Chapter 33, Problem 48GP

(a)

To determine

Q -value of the reaction.

Answer to Problem 48GP

Q -value of the reaction is $9.954\text{ MeV}$

Explanation of Solution

Given:

Reaction

  ${}_8^{16}{\text{O + }}_8^{16}\text{O }\to {}_{14}^{28}\text{Si}+{}_2^4\text{He}$

Formula used:

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

Calculation:

Q -value is calculated as

  $\begin{array}{l}Q=2m_O{c}^2-m_{Si}{c}^2-m_{He}{c}^2\\ Q\text{ = }[2(15.994915\text{ u})-27.976927\text{ u}-\text{4}\text{.002603 u}]{\text{c}}^2(931.5{\text{ MeV/c}}^2)\\ Q\text{ = 9}\text{.954 MeV}\end{array}$

Conclusion:

Thus, Q -value of the reaction is $9.954\text{ MeV}$

(b)

To determine

Kinetic energy of the reactants.

Answer to Problem 48GP

Kinetic energy of the reactants is $7.61\text{ MeV}$ .

Explanation of Solution

Given:

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

Formula Used:

Potential energy is given as

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

Calculation:

Kinetic energy is equal to the electrical potential energy of the two nuclei when they come in contact. Thus, 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}\\ =\frac{1}{2}\frac{1}{4\pi {\epsilon }_0}\frac{{\text{q}}_{\text{nucleus}}^2}{2(1.2\times {10}^{-15}\text{ m}){\text{A}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\\ =\frac{1}{2}\times \left(\frac{1}{4\times 3.14\times 8.85\times {10}^{-12}}\right)\frac{{(8)}^2{(1.60\times {10}^{-19}\text{ C})}^2}{2(1.2\times {10}^{-15}\text{ m}){(16)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\frac{1\text{ eV}}{1.60\times {10}^{-19}\text{ J}}\\ =7.61\text{ MeV}\end{array}$

Conclusion:

Thus, the kinetic energy of each reactant is $7.61\text{ MeV}$ .

(c)

To determine

The temperature of the collision.

Answer to Problem 48GP

The temperature that corresponds to kinetic energy is $5.9\times {10}^{10}\text{ K}$

Explanation of Solution

Given:

Kinetic energy of the reactants is $7.61\text{ MeV}$ .

Formula used:

Kinetic energy is calculated as

  $\text{KE = }\frac{3}{2}\text{kT}$

Calculation:

The temperature that corresponds to kinetic energy is

  $\begin{array}{l}KE=\frac{3}{2}kT\\ T=\frac{2}{3}\frac{KE}{k}\\ T\text{ =}\frac{2}{3}\frac{(7.61\times {10}^6\text{ eV})\left(\frac{1.60\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)}{(1.38\times {10}^{-23}\text{J/K})}\\ T\text{ = 5}\text{.9}\times {10}^{10}\text{K}\end{array}$

Conclusion:

The temperature of the collision is $5.9\times {10}^{10}\text{ K}$