Chapter 21, Problem 21.49QE

(a)

Interpretation Introduction

Interpretation:

Binding energy per nucleon for the three isotopes of aluminium has to be calculated.

Concept Introduction:

Binding energy is a short strong force that is present in the nucleus which holds the protons together by overcoming the electrostatic repulsive forces between them.  Whenever there is a change in energy, a corresponding change in mass is also observed and this can be given by the equation shown below,

    $\Delta E=(\Delta m)c^2$

When more particles combine to form nuclear there is a great change in mass and energy.  The nuclear stabilities can be compared more appropriately by dividing the binding energy of nucleus with the number of nucleons.  The result obtained is the binding energy per nucleon.  Protons and neutrons are known as nucleons.  Binding energy is represented as $E_b$.

Explanation of Solution

Binding energy per nucleon for ${}^{\text{26}}\text{Al}$:

Atomic number of aluminium is 13.  This means there are 13 protons and 13 neutrons in the given isotope.

The change in mass can be calculated as shown below,

  $\begin{array}{c}\Delta m=(13\times 1.007\text{825}\text{g/mol}\text{H}\text{)}\text{+}(13\times 1.008665\text{g/mol}\text{n}\text{)}-\text{25}\text{.9869}\text{g/mol}{}^{\text{26}}\text{Al}\\ \text{=13}\text{.101725}\text{g/mol}\text{+}\text{13}\text{.112645}\text{g/mol}-\text{25}\text{.9869}\text{g/mol}\\ =\text{26}\text{.21437}\text{g/mol}-\text{25}\text{.9869}\text{g/mol}\\ =\text{0}\text{.22747}\text{g/mol}\end{array}$

Nuclear binding energy can be calculated in megaelectron volts as shown below,

  $\begin{array}{c}\Delta E=(\Delta m)\times 931.5MeV\\ =\text{0}\text{.22747}\times 931.5MeV\\ =\text{211}\text{.888305}MeV\\ =211.89MeV\end{array}$

Binding energy per nucleon can be calculated as shown below,

There is a total of 26 nucleons in aluminium-26.  Hence, the binding energy per nucleon can be calculated as,

    $\begin{array}{c}{E}_b=\frac{211.89MeV}{\text{26}\text{mol}\text{nucleons}}\\ =\text{8}\text{.150}MeV\text{/nucleon}\end{array}$

Binding energy per nucleon in ${}^{\text{26}}\text{Al}$ is $\text{8}\text{.150}MeV\text{/nucleon}$.

Binding energy per nucleon for ${}^{\text{27}}\text{Al}$:

Atomic number of aluminium is 13.  This means there are 13 protons and 14 neutrons in the given isotope.

The change in mass can be calculated as shown below,

  $\begin{array}{c}\Delta m=(13\times 1.007\text{825}\text{g/mol}\text{H}\text{)}\text{+}(14\times 1.008665\text{g/mol}\text{n}\text{)}-\text{26}\text{.9815}\text{g/mol}{}^{\text{27}}\text{Al}\\ \text{=13}\text{.101725}\text{g/mol}\text{+}\text{14}\text{.12131}\text{g/mol}-\text{26}\text{.9815}\text{g/mol}\\ =\text{27}\text{.223035}\text{g/mol}-\text{26}\text{.9815}\text{g/mol}\\ =\text{0}\text{.241535}\text{g/mol}\end{array}$

Nuclear binding energy can be calculated in megaelectron volts as shown below,

  $\begin{array}{c}\Delta E=(\Delta m)\times 931.5MeV\\ =\text{0}\text{.241535}\times 931.5MeV\\ =\text{224}\text{.9898525}MeV\\ =224.99MeV\end{array}$

Binding energy per nucleon can be calculated as shown below,

There is a total of 27 nucleons in aluminium-27.  Hence, the binding energy per nucleon can be calculated as,

    $\begin{array}{c}{E}_b=\frac{224.99MeV}{\text{27}\text{mol}\text{nucleons}}\\ =\text{8}\text{.33}MeV\text{/nucleon}\end{array}$

Binding energy per nucleon in ${}^{\text{27}}\text{Al}$ is $\text{8}\text{.33}MeV\text{/nucleon}$.

Binding energy per nucleon for ${}^{\text{28}}\text{Al}$:

Atomic number of aluminium is 13.  This means there are 13 protons and 15 neutrons in the given isotope.

The change in mass can be calculated as shown below,

  $\begin{array}{c}\Delta m=(13\times 1.007\text{825}\text{g/mol}\text{H}\text{)}\text{+}(15\times 1.008665\text{g/mol}\text{n}\text{)}-\text{27}\text{.9819}\text{g/mol}{}^{\text{28}}\text{Al}\\ \text{=13}\text{.101725}\text{g/mol}\text{+}\text{15}\text{.129975}\text{g/mol}-\text{27}\text{.9819}\text{g/mol}\\ =\text{28}\text{.2317}\text{g/mol}-\text{27}\text{.9819}\text{g/mol}\\ =\text{0}\text{.2498}\text{g/mol}\end{array}$

Nuclear binding energy can be calculated in megaelectron volts as shown below,

  $\begin{array}{c}\Delta E=(\Delta m)\times 931.5MeV\\ =\text{0}\text{.2498}\times 931.5MeV\\ =\text{232}\text{.6887}MeV\\ =\text{232}\text{.69}MeV\end{array}$

Binding energy per nucleon can be calculated as shown below,

There is a total of 28 nucleons in aluminium-28.  Hence, the binding energy per nucleon can be calculated as,

    $\begin{array}{c}{E}_b=\frac{232.9MeV}{\text{28}\text{mol}\text{nucleons}}\\ =\text{8}\text{.31}MeV\text{/nucleon}\end{array}$

Binding energy per nucleon in ${}^{\text{28}}\text{Al}$ is $\text{8}\text{.31}MeV\text{/nucleon}$.

(b)

Interpretation Introduction

Interpretation:

The isotope that is stable and the isotopes that are radioactive has to be identified by comparing the binding energy of the three isotopes of aluminium has to be given.

Explanation of Solution

Binding energy per nucleon in ${}^{\text{26}}\text{Al}$ is $\text{8}\text{.150}MeV\text{/nucleon}$.  Binding energy per nucleon in ${}^{\text{27}}\text{Al}$ is $\text{8}\text{.33}MeV\text{/nucleon}$.  Binding energy per nucleon in ${}^{\text{28}}\text{Al}$ is $\text{8}\text{.31}MeV\text{/nucleon}$.

It is known that the nucleus that has greater binding energy will be more stable.  Therefore, the isotope that is stable in the three isotopes of aluminium is ${}^{\text{27}}\text{Al}$.  The two nuclides that are radioactive are ${}^{\text{26}}\text{Al}$ and ${}^{\text{28}}\text{Al}$.