Chapter 18, Problem 45E

Interpretation Introduction

Interpretation: The binding of carbon- $12$ and uranium- $235$ is to be calculated. Justification of whether the binding energy per nucleon of ${}^{\text{56}}\text{Fe}$ be greater or smaller than that of ${}^{\text{12}}\text{C}$ or ${}^{\text{235}}\text{U}$ is to be stated.

Concept introduction: The sum of masses of the component nucleons and the actual mass of a nucleus is known as the mass defect and it can be used to calculate the nuclear binding energy.

To determine: The binding energy per nucleon of carbon- $12$ and uranium- $235$.

Answer to Problem 45E

Answer

The binding of carbon- $12$ is $\underset{\_}{-1.23\times 10^{-12}J/nucleon}$ The binding of uranium- $235$ is $\underset{\_}{-1.21673\times 10^{-12}J/nucleon}$ The binding energy per nucleon for ${}^{\text{56}}\text{Fe}$ is larger than ${}^{\text{12}}\text{C}$ or ${}^{\text{235}}\text{U}$.

Explanation of Solution

Explanation

Given

The atomic mass of ${}_{\text{6}}^{\text{12}}\text{C}$ is $\text{12}\text{.000}\text{amu}$.

Mass of ${}_{\text{1}}^{\text{1}}\text{H}$ $\text{1}\text{.00782}\text{amu}$.

The mass of neutron is $1.00866\text{amu}$.

Number of protons in ${}_{\text{6}}^{\text{12}}\text{C}=\text{6}$

Number of neutrons in ${}_{\text{6}}^{\text{12}}\text{C}=\text{6}$

Mass defect is calculated by the formula,

$\Delta m=\text{Atomic}\text{mass}\text{of}{}_{\text{6}}^{\text{12}}\text{C}\text{nucleus}-\left[\begin{array}{l}\text{Number}\text{of}\text{protons}\times \text{Mass}\text{of}{}_{\text{1}}^{\text{1}}\text{Hproton}-\\ \text{Number}\text{of}\text{neutrons}\times \text{Mass}\text{of}\text{neutron}\end{array}\right]$

Substitute the value of the atomic mass ${}_{\text{6}}^{\text{12}}\text{C}$, the number of protons and mass of the ${}_{\text{1}}^{\text{1}}\text{H}$ proton and that of the neutron in the above equation.

$\begin{array}{l}\Delta m=12.000-\left[\left(6\times 1.0078\right)+\left(6\times 1.00866\right)\right]\\ \Delta m=12.000-6.0469-6.05196\\ \Delta m=12.000-12.09888\\ \Delta m=-0.09888\frac{\text{amu}}{\text{nucleus}}\end{array}$

The conversion of $\text{amu}/\text{nucleus}$ to $\text{Kg/nucleus}$ is done as,

$1\text{amu}=1.66\times {10}^{-27}\text{Kg}$

Therefore, the conversion of $-0.589\text{amu}/\text{nucleus}$ into $\text{kg/nucleus}$ is,

$\begin{array}{c}-0.09888\text{amu}/\text{nucleus}=\left(-0.09888\times 1\times 1.66\times {10}^{-27}\right)\text{kg/nucleus}\\ =\underset{\_}{-0.1641408\times 10^{-27}Kg/nucleus}\end{array}$

Therefore, the mass defect $\left(\Delta m\right)$ of ${}_{\text{6}}^{\text{12}}\text{C}$ is $\underset{\_}{-0.1641408\times 10^{-27}Kg/nucleus}$.

Explanation

The binding energy per nucleon is calculated by Einstein’s mass energy equation, that is,

$\Delta E=\Delta m\cdot C^2$

Where,

• $\Delta E$ is the change in energy.
• $\Delta m$ is the change in mass.
• $C$ is the velocity of light.

Substitute the values of $\Delta m$ and $C$ in the equation.

$\begin{array}{c}\Delta E=\Delta m\cdot C^2\\ \Delta E=\left(-\text{0}{\text{.1641408×10}}^{\text{-27}}\frac{\text{kg}}{\text{nucleus}}\right){\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}^2\\ \Delta E=\underset{\_}{-1.477\times 10^{-11}J/nucleus}\end{array}$

Therefore, the energy released per nucleus is $\underset{\_}{-1.477\times 10^{-11}J/nucleus}$.

