CHEMISTRY (LOOSELEAF) >CUSTOM<
CHEMISTRY (LOOSELEAF) >CUSTOM<
13th Edition
ISBN: 9781264348992
Author: Chang
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 19, Problem 19.62QP

(a)

Interpretation Introduction

Interpretation:

For the given species, binding energy per nucleon should be determined.

Concept Introduction:

  • 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

To determine: Binding energy per nucleon for the given

(a)

Expert Solution
Check Mark

Answer to Problem 19.62QP

The nuclear binding energy for 10B is  1.040×10-12J/nucleon

Explanation of Solution

For the given 10B, there are5 protons and 5 neutrons.

Massofprotons=1.00728amuMassof5protons=1.00728×5=5.0364amuMassofelectrons=5.4858×10-4amuMassof5electrons=5×5.4858×10-4amu=2.7429×103amuMassofneutron=1.008665amuMassof5neutron=5×1.008665amu=5.043325amuSo,predictedmassfor10Bis5.0364amu+2.7429×103amu+5.043325amu=10.0824679amuMassof 10Bis10.0129amuMassdefect=Atomicmass-MP+Me+Mn=10.0129amu-10.0824679amu=-0.0695679amuSince,1kg=6.022×1026amuΔm=-0.0695679amu6.022×1026amu=-1.15522916×1028kg

The binding energy

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,ΔE =(Δm)c2

C=velocity of light is 2.99792458×108m/s

ΔE =(Δm)c2ΔE=-1.15522916 ×10-28kg×(2.99792458×108m/s)2=1.03826819×10-11kg.m2/s2=1.03826819×10-11JNucearbindingenergypernucleon=1.03826819×10-11J10nucleon=1.040×10-12J/nucleon

(b)

Interpretation Introduction

Interpretation:

For the given species, binding energy per nucleon should be determined.

Concept Introduction:

  • 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

To determine: Binding energy per nucleon for the given

(b)

Expert Solution
Check Mark

Answer to Problem 19.62QP

The nuclear binding energy for 11B is 1.111×10-12J/nucleon

Explanation of Solution

For the given 11B, there are5 protons and 6 neutrons.

Massofprotons=1.00728amuMassof5protons=1.00728×5=5.0364amuMassofelectrons=5.4858×10-4amuMassof5electrons=5×5.4858×10-4amu=2.7429×103amuMassofneutron=1.008665amuMassof6neutron=6×1.008665amu=6.05199amuSo,predictedmassfor11Bis5.0364amu+2.7429×103amu+6.05199amu=11.0911329amuMassof 10Bis11.009305amuMassdefect=Atomicmass-MP+Me+Mn=11.009305amu-11.0911329amu=-0.0818279amuSince,1kg=6.022×1026amuΔm=-0.0818279amu6.022×1026amu=-1.358816008×1028kg

The binding energy

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,ΔE =(Δm)c2

C=velocity of light is 2.99792458×108m/s

ΔE =(Δm)c2ΔE=-1.358816008 ×10-28kg×(2.99792458×108m/s)2=1.2212429×10-11kg.m2/s2=1.2212429×10-11JNucearbindingenergypernucleon=1.2212429×10-11J11nucleon=1.11×10-12J/nucleon

(c)

Interpretation Introduction

Interpretation:

For the given species, binding energy per nucleon should be determined.

Concept Introduction:

  • 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

To determine: Binding energy per nucleon for the given

(c)

Expert Solution
Check Mark

Answer to Problem 19.62QP

The nuclear binding energy for 14N   is 1.199×10-12J/nucleon

Explanation of Solution

For the given 14N, there are7 protons and 7 neutrons.

Massofprotons=1.00728amuMassof7protons=1.00728×7=7.05096amuMassofelectrons=5.4858×10-4amuMassof7electrons=5.4858×10-4amu=3.84006×103amuMassofneutron=1.008665amuMassof7neutron=7×1.008665amu=7.060655amuSo,predictedmassfor14Nis7.05096amu+3.84006×103amu+7.060655amu=14.11545506amuMassof 14Nis14.003074amuMassdefect=Atomicmass-MP+Me+Mn=14.003074amu-14.11545506amu=-0.11238106amuSince,1kg=6.022×1026amuΔm=-0.11238106amu6.022×1026amu=-1.866175025×1028kg

The binding energy

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,ΔE =(Δm)c2

C=velocity of light is 2.99792458×108m/s

ΔE =(Δm)c2ΔE=-1.866175 ×10-28kg×(2.99792458×108m/s)2=1.677234468×10-11kg.m2/s2=1.677234468×10-11JNucearbindingenergypernucleon=1.677234468×10-11J14nucleon=1.199×10-12J/nucleon

(d)

Interpretation Introduction

Interpretation: For the given species, binding energy per nucleon should be determined.

Concept Introduction:

  • 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

To determine: Binding energy per nucleon for the given

(d)

Expert Solution
Check Mark

Answer to Problem 19.62QP

The nuclear binding energy for 56Fe is 1.410×10-12J/nucleon

Explanation of Solution

For the given 56Fe, there are26 protons and 30 neutrons.

Massofprotons=1.00728amuMassof26protons=1.00728×26=26.18928amuMassofelectrons=5.4858×10-4amuMassof26electrons=26×5.4858×10-4amu=0.01426308amuMassofneutron=1.008665amuMassof30neutron=30×1.008665amu=30.25995amuSo,predictedmassfor56Feis26.18928amu+0.01426308amu+30.25995amu=56.4634amuMassof 56Feis55.93494amuMassdefect=Atomicmass-MP+Me+Mn=55.93494amu-56.4634amu=-0.52855308amuSince,1kg=6.022×1026amuΔm=-0.52855308amu6.022×1026amu=-8.77698×10-28kg

The binding energy

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,ΔE =(Δm)c2

C=velocity of light is 2.99792458×108m/s

ΔE =(Δm)c2ΔE=-8.77698×10-28kg×(2.99792458×108m/s)2=7.887475×10-11kg.m2/s2=7.887475×10-11JNucearbindingenergypernucleon=7.887475×10-11J56nucleon=1.40×10-12J/nucleon

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Chapter 19 Solutions

CHEMISTRY (LOOSELEAF) >CUSTOM<

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