Chapter 21, Problem 21.49QA

Interpretation Introduction

To calculate:

The binding energy of the two naturally occurring isotopes of chlorine  $Cl and Cl$.

Answer to Problem 21.49QA

Solution:

a) The binding energy of $Cl$ is $2.877 \times {10}^{10} kJ/mol$.

b) The binding energy of $Cl$ is $3.060 \times {10}^{10} kJ/mol$.

Explanation of Solution

1) Concept:

To calculate the binding energy of the nuclei of two isotopes, we need to calculate the mass defect of the nuclei. From the exact mass of the isotopes, we will calculate the exact mass of each isotope. From the mass of each isotope, we will calculate the mass of the nucleons. Calculate the mass of the nucleus by using mass of photons and mass of the neutrons. Calculate the mass defect from the mass of nucleus and mass of nucleons. We will calculate the binding energy from the mass defect.

2) Formula:

Einstein’s equation for binding energy:

$\mathrm{B}\mathrm{E}=(∆\mathrm{m}){\mathrm{c}}^2$

3) Given:

i) $1 J = 1 kg {\left(\frac{m}{s}\right)}^2$

ii) $$ Mass of proton = $1.67262\times {10}^{-27}\mathrm{k}\mathrm{g}$

iii)  Mass of neutrons = $1.67493\times {10}^{-27}\mathrm{k}\mathrm{g}$

iv) Mass of electrons = $9.10939\times {10}^{-31}kg$

v) Speed of light c = $2.998 \times {10}^{8}m/s$

vi) The exact mass of $Cl$ is $34.9689 amu$.

vii) The exact mass of $Cl$ is $36.9659 amu$.

4) Calculation:

First we convert the exact mass of each isotope from amu to kg.

$Cl: 34.9689 amu \times \frac{1.66054\times {10}^{-27}kg}{amu} = 5.806726\times {10}^{-26} kg$

$Cl: 36.9659 amu \times \frac{1.66054\times {10}^{-27} kg}{amu} = 6.138336\times {10}^{-26} kg$

To calculate the mass of nucleons, we have to subtract the mass of $17$ electrons from the exact mass:

$Cl: \left(5.806726\times {10}^{-26} kg\right)- 17\left(9.10939\times {10}^{-31} kg\right)=5.805177\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g}$

$Cl: \left(6.138336\times {10}^{-26} kg\right)- 17\left(9.10939\times {10}^{-31 }kg\right)=6.136787\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g}$

Now, we calculate the mass of neutrons and protons present in one atom of each isotope:

$Cl: \left(17 protons \times \frac{1.67262\times {10}^{-27}kg}{protons}\right)+ \left(18 neutrons \times \frac{1.67493\times {10}^{-27}kg}{neutrons}\right)=\mathrm{ }5.858328\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g}$

$Cl: \left(17protons \times \frac{1.67262\times {10}^{-27}kg}{protons}\right)+ \left(20 neutrons \times \frac{1.67493\times {10}^{-27}kg}{neutrons}\right)=\mathrm{ }6.193314\times {10}^{-26}\mathrm{ }\mathrm{k}\mathrm{g}$

Now, calculate the mass defect $(∆\mathrm{m})$ from the exact mass of the nucleus isotope and its subatomic particles.

$(∆m) = mass of nucleus – mass of nucleons$

$Cl: ∆m= 5.858328\times {10}^{-26} kg – 5.805177\times {10}^{-26} kg=5.3150876 \times {10}^{-28} kg$

$Cl: ∆m= 6.193314\times {10}^{-26} kg – 6.136787\times {10}^{-26} kg=5.6527038 \times {10}^{-28} kg$

Now calculate the binding energy (BE) by using Einstein’s equation:

$\mathrm{B}\mathrm{E}=(∆\mathrm{m}){\mathrm{c}}^2$

$Cl: BE =5.3150876\times {10}^{-28}kg \times {\left(2.998 \times {10}^8\frac{m}{s}\right)}^2$

$Cl: BE =4.7772028 \times {10}^{-11} kg \frac{m^2}{s^2} \times \frac{1 J}{kg \frac{m^2}{s^2} }=4.7772028 \times {10}^{-11}J$

$Cl: BE =\frac{4.7772028 \times {10}^{-11}J}{atoms} \times \frac{1 kJ}{1000 J}\times \frac{6.022 \times {10}^{23} atoms}{mols}$

$Cl: BE = 2.877 \times {10}^{10} kJ/mol$

So the binding energy of $Cl$ is $2.877 \times {10}^{10} kJ/mol$.

$Cl: BE =5.6527038\times {10}^{-28}kg \times {\left(2.998 \times {10}^8\frac{m}{s}\right)}^2$

$Cl: BE =5.0806524 \times {10}^{-11} kg \frac{m^2}{s^2} \times \frac{1 J}{kg \frac{m^2}{s^2} }=5.0806524 \times {10}^{-11}J$

$Cl: BE =\frac{5.0806524 \times {10}^{-11}J}{atoms} \times \frac{1 kJ}{1000 J}\times \frac{6.022 \times {10}^{23} atoms}{mols}$

$Cl: BE = 3.060 \times {10}^{10} kJ/mol$

So the binding energy of $Cl$ is $3.060 \times \frac{{10}^{10}kJ}{mol}.$

Conclusion:

We calculated the binding energy of each isotope from the exact mass and mass defect.