Explanation

The binding energy per nucleon is calculated by the formula,

$\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}}$

Substitute the value of the binding energy per nucleus, the number of protons and the number of neutrons in the above equation.

$\begin{array}{c}\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\left(\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}\right)}\\ =\frac{-1.477\times {10}^{-11}}{12}\text{J/nucleon}\\ \text{=}\underset{\_}{-1.23\times 10^{-12}J/nucleon}\end{array}$

Explanation

The atomic mass of ${}_{\text{92}}^{\text{235}}\text{U}$ $\text{235}\text{.0439}\text{amu}$.

Mass of ${}_{\text{1}}^{\text{1}}\text{H}$ $\text{1}\text{.00782 amu}$.

The mass of neutron is $1.00866\text{amu}$

Number of protons in ${}_{\text{92}}^{\text{235}}\text{U}=\text{92}$

Number of neutrons in ${}_{\text{92}}^{\text{235}}\text{U}=\text{143}$

The mass defect of ${}_{\text{92}}^{\text{235}}\text{U}$ is calculated by the formula,

$\Delta m={\text{Mass of }}_{\text{92}}^{\text{235}}\text{U}\text{nucleus }-\left[\begin{array}{l}\text{Number}\text{of}\text{protons}\times \text{mass}\text{of}{}_{\text{1}}^{\text{1}}\text{H}\text{proton}-\\ \text{Number}\text{of}\text{neutrons}\times \text{mass}\text{of}\text{neutron}\end{array}\right]$

Substitute the value of the mass ${}_{\text{92}}^{\text{235}}\text{U}$, the number of protons and mass of the ${}_{\text{1}}^{\text{1}}\text{H}$ proton and that of the neutron in the above equation.

$\begin{array}{l}\Delta m=235.0439-\left[\left(92\times 1.00782\right)+\left(143\times 1.00866\right)\right]\\ \Delta m=235.0439-92.7194-144.2383\\ \Delta m=-\text{1}\text{.91388}\text{amu/nucleus}\end{array}$

The conversion of $\text{amu}/\text{nucleus}$ to $\text{Kg/nucleus}$ is done as,

$1\text{amu}=1.66\times {10}^{-27}\text{Kg}$

Therefore, the conversion of $-0.589\text{amu}/\text{nucleus}$ into $\text{kg/nucleus}$ is,

$\begin{array}{c}-1.91388\text{amu}/\text{nucleu}=\left(-1.9138\times 1\times 1.66\times {10}^{-27}\right)\text{kg/nucleus}\\ =\underset{\_}{-3.17704\times 10^{-27}Kg/nucleus}\end{array}$

The mass defect $\left(\Delta m\right)$ of ${}_{\text{92}}^{\text{235}}\text{U}$ is. $\underset{\_}{-3.17704\times 10^{-27}Kg/nucleus}$.

Explanation

The binding energy per nucleon is calculated by Einstein’s mass energy equation, that is,

$\Delta E=\Delta m\cdot C^2$

Where,

• $\Delta E$ is the change in energy.
• $\Delta m$ is the change in mass.
• $C$ is the velocity of light.

Substitute the values of $\Delta m$ and $C$ in the equation.

$\begin{array}{l}\text{ΔE}=\Delta \text{m}{\text{.c}}^{\text{2}}\\ \text{ΔE}=\left(-3.17704\times {\text{10}}^{-27}\text{Kg/nucleus}\right){\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}^2\\ \text{ΔE}=-28.5933\times {10}^{-11}\text{J/nucleus}\end{array}$

Therefore, the energy released per nucleus is $-28.5933\times {10}^{-11}\text{J/nucleus}$

The binding energy per nucleon is calculated by the formula,

$\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}}$

Substitute the value of the binding energy per nucleus, the number of protons and the number of neutrons in the above equation.

