Chapter 14, Problem 14.39PAE

Interpretation Introduction

Interpretation: To determine the binding energy of the ${}^{\text{7}}\text{Li}$ nucleus, whose experimentally determined mass is 7.016004 u.

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.

Calculated mass of nucleus:

The calculated mass of nucleus is equal to the sum of masses of its protons and neutrons.

${\text{Calculated mass = m}}_{\text{H}}{\text{+ m}}_{\text{n}}$

m_H = mass of proton

m_n= mass of neutron.

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

Binding energy:

The binding energy is the energy that would be needed to disassemble a nucleus into individual nucleons.

${\text{Calculated mass = m}}_{\text{H}}{\text{+ m}}_{\text{n}}$

The formula of binding energy ${\text{E = mc}}^{\text{2}}$

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 14.39PAE

Solution: The binding energy of ${}^{\text{7}}\text{Li}$ is $\text{3}{\text{.7865 × 10}}^{\text{9}}\text{ kJ/ mol}$

Explanation of Solution

Given information: Experimentally determined mass of ${}^{\text{7}}\text{Li}$ is 7.016004 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: The given ${}^{\text{7}}\text{Li}$ has seven neutrons and three protons.

The calculated mass is equal to the sum of masses o three protons and seven neutrons.

$\begin{array}{l}\text{Mass of proton = 1}\text{.007825 u}\\ \text{Mass of neutron = 1}\text{.008665 u }\end{array}$

Let’s write formula:

$\begin{array}{c}{}^{\text{7}}{\text{Li}}_{\text{cal}}{\text{= m}}_{\text{H}}{\text{+m}}_{\text{n}}\\ \text{Substitue the masses of proton and neutron}\\ \text{= 3(1}\text{.007825 u) + 4(1}\text{.008665 u)}\\ \text{=3}\text{.023475 u + 4}\text{.03466 u}\\ \text{= 7}\text{.058135 u}\end{array}$

Therefore, calculated mass of ${}^{\text{7}}\text{Li}$ is 7.058135 u.

Let’s calculate the mass defect:

$\begin{array}{l}{\text{Δm = }}^{\text{7}}{\text{Li}}_{\text{cal}}{\text{-}}^{\text{7}}{\text{Li}}_{\text{obs}}\\ \text{ = 7}\text{.058135 u - 7}\text{.016004 u}\\ \text{ = 0}\text{.042131 u}\end{array}$

Therefore, mass defect of ${}^{\text{7}}\text{Li}$ is 0.042131 u.

Let’s calculate the binding energy:

$\begin{array}{c}{\text{Binding energy E = (Δm)c}}^{\text{2}}\\ \text{= 0}\text{.042131u × }\left(\frac{\text{1}{\text{.66054×10}}^{\text{-27}}\text{kg}}{\text{u}}\right)\text{ × }{\left(2.99792\times {10}^8\frac{\text{m}}{\text{s}}\right)}^{\text{2}}\\ \text{= 6}{\text{.28769×10}}^{\text{-12}}{\text{ kgm}}^{\text{2}}{\text{s}}^{\text{-2}}\\ \text{= 6}{\text{.28769×10}}^{\text{-12}}\text{ J}\end{array}$

Therefore, $\text{6}{\text{.28769×10}}^{\text{-12}}\text{J}$ of energy is released when one ${}^{\text{7}}\text{Li}$ nucleus forms its constituents nucleons.

Let’s calculate the released energy when one mole of ${}^{\text{7}}\text{Li}$ form its constituent nucleons.

$\begin{array}{c}\text{E = }\left(\frac{\text{6}{\text{.28769×10}}^{\text{-12}}\text{ J}}{{\text{1 atom}}^{\text{7}}\text{Li}}\right)\left(\frac{\text{6}{\text{.02214×10}}^{\text{23}}{\text{ atom}}^{\text{7}}\text{Li}}{{\text{1 mol}}^{\text{7}}\text{Li}}\right)\\ \text{= 3}{\text{.7865 × 10}}^{\text{12}}\text{ J/mol}\\ \text{= 3}{\text{.7865 × 10}}^{\text{9}}\text{ kJ/mol}\end{array}$

The released energy when one mole of ${}^{\text{7}}\text{Li}$ form its constituent nucleons are $\text{3}{\text{.7865 × 10}}^{\text{9}}\text{ kJ/ mol}$

Conclusion

Therefore, The binding energy of ${}^{\text{7}}\text{Li}$ is $\text{3}{\text{.7865 × 10}}^{\text{9}}\text{ kJ/ mol}$