Chapter 19, Problem 43E

Interpretation Introduction

Interpretation: The atomic masses of neutron, proton, electron ${}^{232}\text{Pu}$ and ${}^{231}\text{Pa}$ are given. Using these values, the change in energy when $1$ mole of ${}^{232}\text{Pu}$ nuclei and $1$ mole of ${}^{231}\text{Pa}$ nuclei are formed from their respective number of protons and neutrons is to be calculated.

Concept introduction: The difference between the mass of an atom and the mass of its constituent particles is known as mass defect. This mass can be converted into energy by Einstein’s mass-energy equation as,

$\Delta E=\Delta m{c}^2$

To determine: The change in energy when $1$ mole of ${}^{232}\text{Pu}$ nuclei and $1$ mole of ${}^{231}\text{Pa}$ nuclei are formed from their respective number of protons and neutrons.

Answer to Problem 43E

Solution

The change in energy when $1$ mole of ${}^{232}\text{Pu}$ nuclei formed is $\underset{\_}{-1.6702\times 10^{14}J/mol}$

The change in energy when $1$ mole of ${}^{231}\text{Pa}$ nuclei formed is $\underset{\_}{-1.6704\times 10^{14}J/mol}$

Explanation of Solution

Explanation

Given

The atomic mass of ${}^{232}\text{Pu}$ is $3.85285\times {10}^{-22}\text{g}$.

The mass of a neutron is $1.67493\times {10}^{-24}\text{g}$.

The mass of a proton is $1.67262\times {10}^{-24}\text{g}$.

The atomic number of ${}^{232}\text{Pu}$ is $94$. Therefore, the number of protons in ${}^{232}\text{Pu}$ is 94.

The number of neutrons in ${}^{232}\text{Pu}$ is,

Number of neutrons $\begin{array}{l}=232-94\\ =138\end{array}$

The mass defect is calculated by the formula,

$\Delta m=\text{Atomic}\text{mass of }{}^{232}\text{Pu}-\left[\begin{array}{l}\text{Number}\text{of}\text{protons}\times \text{mass}\text{of}\text{proton}-\\ \text{Number}\text{of}\text{neutrons}\times \text{mass}\text{of}\text{neutron}\end{array}\right]$

Where,

• $\Delta m$ is the mass defect.

Substitute the value of the atomic mass of ${}^{232}\text{Pu}$, the number of protons and neutrons and their respective masses in the above equation.

$\begin{array}{c}\Delta m=3.85285\times {10}^{-22}\text{g}-\left[\left(94\times 1.67262\times {10}^{-24}\text{g}\right)+\left(138\times 1.67493\times {10}^{-24}\text{g}\right)\right]\\ =-0.0308162\times {\text{10}}^{-22}\text{g}\end{array}$

To calculate the mass defect for one mole of ${}^{232}\text{Pu}$, multiply this value by Avogadro’s number $\left(6.022\times {10}^{23}\text{per}\text{mol}\right)$.

$\begin{array}{c}\Delta m\text{'}=-0.0308162\times {\text{10}}^{-22}\text{g}\times 6.022\times {10}^{23}\text{per}\text{mol}\\ =-\underset{\_}{1.85575g/mol}\end{array}$

The change in energy of ${}^{232}\text{Pu}$ is. $\underset{\_}{-1.6702\times 10^{14}J/mol}$.

Explanation

The mass defect is $-1.85575\text{g/mol}$.

The conversion of gram $\left(\text{g}\right)$ into kilogram $\left(\text{kg}\right)$ is done as,

$1\text{g}={10}^{-3}\text{kg}$

Hence, the conversion of $-1.85575\text{g/mol}$ into kilogram/mol is,

$-1.85575\text{g/mol}=-1.85575\times {10}^{-3}\text{kg/mol}$

The energy change is calculated by Einstein’s mass energy equation,

$\Delta E=\Delta m\text{'}{c}^2$

Where,

• $\Delta E$ is the change in energy.
• $\Delta m\text{'}$ is the mass defect.
• $c$ is the speed of light $\left(3.0\times {10}^8\text{m/s}\right)$.

