# Calculate the binding energy in J/nucleon for carbon-12 (atomic mass= 12.0000 u) and uranium-235 (atomic mass= 235.0439 u). The atomic mass of H 1 1 is 1.00782 u and the mass of a neutron is 1.00866 u. The most stable nucleus known is 56 Fe (see Exercise 44). Would the binding energy per nucleon for 56 Fe be larger or smaller than that of 12 C or 235 U? Explain.

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 18, Problem 45E
Textbook Problem
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## Calculate the binding energy in J/nucleon for carbon-12 (atomic mass= 12.0000 u) and uranium-235 (atomic mass= 235.0439 u). The atomic mass of H 1 1 is 1.00782 u and the mass of a neutron is 1.00866 u. The most stable nucleus known is 56Fe (see Exercise 44). Would the binding energy per nucleon for 56Fe be larger or smaller than that of 12C or 235U? Explain.

Interpretation Introduction

Interpretation: The binding of carbon- 12 and uranium- 235 is to be calculated. Justification of whether the binding energy per nucleon of 56Fe be greater or smaller than that of 12C or 235U is to be stated.

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 carbon- 12 and uranium- 235 .

### Explanation of Solution

Explanation

Given

The atomic mass of 612C is 12.000amu .

Mass of 11H 1.00782amu .

The mass of neutron is 1.00866amu .

Number of protons in 612C=6

Number of neutrons in 612C=6

Mass defect is calculated by the formula,

Δm=Atomicmassof612Cnucleus[Numberofprotons×Massof11HprotonNumberofneutrons×Massofneutron]

Substitute the value of the atomic mass 612C , the number of protons and mass of the 11H proton and that of the neutron in the above equation.

Δm=12.000[(6×1.0078)+(6×1.00866)]Δm=12.0006.04696.05196Δm=12.00012.09888Δm=0.09888amunucleus

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,

0.09888amu/nucleus=(0.09888×1×1.66×1027)kg/nucleus=-0.1641408×10-27Kg/nucleus_

Therefore, the mass defect (Δm) of 612C is -0.1641408×10-27Kg/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=(0.1641408×10-27kgnucleus)(3×108ms)2ΔE=-1.477×10-11J/nucleus_

Therefore, the energy released per nucleus is -1.477×10-11J/nucleus_ .

Explanation

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)=1.477×101112J/nucleon=-1.23×10-12J/nucleon_

Explanation

The atomic mass of 92235U 235.0439amu .

Mass of 11H 1.00782 amu .

The mass of neutron is 1.00866amu

Number of protons in 92235U=92

Number of neutrons in 92235U=143

The mass defect of 92235U is calculated by the formula,

Δm=Mass of 92235Unucleus [Numberofprotons×massof11HprotonNumberofneutrons×massofneutron]

Substitute the value of the mass 92235U , the number of protons and mass of the 11H proton and that of the neutron in the above equation.

Δm=235.0439[(92×1.00782)+(143×1.00866)]Δm=235.043992.7194144.2383Δm=1.91388amu/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,

1.91388amu/nucleu=(1.9138×1×1.66×1027)kg/nucleus=-3.17704×10-27Kg/nucleus_

The mass defect (Δm) of 92235U is

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