Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259638138
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 20, Problem 20.16QP
Interpretation Introduction

Interpretation:

The nuclear binding energy and the binding energy per nucleon for the following isotope s should be calculated

Concept introduction:

  • Nuclear binding energy: It is the minimum amount of energy required to disassemble the nucleus of an atom into its component parts.

    The component parts are neutrons and protons, which are collectively called as nucleons.

  • Binding energy per nucleon:
  • The maximum binding energy per nucleon occurs at around mass number A=50.

    Example –Iron nucleolus ( Fe56 ) is located closed to the peak with a binding energy per nucleon value of approximately 8.8 MeV.

  • Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
  • This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
  • Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is, ΔE =(Δm)c2

    Where, (Δm) is called mass defect.

  • The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

Formula:

Einstein mass energy relationship [ ΔE=(ΔM)c2 ]

nuclear binding energy per nucleon =nuclear binding energynumber of nucleons

Expert Solution & Answer
Check Mark

Answer to Problem 20.16QP

Thus the nuclear binding energy is 2.3603×10-10J .

The nuclear binding energy per nucleon is 1.2828×10-12J/nucleon

Explanation of Solution

The binding energy is the energy required for the process

He242p11+2n01

There are two proton and 2 neutron in the helium nucleus.

The mass of 2 proton is

2×1.00728amu=2.01456amu

The mass of 2 electron is

2×5.4858×10-4=0.0010972amu

The mass of 2 neutron is

2×1.008665amu=2.017330amu

Therefore, the predicted mass of helium molecule is

2.01456amu+0.0010972amu+2.017330amu=4.03299amu

So the mass defect is found to be

ΔM=4.002603amu-4.03299amu=-0.03039amu

-0.03039amu×1kg6.0221418×1026amu=-5.046×10-29

The energy change ( ΔE ) for the process is

ΔE=(-5.046×10-29kg)(2.99792458×108m/s)2

ΔE=-4.535×10-12kg.m2/s2=-4.535×10-12J

The nuclear binding energy is 4.535×10 12 J .

So the energy required to break up one helium nucleus into proton and two neutrons.

For helium nucleus

The nuclear binding energy per nucleon= 4.535×10 12 J4 nucleons=1.134×10-12J

The binding energy is the energy required for the process

W741847411p +11001n

There are 74 protons and 110 neutrons in the tungsten nucleus.

The mass of 74 proton is

74 ×1.00728amu=74.5387amu

The mass of 110 neutron is

74×5.4858×10-4amu=0.04059amu

110×1.008665amu=110.9532amu

So the predicted mass of W74184 is

74.5387amu+0.04059amu+110.9532amu=185.5325 amu

The mass defect is found

ΔE=183.950928amu-185.5325amu=-1.5816amu

-1.5816amu×1kg6.0221418×1026amu=-26262×10-27kg

The energy change for the process is

ΔE=(-2.6262×10-27kg)(2.99792458×108m/s)2

ΔE=-2.3603×10-10kg.m2/s2=-2.3603×10-10J

Thus the nuclear binding energy is 2.3603×10-10J .

The nuclear binding energy per nucleon is obtained as follow:

2.3603×10-10J184nucleons=1.2828×10-12J/nucleon

To calculate the binding energy and nuclear binding energy per nucleon by considering the given value

The binding energy is the energy required for the process

He242p11+2n01

There are two protons and 2 neutrons in the helium nucleus.

The mass of 2 protons is

2×1.00728amu=2.01456amu

The mass of 2 electrons is

2×5.4858×10-4=0.0010972amu

The mass of 2 neutrons is

2×1.008665amu=2.017330amu

Therefore, the predicted mass of helium molecule is

2.01456amu+0.0010972amu+2.017330amu=4.03299amu

So the mass defect is found to be

ΔM=4.002603amu-4.03299amu=-0.03039amu

-0.03039amu×1kg6.0221418×1026amu=-5.046×10-29

The energy change ( ΔE ) for the process is

ΔE=(-5.046×10-29kg)(2.99792458×108m/s)2

ΔE=-4.535×10-12kg.m2/s2=-4.535×10-12J

The nuclear binding energy is 4.535×10 12 J .

So the energy required to break up one helium nucleus into proton and two neutrons.

For helium nucleus

The nuclear binding energy per nucleon= 4.535×10 12 J4 nucleons=1.134×10-12J

Formula:

Einstein mass energy relationship [ ΔE=(ΔM)c2 ]

nuclear binding energy per nucleon =nuclear binding energynumber of nucleons

Given:

He24 =4.002603amu

W74184 =183.950928amu

To calculate the binding energy, and the binding energy per nucleons

The binding energy is the energy required for the process

W741847411p +11001n

There are 74 protons and 110 neutrons in the tungsten nucleus.

The mass of 74 proton is

74 ×1.00728amu=74.5387amu

The mass of 110 neutron is

74×5.4858×10-4amu=0.04059amu

110×1.008665amu=110.9532amu

So the predicted mass of W74184 is

74.5387amu+0.04059amu+110.9532amu=185.5325 amu

The mass defect is found

ΔE=183.950928amu-185.5325amu=-1.5816amu

-1.5816amu×1kg6.0221418×1026amu=-26262×10-27kg

The energy change for the process is

ΔE=(-2.6262×10-27kg)(2.99792458×108m/s)2

ΔE=-2.3603×10-10kg.m2/s2=-2.3603×10-10J

Thus the nuclear binding energy is 2.3603×10-10J .

The nuclear binding energy per nucleon is obtained as follow:

2.3603×10-10J184nucleons=1.2828×10-12J/nucleon

Conclusion

The nuclear binding energy and the binding energy per nucleon for the given isotopes were calculated

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Chapter 20 Solutions

Chemistry: Atoms First

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