Substitute the values of $\Delta m\text{'}$ and $c$ in the above equation.

$\begin{array}{c}\Delta E=\Delta m\text{'}{c}^2\\ =\left(-1.85575\times {10}^{-3}\text{kg/mol}\right){\left(3.0\times {10}^8\text{m/s}\right)}^2\\ =\underset{\_}{-1.6702\times 10^{14}J/mol}\end{array}$

The mass defect of ${}^{231}\text{Pa}$ nucleus is $-\underset{\_}{1.85635g/mol}$.

Explanation

Given

The atomic mass of ${}^{231}\text{Pa}$ is $3.83616\times {10}^{-22}\text{g}$.

The mass of a neutron is $1.67493\times {10}^{-24}\text{g}$.

The mass of a proton is $1.67262\times {10}^{-24}\text{g}$.

The atomic number of ${}^{231}\text{Pa}$ is $91$. Therefore, the number of protons in ${}^{231}\text{Pa}$ is $91$.

The number of neutrons in ${}^{232}\text{Pu}$ is,

Number of neutrons $\begin{array}{l}=231-91\\ =140\end{array}$

The mass defect is calculated by the formula,

$\Delta m=\text{Atomic}\text{mass of }{}^{231}\text{Pa}-\left[\begin{array}{l}\text{Number}\text{of}\text{protons}\times \text{mass}\text{of}\text{proton}-\\ \text{Number}\text{of}\text{neutrons}\times \text{mass}\text{of}\text{neutron}\end{array}\right]$

Substitute the value of the atomic mass of ${}^{231}\text{Pa}$, the number of protons and neutrons and their respective masses in the above equation.

$\begin{array}{c}\Delta m=3.83616\times {10}^{-22}\text{g}-\left[\left(91\times 1.67262\times {10}^{-24}\text{g}\right)+\left(140\times 1.67493\times {10}^{-24}\text{g}\right)\right]\\ =-0.0308262\times {\text{10}}^{-22}\text{g}\end{array}$

To calculate the mass defect for one mole of ${}^{231}\text{Pa}$, multiply this value by Avogadro’s number $\left(6.022\times {10}^{23}\text{per}\text{mol}\right)$.

$\begin{array}{c}\Delta m\text{'}=-0.0308262\times {\text{10}}^{-22}\text{g}\times 6.022\times {10}^{23}\text{per}\text{mol}\\ =-\underset{\_}{1.85635g/mol}\end{array}$

The binding energy of ${}^{231}\text{Pa}$ is. $\underset{\_}{-1.6707\times 10^{14}J/mol}$.

Explanation

The mass defect is $-1.85635\text{g/mol}$.

The conversion of gram $\left(\text{g}\right)$ into kilogram $\left(\text{kg}\right)$ is done as,

$1\text{g}={10}^{-3}\text{kg}$

Hence, the conversion of $-1.85635\text{g/mol}$ into kilogram/mol is,

$-1.85635\text{g/mol}=-1.85635\times {10}^{-3}\text{kg/mol}$

The energy change is calculated by Einstein’s mass energy equation,

$\Delta E=\Delta m\text{'}{c}^2$

Where,

• $\Delta E$ is the change in energy.
• $\Delta m\text{'}$ is the mass defect.
• $c$ is the speed of light $\left(3.0\times {10}^8\text{m/s}\right)$.

Substitute the values of $\Delta m\text{'}$ and $c$ in the above equation.

$\begin{array}{c}\Delta E=\Delta m\text{'}{c}^2\\ =\left(-1.85635\times {10}^{-3}\text{kg/mol}\right){\left(3.0\times {10}^8\text{m/s}\right)}^2\\ =\underset{\_}{-1.6707\times 10^{14}J/mol}\end{array}$

Conclusion

Conclusion

The change in energy when $1$ mole of ${}^{232}\text{Pu}$ nuclei formed is $\underset{\_}{-1.6702\times 10^{14}J/mol}$.

The change in energy when $1$ mole of ${}^{231}\text{Pa}$ nuclei formed is $\underset{\_}{-1.6707\times 10^{14}J/mol}$.