Chapter 14, Problem 14.37PAE

Interpretation Introduction

Interpretation: To determine the binding energy of 1 mole of ${}^{\text{14}}\text{C}$ nuclei and experimentally determined mass of a ${}^{\text{14}}\text{C}$ is 14.003242u.

Concept introduction: Mass defect:

The nucleus of each atom (aside from ${}_{\text{1}}^{\text{1}}\text{H}$ ) has a mass lower than expected from adding the masses of its neutrons and protons.

Mass and energy are related by following equation.

$\mathrm{E}=m{c}^2$

E = Energy

M = mass

C = speed of light.

Loss of mass and liberation of energy are related by Einstein’s equations.

$\begin{array}{l}{\text{E = (Δm)c}}^{\text{2}}\\ \text{Δm = Mass defect}\\ \text{c = Velocity of light}\end{array}$

Answer to Problem 14.37PAE

Solution: The binding energy of 1 mole of ${}^{\text{14}}\text{C}$ nuclei is $\text{1}{\text{.015838 × 10}}^{\text{10}}\text{ kJ/mol}$.

Explanation of Solution

Given information: The experimentally determined mass of ${}^{\text{14}}\text{C}$ is 14.003242 u.

Formula used:

$\begin{array}{l}{\text{E = (Δm)c}}^{\text{2}}\\ \text{Δm = Mass defect}\\ \text{c = Velocity of light}\\ \text{E= binding energy}\end{array}$

Calculation: To calculate the binding energy of ${}^{\text{14}}\text{C}$ nuclei by using the Einstein’s equation.

$\begin{array}{l}{\text{E = (Δm)c}}^{\text{2}}\\ \text{Δm = Mass defect}\\ \text{c = Velocity of light}\end{array}$

Let’s calculate the mass which is equal to the sum of masses of six protons and eight neutrons.

$\begin{array}{l}{}^{\text{14}}{\text{C}}_{\text{cal}}{\text{= m}}_{\text{H}}{\text{ + m}}_{\text{N}}\\ \text{Mass of Proton = 1}\text{.007825u}\\ \text{Mass of neutron = 1}\text{.008665u}\\ \text{= 6(1}\text{.007825u) + 8(1}\text{.008665u)}\\ \text{= 6}\text{.04695 u + 8}\text{.06932 u}\\ \text{=14}\text{.11627 u}\end{array}$

Let’s calculate the mass defect:

Here, the mass defect formula is the difference between the calculated and observed masses of carbon -14 nuclei.

Mass defect can be calculated by the following formula.

${\text{Δm = }}^{\text{14}}{\text{C}}_{\text{cal}}{\text{-}}^{\text{14}}{\text{C}}_{\text{obs}}$

${}^{\text{14}}{\text{C}}_{\text{cal}}$ =14.11627 u

${}^{\text{14}}{\text{C}}_{\text{obs}}$ = 14.003242 u

Substitute the values in the above equation.

$\begin{array}{l}\text{= 14}\text{.11627u - 14}\text{.003242u}\\ \text{= 0}\text{.113028u}\end{array}$

Let’s calculate the binding energy:

Here, we have to convert the mass defect from atomic mass units to kilograms.

Then the resulting value for ${\text{mc}}^{\text{2}}$ will be in joules ( ${\text{1J =1Kg m}}^{\text{2}}{\text{s}}^{\text{-2}}$ )

$\begin{array}{c}{\text{E = (Δm)c}}^{\text{2}}\\ \text{Δm = 0}\text{.113028 u}\\ \text{= 0}\text{.113028 u × }\left(\frac{\text{1}{\text{.66054×10}}^{\text{-27}}\text{ kg}}{\text{u}}\right)\left(2.99792{\text{×10}}^{\text{8}}\frac{\text{m}}{\text{s}}\right)\\ \text{=1}{\text{.68684 × 10}}^{\text{-11}}{\text{ kgm}}^{\text{2}}{\text{s}}^{\text{-2}}\\ \text{=1}{\text{.68684 × 10}}^{\text{-11}}\text{J}\end{array}$

Let’s calculate the released energy from the one mole of ${}^{\text{14}}\text{C}$ nuclei.

$\begin{array}{c}\text{E = }\left(\frac{\text{1}{\text{.68684×10}}^{\text{-11}}\text{ J}}{{\text{1 atom}}^{\text{14}}\text{C}}\right)\left(\frac{\text{6}{\text{.02214×10}}^{\text{23}}{\text{ atoms}}^{\text{14}}\text{C}}{{\text{1 mol}}^{\text{14}}\text{C}}\right)\\ \text{= 1}{\text{.015838×10}}^{\text{13}}\text{ J/mol}\\ \text{= 1}{\text{.015838×10}}^{\text{10}}\text{ kJ/mol}\end{array}$

Conclusion

Therefore, the binding energy of 1 mole of ${}^{\text{14}}\text{C}$ nuclei is $\text{1}{\text{.015838 × 10}}^{\text{10}}\text{ kJ/mol}$.