Chapter 30, Problem 52A

(a)

To determine

To Find: The mass defect of the given isotope.

Explanation of Solution

Given:

Nuclear mass of isotope $\text{S}$ is $\text{31}\text{.97207}\text{u}$ .

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.

Calculations:

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

  $\begin{array}{l}\Delta m=\left[\left\{Z{m}_p+\left(A-Z\right)m_n\right\}-M\right]\\ \Delta m=\left[\left\{16\left(1.00727647\text{u}\right)+\left(32-16\right)\left(1.008665\text{u}\right)\right\}-31.97207\text{u}\right]\\ \Delta m=16.11642352+16.13864-31.97207\text{u}\\ \Delta m\text{= 0}\text{.28299352}\text{u}\end{array}$

Conclusion:

Thus, the mass defect of the isotope is $\Delta m\text{= 0}\text{.28299352}\text{u}$ .

(b)

To determine

To Find: The binding energy of the nucleus of the given isotope.

Explanation of Solution

Given:

Nuclear mass of isotope $\text{S}$ is $\text{31}\text{.97207}\text{u}$ .

The mass defect of the isotope is $\Delta m\text{= 0}\text{.28299352}\text{u}$ .

Formula Used:

Binding energy can be obtained by:

  $B.E.=\Delta m{c}^2$

Here, $\Delta m$ is the mass defect and $c$ is the speed of light.

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

Calculations:

Binding energy of the nucleus of the given isotope can be calculated as follows:

  $\begin{array}{l}B.E.\left(\text{MeV}\right)=\left(\frac{931.5\text{MeV}}{\text{1}\text{u}}\right)\left(\text{0}\text{.28299352}\text{u}\right)\\ B.E.\approx 263.61\text{MeV}\end{array}$

Conclusion:

Thus, the binding energy is $263.61\text{MeV}$ .

(c)

To determine

To Find: The binding energy per nucleon of the nucleus of the given isotope.

Explanation of Solution

Given:

The binding energy of the nucleus of the isotope is $263.61\text{MeV}$ .

Formula Used:

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

Here, A is the mass number of the isotope.

Calculations:

Binding energy per nucleon of the nucleus of the given isotope can be calculated as follows:

  $\frac{B.E}{A}=\frac{263.61\text{MeV}}{32}=8.24\text{MeV}$

Conclusion:

Thus, the binding energy per nucleon is $8.24\text{MeV}$ .