Chapter 30.1, Problem 8PP

(a)

To determine

To calculate:The mass defect $\left(\Delta m\right)$ of isotope $O$ and mass defect $\left(\Delta m\right)$ of isotope $O$ .

Answer to Problem 8PP

Mass defect of $\left(\Delta m\right)$ isotope $O=0.137005\text{u}$

Mass defect of $\left(\Delta m\right)$ isotope $O=0.15009\text{u}$

Explanation of Solution

Given:

For isotope $O$ ,

Mass $\left(m\right)=15.994915\text{u}$

Atomic Mass Number $\left(A\right)=16$

Atomic Number $\left(Z\right)=8$

For isotope $O$, Mass $\left(m\right)=17.999160\text{u}$

Atomic Mass Number $\left(A\right)=18$

Atomic Number $\left(Z\right)=8$

Mass of hydrogen $=1.007825\text{u}$

Mass of neutron $=1.008665\text{u}$

  $1\text{u}=931.49\text{MeV}$

Formula used:

Formula to calculate mass defect, $\Delta m=\left[Zm\left(H\right)+\left(A-Z\right)m_n\right]-m_{atom}$

Where,

  $\Delta m:$ Mass defect

  $m\left(H\right)\text{:}$ Mass of hydrogen

  $m_n\text{:}$ Mass of neutron

  $m_{atom}\text{:}$ Mass of overall atom

  $A:$ Atomic mass number

  $Z:$ Atomic number

Calculation:

For isotope $O$ , To obtain the number of protons,

As the number of protons in the nucleus of the atom is equal to the atomic number $Z$ ,

Number of protons $\text{=8}$

Number of neutrons $=A-Z$

Substitute the values in above formula,

  $\begin{array}{l}\text{=16-8}\\ \text{=8}\end{array}$

Number of neutrons $\text{=8}$

Mass defect can be calculated by formula,

  $\Delta m=\left[Zm\left(H\right)+\left(A-Z\right)m_n\right]-m_{atom}$

Substituting the values in above formula,

  $\begin{array}{l}\Delta m=\left[\left(8\times 1.007825\text{u}\right)+\left(8\times 1.008665\text{u}\right)\right]-15.994915\text{u}\\ =\left[8.0626\text{u}+8.06932\text{u}\right]-15.994915\text{u}\\ =16.13192\text{u}-15.99491\text{u}\\ =0.137005\text{u}\end{array}$

For isotope $O$ ,

To obtain the number of protons,

As the number of protons in the nucleus of the atom is equal to the atomic number $Z$ ,

Number of protons $\text{=8}$

Number of neutrons $=A-Z$

Substitute the values in above formula,

  $\begin{array}{l}\text{=18-8}\\ \text{=10}\end{array}$

Number of neutrons $\text{=10}$

Mass defect can be calculated by formula,

  $\Delta m=\left[Zm\left(H\right)+\left(A-Z\right)m_n\right]-m_{atom}$

Substituting the values in above formula,

  $\begin{array}{l}\Delta m=\left[\left(8\times 1.007825\text{u}\right)+\left(10\times 1.008665\text{u}\right)\right]-17.999160\text{u}\\ =\left[8.0626\text{u}+10.08665\text{u}\right]-17.999160\text{u}\\ =18.14925\text{u}-17.999160\text{u}\\ =0.15009\text{u}\end{array}$

Conclusion:

The mass defect of the isotope $O$ is $0.137005\text{u}$ and the mass defect of the isotope $O$ is $0.15009\text{u}$ .

(b)

To determine

To calculate:Binding energy of isotope $O$ and Binding energy of isotope $O$ .

Answer to Problem 8PP

Binding energy of the isotope $O$ is $1.14856{\text{×10}}^{\text{19}}\text{MeV}$ .

Binding energy of the isotope $O$ is $1.25826{\text{×10}}^{\text{19}}\text{MeV}$ .

Explanation of Solution

Given:

Mass defect $(\Delta m)$ of isotope $O$ is $0.137005\text{u}$

Mass defect $(\Delta m)$ of isotope $O$ is $0.15009\text{u}$

Formula used:

Formula to calculate binding energy, $E=\Delta m\times c^2$

Where,

  $\Delta m:$ Mass defect

  $c:$ Speed of light

Calculation:

It is known that $\text{1}\text{u}=\text{931}\text{.49}\text{MeV}$

For isotope $O$ , Binding energy in $\text{MeV}$ can be given as,

  $E=\Delta m\times c^2$

Substituting the values in above formula,

  $\begin{array}{l}\text{=}\left(\text{0}\text{.137005}\text{u}\right)\text{×}{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}\\ \text{=}\left(\text{0}\text{.137005}\text{u}\right)\text{×}\left({\text{9×10}}^{\text{16}}\right)\\ \text{=1}{\text{.23304×10}}^{\text{16}}\text{u}\end{array}$

To convert binding energy in $\text{MeV}$ ,

  $\begin{array}{l}\text{=1}{\text{.23304×10}}^{\text{16}}\text{u}\times \left(931.49\text{MeV}\right)\\ =1.14856{\text{×10}}^{\text{19}}\text{MeV}\end{array}$

For isotope $O$ ,

Binding energy in $\text{MeV}$ can be given as,

  $E=\Delta m\times c^2$

Substituting the values in above formula,

  $\begin{array}{l}\text{=}\left(0.15009\text{u}\right)\text{×}{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}\\ \text{=}\left(0.15009\text{u}\right)\text{×}\left({\text{9×10}}^{\text{16}}\right)\\ \text{=1}{\text{.35081×10}}^{\text{16}}\text{u}\end{array}$

To convert binding energy in $\text{MeV}$ ,

  $\begin{array}{l}\text{=1}{\text{.35081×10}}^{\text{16}}\text{u}\times \left(931.49\text{MeV}\right)\\ =1.25826{\text{×10}}^{\text{19}}\text{MeV}\end{array}$

Conclusion:

The binding energy of the isotope $O$ is $1.14856{\text{×10}}^{\text{19}}\text{MeV}$ and of the isotope $O$ is

  $1.25826{\text{×10}}^{\text{19}}\text{MeV}$ .

(c)

To determine

To Explain:The reason for binding energies are different.

Answer to Problem 8PP

Binding energies are different as they possess different mass defects.

Explanation of Solution

Introduction:

Mass defect can be calculated by formula,

  $\Delta m=\left[Zm\left(H\right)+\left(A-Z\right)m_n\right]-m_{atom}$

Where,

  $\Delta m:$ Mass defect

  $m\left(H\right)\text{:}$ Mass of hydrogen0

  $m_n\text{:}$ Mass of neutron

  $m_{atom}\text{:}$ Mass of overall atom

  $A:$ Atomic mass number

  $Z:$ Atomic number Binding energy in $\text{MeV}$ can be given as,

  $E=\Delta m\times c^2$

Where,

  $\Delta m:$ Mass defect

  $c:$ Speed of light

Although the isotopes $O$ and $O$ have same number of protons but they have different atomic mass number.

Due to different atomic mass number, they differ in the mass defects.

Binding energy is dependent on mass defect hence as the mass defects are different of both the isotopes; the binding energies of the isotopes are different.

Conclusion:

Binding energies are different as they possess different mass defects.