Chapter 19, Problem 19.87AP

Interpretation Introduction

Interpretation: The energy released during the fission process of Uranium $-235$ and fusion reaction involving the fusion of four nuclei of Hydrogen to Helium is given. By finding the energy released in $\text{J}/\text{nucleon}$ the comparison between the energy released during the fission process of Uranium $-235$ and during the given fusion reaction is to be stated.

Concept introduction: The combining of two or more lighter nuclei to form heavy nuclei is known as nuclear fusion reaction while the splitting of heavy nuclei into smaller nuclei is known as nuclear fission.

To determine: The comparison between the energy released during the fission process of Uranium $-235$ and during the given fusion reaction by finding the energy released in $\text{J}/\text{nucleon}$.

Answer to Problem 19.87AP

Solution

The energy released per nucleon in fusion process is $\underset{\_}{-9.01\times 10^{-13}J}$ and the value of energy released per nucleon during fission is $\underset{\_}{-1.36\times 10^{-13}J/nucleon}$.

The energy released is more during the fusion process rather than the fission process.

Explanation of Solution

Explanation

The given reaction is,

$4{}_1^1\text{H}{\to }_2^4\text{He}+2{}_1^0\text{β}$

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

The mass of ${}^1\text{H}$ is $1.00794\text{amu}$.

The mass of ${}_1^0\text{β}$ is $\text{0}\text{.000548756}\text{amu}$.

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,

$\begin{array}{c}\Delta m=\text{Mass of products}-\text{Mass of reactants}\\ =\left(4.00260\text{amu}+\text{2}\times \text{0}\text{.000548756}\text{amu}\right)-\left(4\times 1.00794\text{amu}\right)\\ =4.003697512\text{amu}-4.03176\text{amu}\\ \text{=}-0.028062488\text{amu}\end{array}$

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 $-0.028062488\text{amu}$ into $\text{kg}$ is done as,

$\begin{array}{c}-0.028062488\text{amu}=-0.028062488\times 1.66054\times {10}^{-27}\text{kg}\\ \text{=}-0.0450515182\times {10}^{-27}\text{kg}\end{array}$

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

$\begin{array}{c}\Delta E=\Delta m{c}^2\\ =-0.0450515182\times {10}^{-27}\text{kg}\times {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =-9.01\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}\end{array}$

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 $-9.01\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}$ to $\text{J}$ is done as,

$\begin{array}{c}-9.01\times {10}^{-13}\text{kg}{\text{m}}^2{\text{s}}^{-2}=-9.01\times {10}^{-13}\times 1\text{J}\\ =\underset{\_}{-9.01\times 10^{-13}J}\end{array}$

Therefore, energy released per nucleon in fusion process is $\underset{\_}{-9.01\times 10^{-13}J}$.

The number of nucleons present in Uranium $-235$ are $235$. Therefore, on dividing the energy released during the fission of Uranium $-235$, the value of energy released per nucleon is $\begin{array}{c}=\frac{3.2\times {10}^{-11}}{235}\\ =1.36\times {10}^{-13}\text{J}/\text{nucleon}\end{array}$

As the energy is released, therefore the obtained energy is assigned negative sign. Therefore, the value of energy released per nucleon during fission is $\underset{\_}{-1.36\times 10^{-13}J/nucleon}$.

The energy released is more during the fusion process rather than the fission process.

Conclusion

The energy released per nucleon in fusion process is $\underset{\_}{-9.01\times 10^{-13}J}$ and the value of energy released per nucleon during fission is $\underset{\_}{-1.36\times 10^{-13}J/nucleon}$.

The energy released is more during the fusion process rather than the fission process