Chapter 13, Problem 20P

(a)

To determine

Binding energy per nucleon for  ${}_{29}^{64}\text{Cu}$ and ${}_{30}^{64}\text{Zn}$ using equation 13.7.

Answer to Problem 20P

Binding energy per nucleon for  ${}_{29}^{64}\text{Cu}$ and ${}_{30}^{64}\text{Zn}$ are $\text{8}\text{.79 }\text{MeV}/\text{nucleon}$ and $\text{8}\text{.75 }\text{MeV}/\text{nucleon}$ respectively.

Explanation of Solution

Write the semiempirical binding energy formula.

    ${\text{E}}_b=C_{1}A-C_2{A}^{\frac{2}{3}}-C_3\frac{Z\left(Z-1\right)}{A^{\frac{1}{3}}}-C_4\frac{{\left(A-Z\right)}^2}{A}$

Here, ${\text{E}}_b$ is the binding energy, $Z$ is the number of electrons, $A$ is the mass number and $C_1,C_2,C_3,C_4$ are constants.

Write the equation for binding energy per nucleon.

  $E_{\text{per}\text{nucleon}}\text{=}\frac{E_b}{A}$        (I)

Conclusion:

Consider the case of ${}_{29}^{64}\text{Cu}$.

Substitute $15.7\text{ MeV}$ for $C_1$, $17.8\text{ MeV}$ for $C_2$, $0.71\text{ MeV}$ for $C_3$, $23.6\text{ MeV}$ for $C_4$, 29 for $Z$ and 64 for $A$ in the equation for $E_b$.

  $\begin{array}{c}{E}_b=\left(15.7\text{ MeV}\right)\left(64\right)-\left(17.8\text{ MeV}\right){\left(64\right)}^{2/3}-\left(0.71\text{ MeV}\right)\left[\frac{\left(29\right)\left(28\right)}{{\left(64\right)}^{1/3}}\right]\\ -\left(23.6\text{ MeV}\right)\frac{\left[64-2{\left(29\right)}^2\right]}{64}\\ =562.6\text{ MeV }\end{array}$

Substitute $562.6\text{ MeV }$ for ${\text{E}}_b$ and 64 for $A$ in the equation for $E_{\text{per}\text{nucleon}}$.

  $\begin{array}{c}{E}_b{}_{\text{per}\text{nucleon}}\text{=}\frac{562.6\text{ MeV }}{64}\\ =\text{8}\text{.79 }\text{MeV}/\text{nucleon}\end{array}$

Consider the case of ${}_{30}^{64}\text{Zn}$.

Substitute $15.7\text{ MeV}$ for $C_1$, $17.8\text{ MeV}$ for $C_2$, $0.71\text{ MeV}$ for $C_3$, $23.6\text{ MeV}$ for $C_4$, 30 for $Z$ and 64 for $A$ in the equation for ${\text{E}}_b$.

  $\begin{array}{c}{E}_b=\left(15.7\text{ MeV}\right)\left(64\right)-\left(17.8\text{ MeV}\right){\left(64\right)}^{2/3}-\left(0.71\text{ MeV}\right)\left[\frac{\left(30\right)\left(29\right)}{{\left(64\right)}^{1/3}}\right]\\ -\left(23.6\text{ MeV}\right)\frac{\left[64-2{\left(30\right)}^2\right]}{64}\\ =560\text{ MeV}\end{array}$

Substitute $560\text{ MeV }$ for ${\text{E}}_b$ and 64 for $A$ in the equation for $E_b{}_{\text{per}\text{nucleon}}$.

    $\begin{array}{c}{E}_{\text{per}\text{nucleon}}\text{=}\frac{560\text{ MeV }}{64}\\ =\text{8}\text{.75 }\text{MeV}/\text{nucleon}\end{array}$

Thus, the binding energy per nucleon for  ${}_{29}^{64}\text{Cu}$ and ${}_{30}^{64}\text{Zn}$ are $\text{8}\text{.79 }\text{MeV}/\text{nucleon}$ and $\text{8}\text{.75 }\text{MeV}/\text{nucleon}$ respectively.

(b)

To determine

Binding energy per nucleon for  ${}_{29}^{64}\text{Cu}$ and ${}_{30}^{64}\text{Zn}$ using equation 13.4 and compare with that of part (a).

