Chapter 21, Problem 21.82QP

(a)

Interpretation Introduction

Interpretation: The formation of ${}^{32}\text{S}$ by the fusion of given nuclei is given. The energy released in each of the given reaction is to be calculated.

Concept introduction: The process of formation of a heavy nucleus from two lighter nuclei is known as nuclear fusion.

To determine: The energy released in the given reaction.

Answer to Problem 21.82QP

Solution

The energy released in the given reaction is $\underset{\_}{-2.65\times 10^{-12}J}$.

Explanation of Solution

Explanation

Given

The mass of ${}^{32}\text{S}$ is $31.97207\text{amu}$.

The mass of ${}^{16}\text{O}$ is $15.99491\text{amu}$.

The given reaction is,

${}^{16}\text{O}{+}^{16}\text{O}{\to }^{32}\text{S}$

The formula for the amount of energy released during fusion is given as,

$\Delta E=\Delta m{c}^2$ (1)

Where,

• $\Delta m$ is the change in the mass.
• $c$ is the speed of the light $\left(3.0\times {10}^8\text{m}/\text{s}\right)$.

The change in the mass is calculated as,

$\Delta m=\text{Mass of products}-\text{Mass of reactants}$

Substitute the mass of reactants and products in the above equation as,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\text{31}\text{.97207}\text{amu}-2\left(15.99491\text{amu}\right)\\ =-0.01775\text{amu}\end{array}$

The conversion of $\text{amu}$ into $\text{kg}$ is done as,

$1\text{amu}=1.66054\times {10}^{-27}\text{kg}$

Hence, the conversion of $-0.01775\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.01775\text{amu}=-0.01775\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.029474585\times {10}^{-27}\text{kg}\end{array}$

Substitute the value of $\Delta m$ and $c$ in the equation (1),

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.029474585\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-2.65\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

The conversion of $\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$1\text{kg}{\text{m}}^2{\text{s}}^{-2}=1\text{J}$

Hence, the conversion of $-2.65\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-2.65\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-2.65\times {10}^{-12}\times 1\text{J}\\ =\underset{\_}{-2.65\times 10^{-12}J}\end{array}$

Therefore, energy released in this reaction is $\underset{\_}{-2.65\times 10^{-12}J}$.

(b)

Interpretation Introduction

To determine: The energy released in the given reaction.

Answer to Problem 21.82QP

Solution

The energy released in the given reaction is $\underset{\_}{-1.11\times 10^{-12}J}$.

Explanation of Solution

Explanation

Given

The mass of ${}^{32}\text{S}$ is $31.97207\text{amu}$.

The mass of ${}^4\text{He}$ is $4.00260\text{amu}$.

The mass of ${}^{28}\text{Si}$ is $27.97693\text{amu}$.

The given reaction is,

${}^{28}\text{Si}{+}^4\text{He}{\to }^{32}\text{S}$

The formula for the amount of energy released during fusion is given as,

$\Delta E=\Delta m{c}^2$ (1)

Where,

• $\Delta m$ is the change in the mass.
• $c$ is the speed of the light $\left(3.0\times {10}^8\text{m}/\text{s}\right)$.

The change in the mass is calculated as,

$\Delta m=\text{Mass of products}-\text{Mass of reactants}$

Substitute the mass of reactants and products in the above equation as,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\text{31}\text{.97207}\text{amu}-\left(27.97693+4.00260\right)\text{amu}\\ =\text{31}\text{.97207}\text{amu}-31.97953\text{amu}\\ \text{=}-0.00746\text{amu}\end{array}$

The conversion of $\text{amu}$ into $\text{kg}$ is done as,

$1\text{amu}=1.66054\times {10}^{-27}\text{kg}$

Hence, the conversion of $-0.00746\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.00746\text{amu}=-0.00746\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.012387628\times {10}^{-27}\text{kg}\end{array}$

Substitute the value of $\Delta m$ and $c$ in the equation (1),

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.012387628\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-1.11\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

The conversion of $\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$1\text{kg}{\text{m}}^2{\text{s}}^{-2}=1\text{J}$

Hence, the conversion of $-1.11\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-1.11\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-1.11\times {10}^{-12}\times 1\text{J}\\ =\underset{\_}{-1.11\times 10^{-12}J}\end{array}$

Therefore, energy released in this reaction is $\underset{\_}{-1.11\times 10^{-12}J}$.

(c)

Interpretation Introduction

To determine: The energy released in the given reaction.

Answer to Problem 21.82QP

Solution

The energy released in the given reaction is $\underset{\_}{-6.89\times 10^{-12}J}$.

