Chapter 30, Problem 26E

(a)

To determine

The difference in mass defects of ${}_{82}^{207}\text{Pb}$ and ${}_{82}^{208}\text{Pb}$.

Answer to Problem 26E

The difference in mass defects of ${}_{82}^{207}\text{Pb}$ and ${}_{82}^{208}\text{Pb}$ is $0.00797\text{u}$.

Explanation of Solution

Write the expression for the mass defect of a nucleus.

    $\Delta m=n\left(m_n\right)+p\left(m_p\right)-m$        (I)

Here, $\Delta m$ is the mass defect, $n$ is the number of neutrons, $p$ is the number of protons, $m_n$ is the mass of neutron, $m_p$ is the mass of proton and $m$ is the atomic mass.

Consider that $\Delta m_1$ is the mass defect of ${}_{82}^{208}\text{Pb}$ and $\Delta m_2$ is the mass defect of ${}_{82}^{207}\text{Pb}$.

Conclusion:

Substitute $1.008665\text{u}$ for $m_n$, $1.007825\text{u}$ for $m_p$ and $207.9766\text{u}$ for $m$ in expression (I).

    $\begin{array}{c}\Delta m_1=\left(208-82\right)\left(1.008665\text{u}\right)+82\left(1.007825\text{u}\right)-207.9766\text{u}\\ =1.75684\text{}\text{u}\end{array}$

Substitute $1.008665\text{u}$ for $m_n$, $1.007825\text{u}$ for $m_p$ and $206.9759\text{u}$ for $m$ in expression (I).

    $\begin{array}{c}\Delta m_2=\left(207-82\right)\left(1.008665\text{u}\right)+82\left(1.007825\text{u}\right)-206.9759\text{u}\\ =1.74887\text{}\text{u}\end{array}$

Calculate the difference between $\Delta m_1$ and $\Delta m_2$.

    $\begin{array}{c}\Delta m_1-\Delta m_2=\left(1.75684\text{}\text{u}\right)-\left(1.74887\text{}\text{u}\right)\\ =0.00797\text{u}\end{array}$

Thus, the difference in mass defects of ${}_{82}^{207}\text{Pb}$ and ${}_{82}^{208}\text{Pb}$ is $0.00797\text{u}$.

(b)

To determine

The minimum energy required to remove a neutron from ${}_{82}^{208}\text{Pb}$.

Answer to Problem 26E

The minimum energy required to remove a neutron from ${}_{82}^{208}\text{Pb}$ is $7.86\text{MeV}$.

Explanation of Solution

Write the expression for the binding energy.

    $E=\Delta m\left(\frac{931\text{MeV}}{1\text{u}}\right)$        (II)

Here, $E$ is the total binding energy.

Write the expression of average binding energy.

    $E_{\text{av}}=\frac{E}{A}$        (III)

Here, $A$ is the mass number.

Conclusion:

Substitute $1.75684\text{}\text{u}$ for $\Delta m$ in expression (II).

    $\begin{array}{c}E=\left(1.75684\text{}\text{u}\right)\left(\frac{931\text{MeV}}{1\text{u}}\right)\\ =1635.62\text{}\text{MeV}\end{array}$

Substitute $1635.62\text{}\text{MeV}$ for $E$ and $208$ for $A$ in expression (III).

    $\begin{array}{c}{E}_{\text{av}}=\frac{1635.62\text{}\text{MeV}}{208}\\ =7.86\text{MeV}\end{array}$

Thus, the minimum energy required to remove a neutron from ${}_{82}^{208}\text{Pb}$ is $7.86\text{MeV}$.