Answer to Problem 20P

Binding energy per nucleon for  ${}_{29}^{64}\text{Cu}$ and ${}_{30}^{64}\text{Zn}$ are $\text{8}\text{.74 }\text{MeV}/\text{nucleon}$ and $\text{8}\text{.74 }\text{MeV}/\text{nucleon}$ respectively and result in part (b) is $0.6\%$ and $0.1\%$

Explanation of Solution

Write the equation 13.4 to find the binding energy per nucleon.

    $E_b=\left(Z{m}_{\text{electron}}-\left(A-Z\right)m_p-M\right)\left(931.5\text{MeV}/\text{u}\right)$

Here, $M$ is the atomic mass, $m_{\text{electron}}$ is the mass of electron and $m_p$ is the mass of proton.

Write the equation to find the % difference in values of $E_b$ obtained from two methods.

    $\Delta =\left(\frac{E_a-E_b}{E_a}\right)100\%$

Here, $E_a$ is the binding energy per nucleon obtained in part (a) and $E_b$ is the binding energy per nucleon obtained in part (b)

Conclusion:

Consider the case of ${}_{29}^{64}\text{Cu}$.

Substitute $1.007825\text{ u}$ for $m_{\text{electron}}$, 29 for $Z$ and 64 for $A$, $1.00865\text{ u}$ for $m_p$ and $63.929766\text{ u}$ for $M$ in the equation for $E_b$.

    $\begin{array}{c}{E}_b=\left(\left(29\right)\left(1.007825\text{ u}\right)-\left(64-29\right)\left(1.00865\text{ u}\right)-63.929766\text{ u}\right)\left(931.5\text{MeV}/\text{u}\right)\\ =559.3\text{ MeV}\end{array}$

Substitute $559.3\text{ MeV}$ for ${\text{E}}_b$ and 64 for $A$ in the equation for $E_{\text{per}\text{nucleon}}$.

    $\begin{array}{c}{E}_{\text{per}\text{nucleon}}\text{=}\frac{559.3\text{ MeV }}{64}\\ =\text{8}\text{.74 }\text{MeV}/\text{nucleon}\end{array}$

Substitute $\text{8}\text{.79 }\text{MeV}/\text{nucleon}$ for $E_a$ and $\text{8}\text{.74 }\text{MeV}/\text{nucleon}$ for $E_b$ in the equation for $\Delta$.

    $\begin{array}{c}\Delta =\left(\frac{\text{8}\text{.79 }\text{MeV}/\text{nucleon}-\text{8}\text{.74 }\text{MeV}/\text{nucleon}}{\text{8}\text{.79 }\text{MeV}/\text{nucleon}}\right)100\%\\ \approx 0.6\%\end{array}$

Consider the case of ${}_{30}^{64}\text{Zn}$.

Substitute $1.007825\text{ u}$ for $m_{\text{electron}}$, 30 for $Z$ and 64 for $A$, $1.00865\text{ u}$ for $m_p$ and $63.929766\text{ u}$ for $M$ in the equation for $E_b$.

    $\begin{array}{c}{E}_b=\left(\left(30\right)\left(m_{\text{electron}}\right)-\left(64-30\right)\left(1.00865\text{ u}\right)-63.929766\text{ u}\right)\left(931.5\text{MeV}/\text{u}\right)\\ =559.3\text{ MeV}\end{array}$

Substitute $559.3\text{ MeV}$ for ${\text{E}}_b$ and 64 for $A$ in the equation for $E_{\text{per}\text{nucleon}}$.

    $\begin{array}{c}{E}_{\text{per}\text{nucleon}}\text{=}\frac{559.3\text{ MeV }}{64}\\ =\text{8}\text{.74 }\text{MeV}/\text{nucleon}\end{array}$

Substitute $\text{8}\text{.75 }\text{MeV}/\text{nucleon}$ for $E_a$ and $\text{8}\text{.74 }\text{MeV}/\text{nucleon}$ for $E_b$ in the equation for $\Delta$.

    $\begin{array}{c}\Delta =\left(\frac{\text{8}\text{.75 }\text{MeV}/\text{nucleon}-\text{8}\text{.74 }\text{MeV}/\text{nucleon}}{\text{8}\text{.75 }\text{MeV}/\text{nucleon}}\right)100\%\\ \approx 0.1\%\end{array}$

Thus, the binding energy per nucleon for  ${}_{29}^{64}\text{Cu}$ and ${}_{30}^{64}\text{Zn}$ are $\text{8}\text{.74 }\text{MeV}/\text{nucleon}$ and $\text{8}\text{.74 }\text{MeV}/\text{nucleon}$ respectively and result in part (b) is $0.6\%$ and $0.1\%$