Explanation of Solution

Explanation

Given

The mass of ${}^{32}\text{S}$ is $31.97207\text{amu}$.

The mass of ${}^{14}\text{N}$ is $14.00307\text{amu}$.

The mass of ${}^{12}\text{C}$ is $12.00000\text{amu}$.

The mass of ${}^6\text{Li}$ is $6.01512\text{amu}$.

The given reaction is,

${}^{\text{14}}\text{N}{+}^{12}\text{C}{+}^{\text{6}}\text{Li}{\to }^{32}\text{S}$

The formula for the amount of energy released during fusion is given as,

$\Delta E=\Delta m{c}^2$ (1)

Where,

• $\Delta m$ is the change in the mass.
• $c$ is the speed of the light $\left(3.0\times {10}^8\text{m}/\text{s}\right)$.

The change in the mass is calculated as,

$\Delta m=\text{Mass of products}-\text{Mass of reactants}$

Substitute the mass of reactants and products in the above equation as,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\text{31}\text{.97207}\text{amu}-\left(14.00307+12.00000+6.01512\right)\text{amu}\\ =\text{31}\text{.97207}\text{amu}-32.01819\text{amu}\\ \text{=}-0.04612\text{amu}\end{array}$

The conversion of $\text{amu}$ into $\text{kg}$ is done as,

$1\text{amu}=1.66054\times {10}^{-27}\text{kg}$

Hence, the conversion of $-0.04612\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.04612\text{amu}=-0.04612\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.076584104\times {10}^{-27}\text{kg}\end{array}$

Substitute the value of $\Delta m$ and $c$ in the equation (1),

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.076584104\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-6.89\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

The conversion of $\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$1\text{kg}{\text{m}}^2{\text{s}}^{-2}=1\text{J}$

Hence, the conversion of $-6.89\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-6.89\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-6.89\times {10}^{-12}\text{kg}\times 1\text{J}\\ =\underset{\_}{-6.89\times 10^{-12}J}\end{array}$

Therefore, energy released in this reaction is $\underset{\_}{-6.89\times 10^{-12}J}$.

(d)

Interpretation Introduction

To determine: The energy released in the given reaction.

Answer to Problem 21.82QP

Solution

The energy released in the given reaction is $\underset{\_}{-2.71\times 10^{-12}J}$.

Explanation of Solution

Explanation

Given

The mass of ${}^{24}\text{Mg}$ is $23.98504\text{amu}$.

The mass of ${}^4\text{He}$ is $4.00260\text{amu}$.

The mass of ${}^{32}\text{S}$ is $31.97207\text{amu}$.

The given reaction is,

${}^{24}\text{Mg}+2^4\text{He}{\to }^{32}\text{S}$

The formula for the amount of energy released during fusion is given as,

$\Delta E=\Delta m{c}^2$ (1)

Where,

• $\Delta m$ is the change in the mass.
• $c$ is the speed of the light $\left(3.0\times {10}^8\text{m}/\text{s}\right)$.

The change in the mass is calculated as,

$\Delta m=\text{Mass of products}-\text{Mass of reactants}$

Substitute the mass of reactants and products in the above equation as,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =31.97207\text{amu}-\left(23.98504+2\times 4.00260\right)\text{amu}\\ =31.97207\text{amu}-31.99024\text{amu}\\ \text{=}-0.01817\text{amu}\end{array}$

The conversion of $\text{amu}$ into $\text{kg}$ is done as,

$1\text{amu}=1.66054\times {10}^{-27}\text{kg}$

Hence, the conversion of $-0.01817\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.01817\text{amu}=-0.01817\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.030172011\times {10}^{-27}\text{kg}\end{array}$

Substitute the value of $\Delta m$ and $c$ in the equation (1),

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.030172011\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-2.71\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

The conversion of $\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$1\text{kg}{\text{m}}^2{\text{s}}^{-2}=1\text{J}$

Hence, the conversion of $-2.71\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-2.71\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-2.71\times {10}^{-12}\times 1\text{J}\\ =\underset{\_}{-2.71\times 10^{-12}J}\end{array}$

Therefore, energy released in this reaction is $\underset{\_}{-2.71\times 10^{-12}J}$.

Conclusion

1. a. The energy released in the given reaction is $\underset{\_}{-2.65\times 10^{-12}J}$.
2. b. The energy released in the given reaction is $\underset{\_}{-1.11\times 10^{-12}J}$.
3. c. The energy released in the given reaction is $\underset{\_}{-6.89\times 10^{-12}J}$.
4. d. The energy released in the given reaction is $\underset{\_}{-2.71\times 10^{-12}J}$