Chapter 30, Problem 53A

(a)

To determine

To Find: The mass defect of the given isotope.

Explanation of Solution

Given:

Nuclear mass of isotope $\text{N}$ is $\text{12}\text{.0188}\text{u}$ .

Formula Used:

Mass defect can be obtained by:

  $\Delta m=\left[\left\{Z{m}_p+\left(A-Z\right)m_n\right\}-M\right]$

Here, Z is the atomic number, A is the mass number, $m_p$ is the mass of proton, $m_n$ is the mass of neutron and M is the mass of the isotope.

  $B.E.\left(\text{MeV}\right)=\frac{931.5\text{MeV}}{\text{1}\text{u}}\Delta m\left(\text{u}\right)$

Binding energy per nucleon can be obtained by: $\frac{B.E}{A}$

Here, A is the mass number of the isotope.

Calculations:

The mass defect of the $\text{N}$ can be calculated as follows:

  $\begin{array}{l}\Delta m=\left[\left\{Z{m}_p+\left(A-Z\right)m_n\right\}-M\right]\\ \Delta m=\left[\left\{7\left(1.00727647\text{u}\right)+\left(12-7\right)\left(1.008665\text{u}\right)\right\}-\text{12}\text{.0188}\text{u}\right]\\ \Delta m=12.09426029\text{u}-\text{12}\text{.0188}\text{u}\\ \Delta m\text{= 0}\text{.07546029}\text{u}\end{array}$

Binding energy can be calculated as follows:

  $\begin{array}{l}B.E.\left(\text{MeV}\right)=\frac{931.5\text{MeV}}{\text{1}\text{u}}\Delta m\left(\text{u}\right)\\ B.E.\left(\text{MeV}\right)=\frac{931.5\text{MeV}}{\text{1}\text{u}}\left(\text{ 0}\text{.07546029}\text{u}\right)\\ B.E.\left(\text{MeV}\right)=70.29\text{MeV}\end{array}$

Binding energy per nucleon can be calculated as follows:

  $\frac{B.E}{A}=\frac{70.29\text{MeV}}{12}=5.91\text{MeV}$

Conclusion:

Thus, the binding energy per nucleon of $\text{N}$ is $5.91\text{MeV}$ .

(b)

To determine

To Find: The one which requires more energy to separate a nucleon out of $\text{N}$ and $\text{N}$ .

Explanation of Solution

Given:

Nuclear mass of isotope $\text{N}$ is $\text{14}\text{.00307}\text{u}$ .

The binding energy per nucleon of $\text{N}$ is $5.91\text{MeV}$ .

Formula Used:

Mass defect can be obtained by:

  $\Delta m=\left[\left\{Z{m}_p+\left(A-Z\right)m_n\right\}-M\right]$

Here, Z is the atomic number, A is the mass number, $m_p$ is the mass of proton, $m_n$ is the mass of neutron and M is the mass of the isotope.

  $B.E.\left(\text{MeV}\right)=\frac{931.5\text{MeV}}{\text{1}\text{u}}\Delta m\left(\text{u}\right)$

Binding energy per nucleon can be obtained by: $\frac{B.E}{A}$

Here, A is the mass number of the isotope.

Calculations:

The mass defect of the $\text{N}$ can be calculated as follows:

  $\begin{array}{l}\Delta m=\left[\left\{Z{m}_p+\left(A-Z\right)m_n\right\}-M\right]\\ \Delta m=\left[\left\{7\left(1.00727647\text{u}\right)+\left(14-7\right)\left(1.008665\text{u}\right)\right\}-\text{14}\text{.00307}\text{u}\right]\\ \Delta m=14.11159029\text{u}-\text{14}\text{.00307}\text{u}\\ \Delta m\text{= 0}\text{.10852029}\text{u}\end{array}$

Binding energy can be calculated as follows:

  $\begin{array}{l}B.E.\left(\text{MeV}\right)=\frac{931.5\text{MeV}}{\text{1}\text{u}}\Delta m\left(\text{u}\right)\\ B.E.\left(\text{MeV}\right)=\frac{931.5\text{MeV}}{\text{1}\text{u}}\left(\text{0}\text{.10852029}\text{u}\right)\\ B.E.\left(\text{MeV}\right)=101.09\text{MeV}\end{array}$

Binding energy per nucleon can be calculated as follows:

  $\frac{B.E}{A}=\frac{101.09\text{MeV}}{14}=7.22\text{MeV}$

As the binding energy per nucleon of $\text{N}$ is $5.91\text{MeV}$ and that of $\text{N}$ is $7.22\text{MeV}$ , more energy would be required to separate nucleon from $\text{N}$ than $\text{N}$ .

Conclusion:

More energy would be required to separate nucleon from $\text{N}$ than $\text{N}$ .