Chapter 13, Problem 15P

(a)

To determine

The binding energy per nucleon for the nuclei ${}_{10}^{20}\text{Ne}$.

Answer to Problem 15P

The binding energy per nucleon for the nuclei ${}_{10}^{20}\text{Ne}$ is $\underset{\_}{8.03\text{MeV}/\text{nucleon}}$.

Explanation of Solution

Write the expression for the binding energy of the nuclei.

    $\begin{array}{c}{E}_{\text{b}}=\left[ZM\left(H\right)+\left(A-Z\right)m_n-M\left({}_Z^A\text{X}\right)\right]\\ E_{\text{b}}\left(\text{in eV}\right)=\left[ZM\left(H\right)+\left(A-Z\right)m_n-M\left({}_Z^A\text{X}\right)\right]\left(931.494\text{MeV}/\text{u}\right)\end{array}$        (I)

Here, $E_{\text{b}}$ is the binding energy of the nuclei, $Z$ is the atomic number, $A$ is the mass number, $M\left(H\right)$ is the mass of hydrogen, $M\left({}_Z^A\text{X}\right)$ is the mass of the element.

Using equation (I) to write the expression for the binding energy per nucleon.

    $\frac{E_{\text{b}}}{A}=\frac{\left[ZM\left(H\right)+\left(A-Z\right)m_n-M\left({}_Z^A\text{X}\right)\right]}{A}\left(931.494\text{MeV}/\text{u}\right)$        (II)

Conclusion:

Substitute $20$ for $A$, $10$ for $Z$, $1.007825\text{u}$ for $M\left(H\right)$, $1.008665\text{u}$ for $m_n$, $19.992436\text{u}$ for $M\left({}_{10}^{20}\text{Ne}\right)$ in equation (II) to find $\frac{E_{\text{b}}}{A}$ for ${}_{10}^{20}\text{Ne}$.

    $\begin{array}{c}\frac{E_{\text{b}}}{A}=\frac{\left[\left(10\right)\left(1.007825\text{u}\right)+\left(20-10\right)\left(1.008665\text{u}\right)-\left(19.992436\text{u}\right)\right]}{20}\left(931.494\text{MeV}/\text{u}\right)\\ =\frac{160.650}{20}\text{MeV}/\text{u}\\ =8.03\text{MeV}/\text{u}\end{array}$

Therefore, the binding energy per nucleon for the nuclei ${}_{10}^{20}\text{Ne}$ is $\underset{\_}{8.03\text{MeV}/\text{nucleon}}$.

(b)

To determine

The binding energy per nucleon for the nuclei ${}_{20}^{40}\text{Ca}$.

Answer to Problem 15P

The binding energy per nucleon for the nuclei ${}_{20}^{40}\text{Ca}$ is $\underset{\_}{8.55\text{MeV}/\text{nucleon}}$.

Explanation of Solution

Use equation (II) to solve for the $\frac{E_{\text{b}}}{A}$ for ${}_{20}^{40}\text{Ca}$.

Conclusion:

Substitute $40$ for $A$, $20$ for $Z$, $1.007825\text{u}$ for $M\left(H\right)$, $1.008665\text{u}$ for $m_n$, $39.9625914\text{u}$ for $M\left({}_{20}^{40}\text{Ca}\right)$ in equation (II) to find $\frac{E_{\text{b}}}{A}$ for ${}_{20}^{40}\text{Ca}$.

    $\begin{array}{c}\frac{E_{\text{b}}}{A}=\frac{\left[\left(20\right)\left(1.007825\text{u}\right)+\left(40-20\right)\left(1.008665\text{u}\right)-\left(39.9625914\text{u}\right)\right]}{40}\left(931.494\text{MeV}/\text{u}\right)\\ =\frac{342.053}{40}\text{MeV}/\text{u}\\ =8.55\text{MeV}/\text{u}\end{array}$

Therefore, the binding energy per nucleon for the nuclei ${}_{20}^{40}\text{Ca}$ is $\underset{\_}{8.55\text{MeV}/\text{nucleon}}$.

(c)

To determine

The binding energy per nucleon for the nuclei ${}_{41}^{93}\text{Nb}$.

Answer to Problem 15P

The binding energy per nucleon for the nuclei ${}_{41}^{93}\text{Nb}$ is $\underset{\_}{8.66\text{MeV}/\text{nucleon}}$.

Explanation of Solution

Use equation (II) to solve for the $\frac{E_{\text{b}}}{A}$ for ${}_{41}^{93}\text{Nb}$.

Conclusion:

Substitute $93$ for $A$, $41$ for $Z$, $1.007825\text{u}$ for $M\left(H\right)$, $1.008665\text{u}$ for $m_n$, $92.906377\text{u}$ for $M\left({}_{41}^{93}\text{Nb}\right)$ in equation (II) to find $\frac{E_{\text{b}}}{A}$ for ${}_{41}^{93}\text{Nb}$.

    $\begin{array}{c}\frac{E_{\text{b}}}{A}=\frac{\left[\left(41\right)\left(1.007825\text{u}\right)+\left(93-41\right)\left(1.008665\text{u}\right)-\left(92.906377\text{u}\right)\right]}{93}\left(931.494\text{MeV}/\text{u}\right)\\ =\frac{805.768}{93}\text{MeV}/\text{u}\\ =8.66\text{MeV}/\text{u}\end{array}$

Therefore, the binding energy per nucleon for the nuclei ${}_{41}^{93}\text{Nb}$ is $\underset{\_}{8.66\text{MeV}/\text{nucleon}}$.

(d)

To determine

The binding energy per nucleon for the nuclei ${}_{79}^{197}\text{Au}$.

Answer to Problem 15P

The binding energy per nucleon for the nuclei ${}_{79}^{197}\text{Au}$ is $\underset{\_}{7.92\text{MeV}/\text{nucleon}}$.

Explanation of Solution

Use equation (II) to solve for the $\frac{E_{\text{b}}}{A}$ for ${}_{79}^{197}\text{Au}$.

Conclusion:

Substitute $197$ for $A$, $79$ for $Z$, $1.007825\text{u}$ for $M\left(H\right)$, $1.008665\text{u}$ for $m_n$, $196.9665431\text{u}$ for $M\left({}_{79}^{197}\text{Au}\right)$ in equation (II) to find $\frac{E_{\text{b}}}{A}$ for ${}_{79}^{197}\text{Au}$.

    $\begin{array}{c}\frac{E_{\text{b}}}{A}=\frac{\left[\left(79\right)\left(1.007825\text{u}\right)+\left(197-79\right)\left(1.008665\text{u}\right)-\left(196.9665431\text{u}\right)\right]}{197}\left(931.494\text{MeV}/\text{u}\right)\\ =\frac{1559.416}{197}\text{MeV}/\text{u}\\ =7.92\text{MeV}/\text{u}\end{array}$

Therefore, the binding energy per nucleon for the nuclei ${}_{79}^{197}\text{Au}$ is $\underset{\_}{7.92\text{MeV}/\text{nucleon}}$.