Chapter 19, Problem 46E

Interpretation Introduction

Interpretation: The binding energy per nucleon of ${}_1^2\text{H}$ and ${}_1^3\text{H}$ is to be calculated.

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 ${}_1^2\text{H}$ and ${}_1^3\text{H}$.

Answer to Problem 46E

Answer

The binding energy per nucleon of ${}_1^2\text{H}$ is $\underset{\_}{17.78\times 10^{-14}}\text{J/nucleon}$.

The binding energy per nucleon of ${}_1^3\text{H}$ is $\underset{\_}{45.26\times 10^{-14}}\text{J/nucleon}$.

Explanation of Solution

Explanation

The atomic mass of ${}_1^1\text{H}=\text{1}\text{.0078}\text{amu}$

The mass of a neutron is $1.0087\text{amu}$.

Mass of ${}_1^2\text{H}=\text{2}\text{.01410}\text{amu}$

Number of protons in ${}_1^2\text{H}=\text{1}$

Number of neutrons in ${}_1^2\text{H}=\text{1}$

The mass defect is calculated by the formula,

$\Delta m=\text{Atomic}\text{mass of }{}_1^2\text{H}-\left[\begin{array}{l}\text{Number}\text{of}\text{protons}\times \text{mass}\text{of}{}_{\text{1}}^{\text{1}}\text{H}\text{proton}-\\ \text{Number}\text{of}\text{neutrons}\times \text{mass}\text{of}\text{neutron}\end{array}\right]$

Substitute the value of the atomic mass ${}_1^2\text{H}$, the number of protons and mass of the ${}_{\text{1}}^{\text{1}}\text{H}$ proton and that of the neutron in the above equation.

$\begin{array}{l}\Delta m=\text{2}\text{.01410}-\left[\left(1\times 1.00782\right)+\left(1\times 1.00866\right)\right]\\ \Delta m=-\text{2}\text{.38}\times {\text{10}}^{-3}\text{amu}/\text{nucleus}\end{array}$

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,

$\begin{array}{c}-\text{2}\text{.38}\times {\text{10}}^{-3}\text{amu}/\text{nucleus}=\left(-\text{2}\text{.38}\times {\text{10}}^{-3}\times 1.66\times {10}^{-27}\right)\text{kg/nucleus}\\ =\underset{\_}{-3.9508\times 10^{-30}Kg/nucleus}\end{array}$

Therefore, the mass defect $\left(\Delta m\right)$ of ${}_1^2\text{H}$ is $\underset{\_}{-3.9508\times 10^{-30}Kg/nucleus}$.

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.

$\begin{array}{l}\Delta E=\Delta m\cdot c^2\\ \Delta E=\left(-3.9508\times {10}^{-30}\right){\left(3\times {10}^8\right)}^2\\ \Delta E=\text{35}{\text{.55×10}}^{-\text{14}}\text{J/nucleus}\end{array}$

Therefore, the energy released per nucleus is $35.55\times {10}^{-14}\text{J/nucleus}$.

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.

$\begin{array}{c}\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\left(\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}\right)}\\ =\frac{35.55\times {10}^{-14}}{2}\text{J/nucleon}\\ =\underset{\_}{17.78\times 10^{-14}J/nucleon}\end{array}$

Explanation

The atomic mass of ${}_1^1\text{H=1}\text{.0078}\text{amu}$ and mass of neutron is $1.0087\text{amu}$

The atomic mass of ${}_1^3\text{H}=\text{3}\text{.01605}\text{amu}$

The mass of a neutron is $1.0087\text{amu}$.

Mass of ${}_1^2\text{H}=\text{2}\text{.01410}\text{amu}$

Number of protons in ${}_1^3\text{H}=\text{1}$

Number of neutrons in ${}_1^3\text{H}=\text{2}$

The mass defect is calculated by the formula,

$\Delta m=\text{Atomic}\text{mass of }{}_1^3\text{H}-\left[\begin{array}{l}\text{Number}\text{of}\text{protons}\times \text{mass}\text{of}{}_{\text{1}}^{\text{1}}\text{H}\text{proton}-\\ \text{Number}\text{of}\text{neutrons}\times \text{mass}\text{of}\text{neutron}\end{array}\right]$

Substitute the value of the atomic mass ${}_1^3\text{H}$, the number of protons and mass of the ${}_{\text{1}}^{\text{1}}\text{H}$ proton and that of the neutron in the above equation.

$\begin{array}{l}\Delta m=\text{3}\text{.01605}-\left[\left(1\times 1.00782\right)+\left(2\times 1.00866\right)\right]\\ \Delta m=-0.00909\text{amu}/\text{nucleus}\end{array}$

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,

$\begin{array}{c}-\text{2}\text{.38}\times {\text{10}}^{-3}\text{amu}/\text{nucleus}=-\text{2}\text{.38}\times {\text{10}}^{-3}\times 1.66\times {10}^{-27}\text{kg/nucleus}\\ =\underset{\_}{-3.9508\times 10^{-30}Kg/nucleus}\end{array}$

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.

$\begin{array}{l}\Delta E=\Delta m\cdot c^2\\ \Delta E=\left(-0.0150894\times {10}^{-27}\right)\times 9\times {\left({10}^8\right)}^2\\ \Delta E=-0.1358046\times {10}^{-11}\text{J/nucleus}\end{array}$

Therefore, the energy released per nucleus is $0.1358046\times {10}^{-11}\text{J/nucleus}$.

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.

$\begin{array}{c}\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\left(\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}\right)}\\ =\frac{0.1358046\times {10}^{-11}}{3}\text{J/nucleon}\\ =\underset{\_}{45.26\times 10^{-14}J/nucleon}\end{array}$

Conclusion

Conclusion

The binding energy per nucleon of ${}_1^2\text{H}$ is $\underset{\_}{17.78\times 10^{-14}}\text{J/nucleon}$.

The binding energy per nucleon of   ${}_1^3\text{H}$ is $\underset{\_}{45.26\times 10^{-14}}\text{J/nucleon}$.