Chapter 19, Problem 19.81QP

Interpretation Introduction

Interpretation: The formation of ${}^{28}\text{Si}$ 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 19.81QP

Solution

The energy released in the given reaction is $\underset{\_}{-4.37\times 10^{-12}J}$.

Explanation of Solution

Explanation

Given

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

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

The given reaction is,

${}^{14}\text{N}{+}^{14}\text{N}{\to }^{28}\text{Si}$

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,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\text{27}\text{.97693}\text{amu}-2\left(14.00307\text{amu}\right)\\ =-0.02921\text{amu}\end{array}$

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 $-0.02921\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.02921\text{amu}=-0.02921\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.0485043734\times {10}^{-27}\text{kg}\end{array}$

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

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.0485043734\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-4.37\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

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 $-4.37\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-4.37\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-4.37\times {10}^{-12}\times 1\text{J}\\ =\underset{\_}{-4.37\times 10^{-12}J}\end{array}$

Therefore, energy released in this reaction is $\underset{\_}{-4.37\times 10^{-12}J}$.

Interpretation Introduction

To determine: The energy released in the given reaction.

Answer to Problem 19.81QP

Solution

The energy released in the given reaction is $\underset{\_}{-6.8\times 10^{-12}J}$.

Explanation of Solution

Explanation

Given

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

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

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

The mass of ${}^2\text{H}$ is $2.0146\text{amu}$.

The given reaction is,

${}^{10}\text{B}{+}^{16}\text{O}{+}^{\text{2}}\text{H}{\to }^{28}\text{Si}$

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,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\text{27}\text{.97693}\text{amu}-\left(10.0129+15.99491+2.0146\right)\text{amu}\\ =\text{27}\text{.97693}\text{amu}-28.02241\text{amu}\\ \text{=}-0.04548\text{amu}\end{array}$

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 $-0.04548\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.04548\text{amu}=-0.04548\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.0755213592\times {10}^{-27}\text{kg}\end{array}$

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

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.0755213592\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-6.8\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

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 $-6.8\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-6.8\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-6.8\times {10}^{-12}\times 1\text{J}\\ =\underset{\_}{-6.8\times 10^{-12}J}\end{array}$

Therefore, energy released in this reaction is $\underset{\_}{-6.8\times 10^{-12}J}$.

Interpretation Introduction

To determine: The energy released in the given reaction.

Answer to Problem 19.81QP

Solution

The energy released in the given reaction is $\underset{\_}{-2.69\times 10^{-12}J}$.

Explanation of Solution

Explanation

Given

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

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

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

The given reaction is,

${}^{16}\text{O}{+}^{\text{12}}\text{C}{\to }^{28}\text{Si}$

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,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\text{27}\text{.97693}\text{amu}-\left(15.99491+12.00000\right)\text{amu}\\ =\text{27}\text{.97693}\text{amu}-27.99491\text{amu}\\ \text{=}-0.01798\text{amu}\end{array}$

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 $-0.01798\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.01798\text{amu}=-0.01798\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.0298565092\times {10}^{-27}\text{kg}\end{array}$

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

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.0298565092\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-2.69\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

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 $-2.69\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-2.69\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-2.69\times {10}^{-12}\times 1\text{J}\\ =\underset{\_}{-2.69\times 10^{-12}J}\end{array}$

Therefore, energy released in this reaction is $\underset{\_}{-2.69\times 10^{-12}J}$.

Interpretation Introduction

To determine: The energy released in the given reaction.

Answer to Problem 19.81QP

Solution

The energy released in the given reaction is $\underset{\_}{-1.69\times 10^{-12}J}$.

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

The given reaction is,

${}^{24}\text{Mg}{+}^4\text{He}{\to }^{28}\text{Si}$

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,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\text{27}\text{.97693}\text{amu}-\left(23.98504+4.00260\right)\text{amu}\\ =\text{27}\text{.97693}\text{amu}-27.98764\text{amu}\\ \text{=}-0.01134\text{amu}\end{array}$

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 $-0.01134\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.01134\text{amu}=-0.01134\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.018830523\times {10}^{-27}\text{kg}\end{array}$

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

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =--0.018830523\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-1.69\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

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 $-1.69\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-1.69\times {10}^{-12}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-1.69\times {10}^{-12}\times 1\text{J}\\ =\underset{\_}{-1.69\times 10^{-12}J}\end{array}$

Therefore, energy released in this reaction is $\underset{\_}{-1.69\times 10^{-12}J}$.

Conclusion

1. a. The energy released in the given reaction is $\underset{\_}{-4.37\times 10^{-12}J}$.
2. b. The energy released in the given reaction is $\underset{\_}{-6.8\times 10^{-12}J}$.
3. c. The energy released in the given reaction is $\underset{\_}{-2.69\times 10^{-12}J}$.
4. d. The energy released in the given reaction is $\underset{\_}{-1.69\times 10^{-12}J}$