Chapter 44, Problem 15P

(a)

To determine

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

Answer to Problem 15P

The binding energy per nucleon for ${}^2\text{H}$ is $\text{1}\text{.11 MeV}$.

Explanation of Solution

Write the equation for binding energy per nucleon.

    $\frac{E_b}{A}=\frac{\left[ZM\left(\text{H}\right)+N{m}_n-M\left(\text{X}\right)\right]}{A}\left(\frac{931.5\text{ MeV}}{\text{u}}\right)$                                                    (I)

Here, $E_b$ is the binding energy, $A$ is the mass number, $Z$ is the atomic number, $M\left(\text{H}\right)$ is the atomic mass of the neutral hydrogen atom, $N$ is the number of neutrons, $m_n$ is the mass of neutron and $M\left(\text{X}\right)$ is the mass of the isotope.

Conclusion:

Substitute $2$ for $A$, $1$ for $Z$, $1.007 825 \text{u}$ for $M\left(\text{H}\right)$, $1$ for $N$, $1.008 665 \text{u}$ for $m_n$ and $2.014 102 \text{u}$ for $M\left(\text{X}\right)$ in equation I.

    $\begin{array}{c}\frac{E_b}{A}=\frac{\left[\left(1\right)\left(1.007 825 \text{u}\right)+\left(1\right)\left(1.008 665 \text{u}\right)-2.014 102 \text{u}\right]}{2}\left(\frac{931.5\text{ MeV}}{\text{u}}\right)\\ =1.11\text{ MeV}\end{array}$

Therefore, the binding energy per nucleon for ${}^2\text{H}$ is $\text{1}\text{.11 MeV}$.

(b)

To determine

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

Answer to Problem 15P

The binding energy per nucleon for ${}^4\text{He}$ is $\text{7}\text{.07 MeV}$.

Explanation of Solution

Use the previous section equation.

Conclusion:

Substitute $4$ for $A$, $2$ for $Z$, $1.007 825 \text{u}$ for $M\left(\text{H}\right)$, $2$ for $N$, $1.008 665 \text{u}$ for $m_n$ and $4.002 603 \text{u}$ for $M\left(\text{X}\right)$ in equation I.

    $\begin{array}{c}\frac{E_b}{A}=\frac{\left[\left(2\right)\left(1.007 825 \text{u}\right)+\left(2\right)\left(1.008 665 \text{u}\right)-4.002 603 \text{u}\right]}{4}\left(\frac{931.5\text{ MeV}}{\text{u}}\right)\\ =7.07\text{ MeV}\end{array}$

Therefore, the binding energy per nucleon for ${}^4\text{He}$ is $\text{7}\text{.07 MeV}$.

(c)

To determine

Find the binding energy per nucleon for ${}^{56}\text{Fe}$.

Answer to Problem 15P

The binding energy per nucleon for ${}^{56}\text{Fe}$ is $\text{8}\text{.79 MeV}$.

Explanation of Solution

Use the previous section equation.

Conclusion:

Substitute $56$ for $A$, $26$ for $Z$, $1.007 825 \text{u}$ for $M\left(\text{H}\right)$, $30$ for $N$, $1.008 665 \text{u}$ for $m_n$ and $55.934 942 \text{u}$ for $M\left(\text{X}\right)$ in equation I.

    $\begin{array}{c}\frac{E_b}{A}=\frac{\left[\left(26\right)\left(1.007 825 \text{u}\right)+\left(30\right)\left(1.008 665 \text{u}\right)-55.934 942 \text{u}\right]}{56}\left(\frac{931.5\text{ MeV}}{\text{u}}\right)\\ =8.79\text{ MeV}\end{array}$

Therefore, the binding energy per nucleon for ${}^{56}\text{Fe}$ is $\text{8}\text{.79 MeV}$.

(d)

To determine

Find the binding energy per nucleon for ${}^{238}\text{U}$.

Answer to Problem 15P

The binding energy per nucleon for ${}^{238}\text{U}$ is $\text{7}\text{.57 MeV}$.

Explanation of Solution

Use the previous section equation.

Conclusion:

Substitute $238$ for $A$, $92$ for $Z$, $1.007 825 \text{u}$ for $M\left(\text{H}\right)$, $146$ for $N$, $1.008 665 \text{u}$ for $m_n$ and $238.050 783\text{ u}$ for $M\left(\text{X}\right)$ in equation I.

    $\begin{array}{c}\frac{E_b}{A}=\frac{\left[\left(92\right)\left(1.007 825 \text{u}\right)+\left(146\right)\left(1.008 665 \text{u}\right)-238.050 783\text{ u}\right]}{238}\left(\frac{931.5\text{ MeV}}{\text{u}}\right)\\ =7.57\text{ MeV}\end{array}$

Therefore, the binding energy per nucleon for ${}^{238}\text{U}$ is $\text{7}\text{.57 MeV}$.