Chapter 30, Problem 11P

(a)

To determine

The binding energy per nucleon for ${}_1^2\text{H}$.

Answer to Problem 11P

The binding energy per nucleon for ${}_1^2\text{H}$ is $\underset{\_}{1.11\text{MeV}}$.

Explanation of Solution

Write the expression for the binding energy of the nucleus.

    $E_b\left(\text{MeV}\right)=\left[ZM\left(\text{H}\right)+N{m}_n-M\left({}_Z^{A}X\right)\right]\times 931.494\text{MeV}/\text{u}$        (I)

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

Write the expression for the binding energy per nucleon.

    $\text{Binding}\text{energy}\text{per}\text{nucleon}=\frac{E_b}{A}$        (II)

Use equation (I) in (II) to solve for binding energy per nucleon.

    $\text{Binding}\text{energy}\text{per}\text{nucleon}=\frac{\left[ZM\left(\text{H}\right)+N{m}_n-M\left({}_Z^{A}X\right)\right]\times 931.494\text{MeV}/\text{u}}{A}$        (III)

Write the expression for $N$.

    $N=A-Z$        (IV)

Use equation (IV) in (III) to solve for binding energy per nucleon.

    $\text{Binding}\text{energy}\text{per}\text{nucleon}=\frac{\left[ZM\left(\text{H}\right)+\left(A-Z\right)m_n-M\left({}_Z^{A}X\right)\right]\times 931.494\text{MeV}/\text{u}}{A}$        (V)

Conclusion:

Substitute $1$ for $Z$, $1.007825\text{u}$ for $M\left(\text{H}\right)$, $1.008665\text{u}$ for $m_n$, $2.014102\text{u}$ for $M\left({}_1^2\text{H}\right)$, $2$ for $A$ in equation (V) to find binding energy per nucleon for ${}_1^2\text{H}$.

    $\begin{array}{c}\text{Binding}\text{energy}\text{per}\text{nucleon}=\frac{\left[1\left(1.007825\text{u}\right)+\left(2-1\right)\left(1.008665\text{u}\right)-\left(2.014102\text{u}\right)\right]\times 931.494\text{MeV}/\text{u}}{2}\\ =\left(\frac{0.002388\text{u}}{2}\right)931.494\text{MeV}/\text{u}\\ =1.11\text{MeV}\end{array}$

Therefore, the binding energy per nucleon for ${}^2\text{H}$ is $\underset{\_}{1.11\text{MeV}}$.

(b)

To determine

The binding energy per nucleon for ${}_2^4\text{He}$.

Answer to Problem 11P

The binding energy per nucleon for ${}_2^4\text{He}$ is $\underset{\_}{7.07\text{MeV}}$.

Explanation of Solution

Use equation (III) to solve for binding energy per nucleon for ${}_2^4\text{He}$.

Conclusion:

Substitute $2$ for $Z$, $1.007825\text{u}$ for $M\left(\text{H}\right)$, $1.008665\text{u}$ for $m_n$, $4.002603\text{u}$ for $M\left({}_2^4\text{He}\right)$, $4$ for $A$ in equation (III) to find binding energy per nucleon for ${}_2^4\text{He}$.

    $\begin{array}{c}\text{Binding}\text{energy}\text{per}\text{nucleon}=\frac{\left[2\left(1.007825\text{u}\right)+\left(4-2\right)\left(1.008665\text{u}\right)-\left(4.002603\text{u}\right)\right]\times 931.494\text{MeV}/\text{u}}{4}\\ =\left(\frac{0.030377\text{u}}{4}\right)931.494\text{MeV}/\text{u}\\ =7.07\text{MeV}\end{array}$

Therefore, the binding energy per nucleon for ${}_2^4\text{He}$ is $\underset{\_}{7.07\text{MeV}}$.

(c)

To determine

The binding energy per nucleon for ${}_{26}^{56}\text{Fe}$.

Answer to Problem 11P

The binding energy per nucleon for ${}_{26}^{56}\text{Fe}$ is $\underset{\_}{8.79\text{MeV}}$.

Explanation of Solution

Use equation (V) to solve for binding energy per nucleon for ${}_{26}^{56}\text{Fe}$.

Conclusion:

Substitute $26$ for $Z$, $1.007825\text{u}$ for $M\left(\text{H}\right)$, $1.008665\text{u}$ for $m_n$, $55.934942\text{u}$ for $M\left({}_{26}^{56}\text{Fe}\right)$, $56$ for $A$ in equation (III) to find binding energy per nucleon for ${}_{26}^{56}\text{Fe}$.

    $\begin{array}{c}\text{Binding}\text{energy}\text{per}\text{nucleon}=\frac{\left[26\left(1.007825\text{u}\right)+\left(56-26\right)\left(1.008665\text{u}\right)-\left(55.934942\text{u}\right)\right]\times 931.494\text{MeV}/\text{u}}{56}\\ =\left(\frac{0.528458\text{u}}{56}\right)931.494\text{MeV}/\text{u}\\ =8.79\text{MeV}\end{array}$

Therefore, the binding energy per nucleon for ${}_{26}^{56}\text{Fe}$ is $\underset{\_}{8.79\text{MeV}}$.

(d)

To determine

The binding energy per nucleon for ${}_{92}^{238}\text{U}$.

Answer to Problem 11P

The binding energy per nucleon for ${}_{92}^{238}\text{U}$ is $\underset{\_}{7.57\text{MeV}}$.

Explanation of Solution

Use equation (V) to solve for binding energy per nucleon for ${}_{92}^{238}\text{U}$.

Conclusion:

Substitute $92$ for $Z$, $1.007825\text{u}$ for $M\left(\text{H}\right)$, $1.008665\text{u}$ for $m_n$, $238.050783\text{u}$ for $M\left({}_{92}^{238}\text{U}\right)$, $238$ for $A$ in equation (III) to find binding energy per nucleon for ${}_{92}^{238}\text{U}$.

    $\begin{array}{c}\text{Binding}\text{energy}\text{per}\text{nucleon}=\frac{\left[92\left(1.007825\text{u}\right)+\left(238-92\right)\left(1.008665\text{u}\right)-\left(238.050783\text{u}\right)\right]\times 931.494\text{MeV}/\text{u}}{238}\\ =\left(\frac{1.934207\text{u}}{238}\right)931.494\text{MeV}/\text{u}\\ =7.57\text{MeV}\end{array}$

Therefore, the binding energy per nucleon for ${}_{92}^{238}\text{U}$ is $\underset{\_}{7.57\text{MeV}}$.