Chapter 44, Problem 16P

(a)

To determine

The difference in binding energy per nucleon for nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$.

Answer to Problem 16P

The difference in binding energy per nucleon for nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$ is $0.210\text{ MeV}$.

Explanation of Solution

Write the expression to find the binding energy of a nucleus.

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

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

Re-write the expression to find the binding energy of a nucleus for $\text{N}\text{a}$.

  $E_b\left(\text{N}\text{a}\right)=\left[11M\left(\text{H}\right)+12m_n-M\left(\text{N}\text{a}\right)\right]\left(931.494\text{MeV/u}\right)$

Substitute $1.007825\text{ u}$ for $M\left(\text{H}\right)$, $1.008665\text{ u}$ for $m_n$, $22.989769\text{ u}$ for $M\left(\text{N}\text{a}\right)$ to find the binding energy of $\text{N}\text{a}$.

  $\begin{array}{c}{E}_b\left(\text{N}\text{a}\right)=\left[11\left(1.007825\text{ u}\right)+12\left(1.008665\text{ u}\right)-22.989769\text{ u}\right]\\ \times \left(931.494\text{ MeV/u}\right)\\ =186.565\text{ MeV}\end{array}$

Divide binding energy with mass number to find the difference in binding energy per nucleon for $\text{N}\text{a}$.

  $\begin{array}{c}\frac{E_b}{A}=\frac{186.565\text{ MeV}}{23}\\ =8.11\text{ MeV}\end{array}$

Re-write the expression to find the binding energy of a nucleus for $\text{M}\text{g}$.

  $E_b\left(\text{M}\text{g}\right)=\left[12M\left(\text{H}\right)+11m_n-M\left(\text{M}\text{g}\right)\right]\left(931.494\text{MeV/u}\right)$

Substitute $1.007825\text{ u}$ for $M\left(\text{H}\right)$, $1.008665\text{ u}$ for $m_n$, $22.994124\text{ u}$ for $M\left(\text{M}\text{g}\right)$ to find the binding energy of $\text{M}\text{g}$.

  $\begin{array}{c}{E}_b=E_b\left(\text{M}\text{g}\right)\\ =\left[12M\left(\text{H}\right)+11m_n-M\left(\text{M}\text{g}\right)\right]\left(931.494\text{MeV/u}\right)\\ =\left[12\left(1.007825\text{ u}\right)+11\left(1.008665\text{ u}\right)-22.994124\text{ u}\right]\times \left(931.494\text{MeV/u}\right)\\ =181.726\text{ MeV}\end{array}$

Divide binding energy with mass number to find the difference in binding energy per nucleon for $\text{M}\text{g}$.

  $\begin{array}{c}\frac{E_b}{A}=\frac{181.726\text{ MeV}}{23}\\ =7.90\text{ MeV}\end{array}$

Write the expression to find the difference in binding energy per nucleon for nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$.

  $\frac{\Delta E_b}{A}=\frac{E_b\left(\text{N}\text{a}\right)-E_b\left(\text{M}\text{g}\right)}{A}$

Here, $\frac{\Delta E_b}{A}$ is the difference in binding energy per nucleon.

Conclusion:

Substitute $8.11\text{ MeV}$ for $\frac{E_b\left(\text{N}\text{a}\right)}{A}$ and $7.90\text{ MeV}$ for $\frac{E_b\left(\text{M}\text{g}\right)}{A}$ to find the difference in binding energy per nucleon for nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$.

  $\begin{array}{c}\frac{\Delta E_b}{A}=8.11\text{ MeV}-7.90\text{ MeV}\\ =0.210\text{ MeV}\end{array}$

Therefore, the difference in binding energy per nucleon for nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$ is $0.210\text{ MeV}$.

(b)

To determine

The findings from the difference in binding energy per nucleon for the nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$

Answer to Problem 16P

The $\text{N}\text{a}$ is more stable nucleus than $\text{M}\text{g}$.

Explanation of Solution

Write the expression to find the difference in binding energy per nucleon for nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$.

  $\frac{\Delta E_b}{A}=\frac{E_b\left(\text{N}\text{a}\right)-E_b\left(\text{M}\text{g}\right)}{A}$

Here, $\frac{\Delta E_b}{A}$ is the difference in binding energy per nucleon.

Substitute $8.11\text{ MeV}$ for $\frac{E_b\left(\text{N}\text{a}\right)}{A}$ and $7.90\text{ MeV}$ for $\frac{E_b\left(\text{M}\text{g}\right)}{A}$ to find the difference in binding energy per nucleon for nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$.

  $\begin{array}{c}\frac{\Delta E_b}{A}=8.11\text{ MeV}-7.90\text{ MeV}\\ =0.210\text{ MeV}\end{array}$

Conclusion:

Therefore, the difference in binding energy per nucleon for nuclei $\text{N}\text{a}$ and $\text{M}\text{g}$ is $0.210\text{ MeV}$. It indicates that the binding energy per nucleon is higher in $\text{N}\text{a}$ by $0.210\text{ MeV}$.

Thus, In $\text{N}\text{a}$ the proton repulsion will be less compared to $\text{M}\text{g}$. therefore, the $\text{N}\text{a}$ is more stable nucleus than $\text{M}\text{g}$.