Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN: 9781337398909
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Chapter 14, Problem 14.39PAE
Interpretation Introduction

Interpretation: To determine the binding energy of the 7Li nucleus, whose experimentally determined mass is 7.016004 u.

Concept introduction:Mass defect:

The nucleus of each atom (aside from 11H ) has a mass lower than expected from adding the masses of its neutrons and protons.

Mass and energy are related by following equation.

E=mc2

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.

Calculated mass = mH+ mn

mH = mass of proton

mn= 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.

Calculated mass = mH+ mn

The formula of binding energy E = mc2

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

E = (Δm)c2Δm = Mass defectc = Velocity of light

Expert Solution & Answer
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Answer to Problem 14.39PAE

Solution: The binding energy of 7Li is 3.7865 × 109 kJ/ mol

Explanation of Solution

Given information: Experimentally determined mass of 7Li is 7.016004 u.

Formula used:

E = (Δm)c2Δm = Mass defectc = Velocity of lightE =binding energy

Calculation: The given 7Li has seven neutrons and three protons.

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

Mass of proton = 1.007825 uMass of neutron = 1.008665 u 

Let’s write formula:

7Lical= mH+mnSubstitue the masses of proton and neutron= 3(1.007825 u) + 4(1.008665 u)=3.023475 u + 4.03466 u= 7.058135 u

Therefore, calculated mass of 7Li is 7.058135 u.

Let’s calculate the mass defect:

Δm  = 7Lical-7Liobs       = 7.058135 u - 7.016004 u       = 0.042131 u

Therefore, mass defect of 7Li is 0.042131 u.

Let’s calculate the binding energy:

Binding energy E = (Δm)c2= 0.042131u × ( 1 .66054×10 -27 kgu) ×  (2.99792× 10 8 m s)2= 6.28769×10-12 kgm2s-2= 6.28769×10-12 J

Therefore, 6.28769×10-12J of energy is released when one 7Li nucleus forms its constituents nucleons.

Let’s calculate the released energy when one mole of 7Li form its constituent nucleons.

E = ( 6 .28769×10 -12  J 1 atom 7 Li)( 6 .02214×10 23  atom 7 Li 1 mol 7 Li)= 3.7865 × 1012 J/mol= 3.7865 × 109 kJ/mol

The released energy when one mole of 7Li form its constituent nucleons are 3.7865 × 109 kJ/ mol

Conclusion

Therefore, The binding energy of 7Li is 3.7865 × 109 kJ/ mol

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Chapter 14 Solutions

Chemistry for Engineering Students

Ch. 14 - Prob. 14.1PAECh. 14 - Prob. 14.2PAECh. 14 - Prob. 14.3PAECh. 14 - Prob. 14.4PAECh. 14 - (a) How does 14C enter a living plant? (b) Write...Ch. 14 - Prob. 14.6PAECh. 14 - Prob. 14.7PAECh. 14 - Prob. 14.8PAECh. 14 - Prob. 14.9PAECh. 14 - Prob. 14.10PAECh. 14 - Prob. 14.11PAECh. 14 - Prob. 14.12PAECh. 14 - Prob. 14.13PAECh. 14 - Prob. 14.14PAECh. 14 - Prob. 14.15PAECh. 14 - Prob. 14.16PAECh. 14 - Prob. 14.17PAECh. 14 - Prob. 14.18PAECh. 14 - Prob. 14.19PAECh. 14 - Prob. 14.20PAECh. 14 - Prob. 14.21PAECh. 14 - Prob. 14.22PAECh. 14 - Prob. 14.23PAECh. 14 - Prob. 14.24PAECh. 14 - Prob. 14.25PAECh. 14 - Prob. 14.26PAECh. 14 - Prob. 14.27PAECh. 14 - Prob. 14.28PAECh. 14 - Prob. 14.29PAECh. 14 - Prob. 14.30PAECh. 14 - Prob. 14.31PAECh. 14 - Prob. 14.32PAECh. 14 - Prob. 14.33PAECh. 14 - Prob. 14.34PAECh. 14 - Prob. 14.35PAECh. 14 - Prob. 14.36PAECh. 14 - Prob. 14.37PAECh. 14 - Prob. 14.38PAECh. 14 - Prob. 14.39PAECh. 14 - Prob. 14.40PAECh. 14 - Prob. 14.41PAECh. 14 - Prob. 14.42PAECh. 14 - Prob. 14.43PAECh. 14 - Prob. 14.44PAECh. 14 - Prob. 14.45PAECh. 14 - Prob. 14.46PAECh. 14 - Prob. 14.47PAECh. 14 - Prob. 14.48PAECh. 14 - Prob. 14.49PAECh. 14 - Prob. 14.50PAECh. 14 - Prob. 14.51PAECh. 14 - Prob. 14.52PAECh. 14 - Prob. 14.53PAECh. 14 - Prob. 14.54PAECh. 14 - Prob. 14.55PAECh. 14 - Prob. 14.56PAECh. 14 - Prob. 14.57PAECh. 14 - Prob. 14.58PAECh. 14 - Prob. 14.59PAECh. 14 - Prob. 14.60PAECh. 14 - Prob. 14.61PAECh. 14 - Prob. 14.62PAECh. 14 - Prob. 14.63PAECh. 14 - Prob. 14.64PAECh. 14 - Prob. 14.65PAECh. 14 - Prob. 14.66PAECh. 14 - Prob. 14.67PAECh. 14 - Prob. 14.68PAECh. 14 - Prob. 14.69PAECh. 14 - Prob. 14.70PAECh. 14 - Prob. 14.71PAECh. 14 - Prob. 14.72PAECh. 14 - Prob. 14.73PAECh. 14 - Prob. 14.74PAECh. 14 - Prob. 14.75PAECh. 14 - Prob. 14.76PAECh. 14 - Prob. 14.77PAECh. 14 - Prob. 14.78PAECh. 14 - Prob. 14.79PAECh. 14 - Prob. 14.80PAECh. 14 - Prob. 14.81PAECh. 14 - Prob. 14.82PAECh. 14 - Prob. 14.83PAECh. 14 - Prob. 14.84PAECh. 14 - Prob. 14.85PAECh. 14 - Prob. 14.86PAECh. 14 - Prob. 14.87PAECh. 14 - Prob. 14.88PAECh. 14 - Prob. 14.89PAECh. 14 - Prob. 14.90PAECh. 14 - Prob. 14.91PAECh. 14 - Prob. 14.92PAECh. 14 - Prob. 14.93PAECh. 14 - Prob. 14.94PAECh. 14 - Prob. 14.95PAECh. 14 - Prob. 14.96PAECh. 14 - Prob. 14.97PAECh. 14 - Prob. 14.98PAECh. 14 - Prob. 14.99PAECh. 14 - Prob. 14.100PAECh. 14 - Prob. 14.101PAECh. 14 - Prob. 14.102PAECh. 14 - Prob. 14.103PAECh. 14 - Prob. 14.104PAECh. 14 - Prob. 14.105PAECh. 14 - Prob. 14.106PAECh. 14 - Prob. 14.107PAE
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