.2\times {10}^{-15}\text{ m}){(12)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\frac{1\text{ eV}}{1.60\times {10}^{-19}\text{ J}}\\ =4.71\text{ MeV}\end{array}$

Conclusion:

Thus, the kinetic energy of each carbon nucleus is $4.71\text{ MeV}$ .

(c)

To determine

The temperature corresponds to kinetic energy.

Answer to Problem 47GP

The temperature that corresponds to kinetic energy is $3.6\times {10}^{10}\text{ K}$ .

Explanation of Solution

Given:

Kinetic energy of the reactants is $4.71\text{ MeV}$ .

Formula used:

Kinetic energy is calculated as

  $KE=\frac{3}{2}kT$

Calculation:

The temperature that corresponds to kinetic energy is

  $\begin{array}{c}KE=\frac{3}{2}kT\\ T=\frac{2}{3}\frac{KE}{k}\\ T\text{ =}\frac{2}{3}\frac{(4.71\times {10}^6\text{ eV})\left(\frac{1.60\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)}{(1.38\times {10}^{-23}\text{J/K})}\\ \text{ = 3}\text{.6}\times {10}^{10}\text{K}\end{array}$

Conclusion:

The temperature that corresponds to kinetic energy is $3.6\times {10}^{10}\text{ K}$ .