$\begin{array}{c}\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\left(\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}\right)}\\ =\frac{-28.5933\times {10}^{-11}}{235}\text{J/nucleon}\\ \text{=}\underset{\_}{-1.21673\times 10^{-12}J/nucleon}\end{array}$

The mass defect $\left(\Delta \text{m}\right)$ of ${}^{56}\text{Fe}$ is. $\underset{\_}{-0.8779\times 10^{-27}Kg/nucleus}$

Explanation

The atomic mass of ${}^{56}\text{Fe}=\text{55}\text{.9349}\text{amu}$.

Mass of ${}_{\text{1}}^{\text{1}}\text{H}=\text{1}\text{.00782 amu}$.

The mass of neutron is $1.00866\text{amu}$.

Number of protons in ${}^{56}\text{Fe}=\text{26}$

Number of neutrons in ${}^{56}\text{Fe}=\text{30}$

The mass defect is calculated by the formula,

$\Delta m=\text{Mass of }{}^{56}\text{Fe }-\left[\begin{array}{l}\text{Number}\text{of}\text{protons}\times \text{mass}\text{of}{}_{\text{1}}^{\text{1}}\text{H}\text{proton}-\\ \text{Number}\text{of}\text{neutrons}\times \text{mass}\text{of}\text{neutron}\end{array}\right]$

Substitute the value of the mass ${}^{56}\text{Fe }$, the number of protons and mass of the ${}_{\text{1}}^{\text{1}}\text{H}$ proton and that of the neutron in the above equation.

$\begin{array}{c}\Delta m=\text{55}\text{.9349}-\left[\left(26\times 1.00782\right)+\left(30\times 1.00866\right)\right]\\ \Delta m=-0.589\text{amu}/\text{nucleus}\end{array}$

The conversion of $\text{amu}/\text{nucleus}$ to $\text{Kg/nucleus}$ is done as,

$1\text{amu}=1.66\times {10}^{-27}\text{Kg}$

Therefore, the conversion of $-0.589\text{amu}/\text{nucleus}$ into $\text{kg/nucleus}$ is,

$\begin{array}{c}-0.5289\text{amu}/\text{nucleus}=\left(-0.5289\times 1\times 1.66\times {10}^{-27}\right)\text{kg/nucleus}\\ =\underset{\_}{-0.8779\times 10^{-27}Kg/nucleus}\end{array}$

Therefore, the mass defect $\left(\Delta m\right)$ of ${}^{56}\text{Fe}$ is $\underset{\_}{-0.8779\times 10^{-27}Kg/nucleus}$.

Explanation

The binding energy per nucleon is calculated by Einstein’s mass energy equation, that is,

$\Delta E=\Delta m\cdot C^2$

Where,

• $\Delta E$ is the change in energy.
• $\Delta m$ is the change in mass.
• $C$ is the velocity of light.

Substitute the values of $\Delta m$ and $C$ in the equation.

$\begin{array}{l}\Delta E=\Delta m\cdot c^2\\ \Delta E=\left(-\text{0}\text{.8779}\times {\text{10}}^{-27}\frac{\text{kg}}{\text{nucleus}}\right){\left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}^2\\ \Delta E=-7.90\times {10}^{-11}\text{J/nucleus}\end{array}$

Therefore, the energy released per nucleus is $-7.90\times {10}^{-11}\text{J/nucleus}$.

The binding energy per nucleon is calculated by the formula,

$\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}}$

Substitute the value of the binding energy per nucleus, the number of protons and the number of neutrons in the above equation.

$\begin{array}{c}\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\left(\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}\right)}\\ =\frac{7.90\times {10}^{-11}}{56}\text{J/nucleon}\\ \text{=}\underset{\_}{14.10\times 10^{-13}J/nucleon}\end{array}$

Conclusion

Conclusion

The binding of carbon- $12$ is $\underset{\_}{-1.23\times 10^{-12}J/nucleon}$ The binding of uranium- $235$ is $\underset{\_}{-1.21673\times 10^{-12}J/nucleon}$ The binding energy per nucleon for ${}^{\text{56}}\text{Fe}$ is larger than ${}^{\text{12}}\text{C}$ or ${}^{\text{235}}\text{U}$.