Chapter 21, Problem 21.83QP

(a)

Interpretation Introduction

Interpretation: The energy released per nucleus of Tritium in the given reactions 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 per nucleus of Tritium in this reaction.

Answer to Problem 21.83QP

Solution

The energy released per nucleus of Tritium in this reaction is $\underset{\_}{-7.68\times 10^{-13}J}$.

Explanation of Solution

Explanation

The given reaction is,

${}_0^1\text{n}{+}_3^6\text{Li}{\to }_2^4\text{He}{+}_1^3\text{H}$

The mass of ${}^1\text{n}$ is $1.00866\text{amu}$.

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

The mass of ${}^3\text{H}$ is $3.0160492\text{amu}$.

The mass of ${}^6\text{Li}$ is $6.015122794\text{amu}$.

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,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\left(4.00260\text{amu}+\text{3}\text{.0160492}\text{amu}\right)-\left(1.00866\text{amu}+\text{6}\text{.015122794}\text{amu}\right)\\ =7.0186492\text{amu}-7.02378794\text{amu}\\ \text{=}-0.00513874\text{amu}\end{array}$

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 $-0.00513874\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.00513874\text{amu}=-0.00513874\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.0085330833\times {10}^{-27}\text{kg}\end{array}$

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

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.0085330833\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-7.68\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

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 $-7.68\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-7.68\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-7.68\times {10}^{-13}\times 1\text{J}\\ =\underset{\_}{-7.68\times 10^{-13}J}\end{array}$

Therefore, energy released per nucleus of Tritium in this reaction is $\underset{\_}{-7.68\times 10^{-13}J}$.

(b)

Interpretation Introduction

To determine: The energy released per nucleus of Tritium in this reaction.

Answer to Problem 21.83QP

Solution

The energy released per nucleus of Tritium in this reaction is $\underset{\_}{3.95\times \times 10^{-13}J}$.

Explanation of Solution

Explanation

The given reaction is,

${}_0^1\text{n}{+}_3^7\text{Li}{\to }_2^4\text{He}{+}_1^3\text{H}{+}_0^1\text{n}$

The mass of ${}^1\text{n}$ is $1.00866\text{amu}$.

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

The mass of ${}^3\text{H}$ is $3.0160492\text{amu}$.

The mass of ${}^7\text{Li}$ is $7.016004548\text{amu}$.

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,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\left(4.00260\text{amu}+\text{3}\text{.0160492}\text{amu+1}\text{.00866}\text{amu}\right)-\left(1.00866\text{amu}+7.016004548\text{amu}\right)\\ =8.0273092\text{amu}-8.024664548\text{amu}\\ =\text{0}\text{.002664652}\text{amu}\end{array}$

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 $\text{0}\text{.002664652}\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}\text{0}\text{.002664652}\text{amu}=\text{0}\text{.002664652}\times 1.66054\times {10}^{-27}\text{kg}\\ =0.0043915504\times {10}^{-27}\text{kg}\end{array}$

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

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =0.0043915504\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =3.95\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

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 $3.95\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}3.95\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}=3.95\times {10}^{-13}\times {10}^{-13}\times 1\text{J}\\ =\underset{\_}{3.95\times \times 10^{-13}J}\end{array}$

Therefore, energy released per nucleus of Tritium in this reaction is $\underset{\_}{3.95\times \times 10^{-13}J}$.

Conclusion

1. a. The energy released per nucleus of Tritium in this reaction is $\underset{\_}{-7.68\times 10^{-13}J}$.
2. b. The energy released per nucleus of Tritium in this reaction is $\underset{\_}{3.95\times \times 10^{-13}J}$