Chemistry with Access Code, Hybrid Edition
Chemistry with Access Code, Hybrid Edition
9th Edition
ISBN: 9781285188492
Author: Steven S. Zumdahl
Publisher: CENGAGE L
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Chapter 19, Problem 46E
Interpretation Introduction

Interpretation: The binding energy per nucleon of 12H and 13H is to be calculated.

Concept introduction: The sum of masses of the component nucleons and the actual mass of a nucleus is known as the mass defect and it can be used to calculate the nuclear binding energy.

To determine: The binding energy per nucleon of 12H and 13H .

Expert Solution & Answer
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Answer to Problem 46E

Answer

The binding energy per nucleon of 12H is 17.78×10-14_J/nucleon .

The binding energy per nucleon of 13H is 45.26×10-14_J/nucleon .

Explanation of Solution

Explanation

The atomic mass of 11H=1.0078amu

The mass of a neutron is 1.0087amu .

Mass of 12H=2.01410amu

Number of protons in 12H=1

Number of neutrons in 12H=1

The mass defect is calculated by the formula,

Δm=Atomicmass of 12H[Numberofprotons×massof11HprotonNumberofneutrons×massofneutron]

Substitute the value of the atomic mass 12H , the number of protons and mass of the 11H proton and that of the neutron in the above equation.

Δm=2.01410[(1×1.00782)+(1×1.00866)]Δm=2.38×103amu/nucleus

The conversion of amu/nucleus to Kg/nucleus is done as,

1amu=1.66×1027Kg

Therefore, the conversion of 0.589amu/nucleus into kg/nucleus is,

2.38×103amu/nucleus=(2.38×103×1.66×1027)kg/nucleus=-3.9508×10-30Kg/nucleus_

Therefore, the mass defect (Δm) of 12H is -3.9508×10-30Kg/nucleus_ .

Explanation

The binding energy per nucleon is calculated by Einstein’s mass energy equation, that is,

ΔE=ΔmC2

Where,

  • ΔE is the change in energy.
  • Δm is the change in mass.
  • C is the velocity of light.

Substitute the values of Δm and C in the equation.

ΔE=Δmc2ΔE=(3.9508×1030)(3×108)2ΔE=35.55×1014J/nucleus

Therefore, the energy released per nucleus is 35.55×1014J/nucleus .

The binding energy per nucleon is calculated by the formula,

Binding energy per nucleon=Bindingenergy per nucleusNumberofprotons+Numberofneutrons

Substitute the value of the binding energy per nucleus, the number of protons and the number of neutrons in the above equation.

Binding energy per nucleon=Bindingenergy per nucleus(Numberofprotons+Numberofneutrons)=35.55×10142J/nucleon=17.78×10-14J/nucleon_

Explanation

The atomic mass of 11H=1.0078amu and mass of neutron is 1.0087amu

The atomic mass of 13H=3.01605amu

The mass of a neutron is 1.0087amu .

Mass of 12H=2.01410amu

Number of protons in 13H=1

Number of neutrons in 13H=2

The mass defect is calculated by the formula,

Δm=Atomicmass of 13H[Numberofprotons×massof11HprotonNumberofneutrons×massofneutron]

Substitute the value of the atomic mass 13H , the number of protons and mass of the 11H proton and that of the neutron in the above equation.

Δm=3.01605[(1×1.00782)+(2×1.00866)]Δm=0.00909amu/nucleus

The conversion of amu/nucleus to Kg/nucleus is done as,

1amu=1.66×1027Kg

Therefore, the conversion of 0.589amu/nucleus into kg/nucleus is,

2.38×103amu/nucleus=2.38×103×1.66×1027kg/nucleus=-3.9508×10-30Kg/nucleus_

Explanation

The binding energy per nucleon is calculated by Einstein’s mass energy equation, that is,

ΔE=ΔmC2

Where,

  • ΔE is the change in energy.
  • Δm is the change in mass.
  • C is the velocity of light.

Substitute the values of Δm and C in the equation.

ΔE=Δmc2ΔE=(0.0150894×1027)×9×(108)2ΔE=0.1358046×1011J/nucleus

Therefore, the energy released per nucleus is 0.1358046×1011J/nucleus .

Therefore, the energy released per nucleus is 7.90×1011J/nucleus .

The binding energy per nucleon is calculated by the formula,

Binding energy per nucleon=Bindingenergy per nucleusNumberofprotons+Numberofneutrons

Substitute the value of the binding energy per nucleus, the number of protons and the number of neutrons in the above equation.

Binding energy per nucleon=Bindingenergy per nucleus(Numberofprotons+Numberofneutrons)=0.1358046×10113J/nucleon=45.26×10-14J/nucleon_

Conclusion

Conclusion

The binding energy per nucleon of 12H is 17.78×10-14_J/nucleon .

The binding energy per nucleon of   13H is 45.26×10-14_J/nucleon .

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Chapter 19 Solutions

Chemistry with Access Code, Hybrid Edition

Ch. 19 - Prob. 1QCh. 19 - Prob. 2QCh. 19 - Prob. 3QCh. 19 - Prob. 4QCh. 19 - Prob. 5QCh. 19 - Prob. 6QCh. 19 - Prob. 7QCh. 19 - Prob. 8QCh. 19 - Prob. 9QCh. 19 - Prob. 10QCh. 19 - Prob. 11ECh. 19 - Prob. 12ECh. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - Prob. 59AECh. 19 - Prob. 60AECh. 19 - Prob. 61AECh. 19 - Prob. 62AECh. 19 - Prob. 63AECh. 19 - Prob. 64AECh. 19 - Prob. 65AECh. 19 - Prob. 66AECh. 19 - Prob. 67AECh. 19 - Prob. 68AECh. 19 - Prob. 69AECh. 19 - Prob. 70AECh. 19 - Prob. 71AECh. 19 - Prob. 72AECh. 19 - Prob. 73CWPCh. 19 - Prob. 74CWPCh. 19 - Prob. 75CWPCh. 19 - Prob. 76CWPCh. 19 - Prob. 77CWPCh. 19 - Prob. 78CWPCh. 19 - Prob. 79CPCh. 19 - Prob. 80CPCh. 19 - Prob. 81CPCh. 19 - Prob. 82CPCh. 19 - Prob. 83CPCh. 19 - Prob. 84CPCh. 19 - Prob. 85CPCh. 19 - Prob. 86CPCh. 19 - Prob. 87IPCh. 19 - Prob. 88IP
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