Chapter 13, Problem 13P

(a)

To determine

Among the three isobars which has the highest neutron-proton ratio.

Answer to Problem 13P

Among the three isobars ${}_{55}^{139}\text{Cs}$ has the highest neutron-proton ratio equal to $\underset{\_}{1.53}$.

Explanation of Solution

Write the expression for neutron to proton ratio.

    $\text{neutron to proton ratio}=\frac{A-Z}{Z}$        (I)

Here, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $55$ for $Z$, and $139$ for $A$ in equation (I), to find neutron to proton ratio of ${}_{55}^{139}\text{Cs}$.

    $\begin{array}{c}\text{neutron to proton ratio}=\frac{139-55}{55}\\ =1.53\end{array}$

Substitute $57$ for $Z$, and $139$ for $A$ in equation (I), to find neutron to proton ratio of ${}_{57}^{139}\text{La}$.

    $\begin{array}{c}\text{neutron to proton ratio}=\frac{139-57}{57}\\ =1.43\end{array}$

Substitute $59$ for $Z$, and $139$ for $A$ in equation (I), to find neutron to proton ratio of ${}_{59}^{139}\text{Pr}$.

    $\begin{array}{c}\text{neutron to proton ratio}=\frac{139-59}{59}\\ =1.35\end{array}$

Therefore, among the three isobars ${}_{55}^{139}\text{Cs}$ has the highest neutron-proton ratio equal to $\underset{\_}{1.53}$.

(b)

To determine

Among the three isobars which has the highest binding energy per nucleon.

Answer to Problem 13P

Among the three isobars ${}_{57}^{139}\text{La}$ has the highest binding energy per nucleon equal to $\underset{\_}{8.353\text{MeV}}$.

Explanation of Solution

Write the expression for binding energy per nucleon.

    $\frac{E_b}{A}=\frac{1}{A}\left[C_{1}A-C_2{A}^{2/3}-C_3\left(Z\left(Z-1\right)\right)A^{-1/3}-C_4\frac{{\left(N-Z\right)}^2}{A}\right]$

Here, $\frac{E_b}{A}$ is the binding energy per nucleon, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $15.7$ for $C_1$, $17.8$ for $C_2$, $0.71$ for $C_3$, $23.6$ for $C_4$, $139$ for $A$, $59$ for $Z$  and $80$ for $N$ in equation (I), to find $\frac{E_b}{A}$ for ${}_{59}^{139}\text{Pr}$.

    $\begin{array}{c}\frac{E_b}{A}=\frac{1}{139}\left[\left(15.7\right)\left(139\right)-\left(17.8\right){\left(139\right)}^{2/3}-0.71\left(59\right)\left(58\right){\left(139\right)}^{1/3}-\frac{23.6{\left(80-59\right)}^2}{139}\right]\\ \frac{1}{139}\left[\left(15.7\right)\left(139\right)-\left(17.8\right){\left(139\right)}^{2/3}-0.71\left(59\right){\left(139\right)}^{1/3}-\frac{23.6{\left(21\right)}^2}{139}\right]\\ =\frac{1160.8\text{MeV}}{139}\\ =8.351\text{MeV}\end{array}$

Substitute $15.7$ for $C_1$, $17.8$ for $C_2$, $0.71$ for $C_3$, $23.6$ for $C_4$, $139$ for $A$, $59$ for $Z$  and $82$ for $N$ in equation (I), to find $\frac{E_b}{A}$ for ${}_{57}^{139}\text{La}$.

    $\begin{array}{c}\frac{E_b}{A}=\frac{1}{139}\left[\left(15.7\right)\left(139\right)-\left(17.8\right){\left(139\right)}^{2/3}-0.71\left(55\right)\left(54\right){\left(139\right)}^{1/3}-\frac{23.6{\left(82-57\right)}^2}{139}\right]\\ \frac{1}{139}\left[\left(15.7\right)\left(139\right)-\left(17.8\right){\left(139\right)}^{2/3}-0.71\left(59\right){\left(139\right)}^{1/3}-\frac{23.6{\left(25\right)}^2}{139}\right]\\ =\frac{1161.1\text{MeV}}{139}\\ =8.355\text{MeV}\end{array}$

Substitute $15.7$ for $C_1$, $17.8$ for $C_2$, $0.71$ for $C_3$, $23.6$ for $C_4$, $139$ for $A$, $59$ for $Z$  and $82$ for $N$ in equation (I), to find $\frac{E_b}{A}$ for ${}_{55}^{139}\text{Cs}$.

    $\begin{array}{c}\frac{E_b}{A}=\frac{1}{139}\left[\left(15.7\right)\left(139\right)-\left(17.8\right){\left(139\right)}^{2/3}-0.71\left(55\right)\left(54\right){\left(139\right)}^{1/3}-\frac{23.6{\left(84-55\right)}^2}{139}\right]\\ \frac{1}{139}\left[\left(15.7\right)\left(139\right)-\left(17.8\right){\left(139\right)}^{2/3}-0.71\left(59\right){\left(139\right)}^{1/3}-\frac{23.6{\left(29\right)}^2}{139}\right]\\ =\frac{1154.9\text{MeV}}{139}\\ =8.308\text{MeV}\end{array}$

Therefore, among the three isobars ${}_{57}^{139}\text{La}$ has the highest binding energy per nucleon equal to $\underset{\_}{8.353\text{MeV}}$.

(c)

To determine

Among the two radioactive nuclei $\left({}^{139}\mathrm{Pr}{\text{and}}^{139}\text{Cs}\right)$ which one is heavier.

Answer to Problem 13P

Among the two radioactive nuclei ${}^{139}\text{Cs}$ is more heavier.

Explanation of Solution

The mass of neutron is greater than the mass of proton , thus the atom with more number of nucleon $\left(N\right)$  and smallest atomic number $\left(Z\right)$ is more heavier.

Therefore ${}_{55}^{139}\text{Cs}$ is more heavier having a mass of $138.913\text{u}$.

Conclusion:

Therefore, among the two radioactive nuclei ${}^{139}\text{Cs}$ is more heavier.