Chapter 18, Problem 24QRT

Interpretation Introduction

Interpretation:

Binding energy per nucleon for the two isotopes of phosphorus has to be calculated and their stability has to be compared.

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

Given equations in the problem statement is,

    $\begin{array}{l}15\text{H}\text{+}15\text{n}\stackrel{}{\to }\text{P}\\ 15\text{H}\text{+}16\text{n}\stackrel{}{\to }\text{P}\end{array}$

For $\text{P}$:

The change in mass can be calculated as shown below,

$\begin{array}{l}\Delta m=(15\times 1.007\text{83}\text{g/mol}\text{H}\text{)}\text{+}(15\times 1.00867\text{g/mol}\text{n}\text{)}-\text{29}\text{.97832}\text{g/mol}\text{P}\\ \text{=}1\text{5}\text{.11745}\text{g/mol}\text{+}1\text{5}\text{.13005}\text{g/mol}-\text{29}\text{.97832}\text{g/mol}\\ =30.2475\text{g/mol}-\text{29}\text{.97832}\text{g/mol}\\ =0.26918\text{g/mol}\end{array}$

Nuclear binding energy can be calculated as shown below,

$\begin{array}{l}\Delta E=(\Delta m)c^2\\ =\left(\frac{0.26918\text{g}}{\text{1}\text{mol}}\times \frac{1\text{kg}}{\text{1000}\text{g}}\right)\times {(2.99792\times {10}^8\text{m/s})}^2\times \frac{\text{1}\text{J}}{\text{1}{\text{kgm}}^{\text{2}}{\text{s}}^{\text{-2}}}\times \frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\\ =2.4192\times {10}^{10}\text{kJ/mol}\end{array}$

Binding energy per nucleon can be calculated as shown below,

There is a total of thirty nucleons in phosphorus-30.  Hence, the binding energy per nucleon can be calculated as,

    $\begin{array}{l}{E}_b=\frac{2.4192\times {10}^{10}\text{kJ}}{\text{30}\text{mol}\text{nucleons}}\\ =8.064\times {10}^8\text{kJ/mol}\end{array}$

Binding energy per nucleon in $\text{P}$ is $8.064\times {10}^8\text{kJ/mol}$.

For $\text{P}$:

The change in mass can be calculated as shown below,

$\begin{array}{l}\Delta m=(15\times 1.007\text{83}\text{g/mol}\text{H}\text{)}\text{+}(16\times 1.00867\text{g/mol}\text{n}\text{)}-\text{30}\text{.97376}\text{g/mol}\text{P}\\ \text{=}1\text{5}\text{.11745}\text{g/mol}\text{+}1\text{6}\text{.13872}\text{g/mol}-\text{30}\text{.97376}\text{g/mol}\\ =31.25617\text{g/mol}-\text{30}\text{.97376}\text{g/mol}\\ =0.28241\text{g/mol}\end{array}$

Nuclear binding energy can be calculated as shown below,

$\begin{array}{l}\Delta E=(\Delta m)c^2\\ =\left(\frac{0.28241\text{g}}{\text{1}\text{mol}}\times \frac{1\text{kg}}{\text{1000}\text{g}}\right)\times {(2.99792\times {10}^8\text{m/s})}^2\times \frac{\text{1}\text{J}}{\text{1}{\text{kgm}}^{\text{2}}{\text{s}}^{\text{-2}}}\times \frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\\ =2.5381\times {10}^{10}\text{kJ/mol}\end{array}$

Binding energy per nucleon can be calculated as shown below,

There is a total of thirty one nucleons in phosphorus-11.  Hence, the binding energy per nucleon can be calculated as,

    $\begin{array}{l}{E}_b=\frac{2.5381\times {10}^{10}\text{kJ}}{\text{31}\text{mol}\text{nucleons}}\\ =0.0846\times {10}^{10}\text{kJ/mol}\\ \text{=}\text{8}\text{.46}\times {10}^8\text{kJ/mol}\end{array}$

Binding energy per nucleon in $\text{P}$ is $\text{8}\text{.46}\times {10}^8\text{kJ/mol}$.

On comparing the binding energy per nucleon for $\text{P}$ and $\text{P}$, it is found that $\text{P}$ has more binding energy per nucleon than $\text{P}$.  Hence, $\text{P}$ is more stable than $\text{P}$.