Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 20, Problem 20QP
Interpretation Introduction

Interpretation:

The nuclear binding energy and the nuclear binding energyper nucleon, in joules, for a given nucleus is to be determined.

Concept Introduction:

The nuclear binding energy is calculated by using the relation represented as follows:

ΔE=(Δm)c2

Here, Δm is the mass defect and c is the speed of light.

The mass defect is defined as the difference between the given mass and the predicted mass. Thus, the formula of mass defect is as follows:

Δm= Given massPredicted mass

Here, Δm is the mass defect.

Expert Solution & Answer
Check Mark

Answer to Problem 20QP

Solution:

4.62×1012 J , 1.15×1012 J/nucleon.

2.36×1010 J , 1.28×1012 J/nucleon.

Explanation of Solution

a) 24He (4.002603  amu).

In the given nucleus, 24He, the number of protons and neutrons are 2 and 2, respectively.

The mass of 2 11H atoms is as follows:

Mass of 2 11=(2×mass of hydrogen)=(2×1.007825 amu)=2.01565 amu

Mass of 2 neutrons =2×mass of neutron=(2×1.008665 amu)=2.01733 amu

The predicted mass of 24He is calculated as follows:

Predicted mass = Mass of 2 11H+Mass of 2 neutrons=(2.01565 amu+2.01733 amu)=4.032980 amu

The mass defect is calculated by using the relation as follows:.

Δm=Given mass Predicted mass

Substitute 4.032980 amu for predicted mass and 4.002603  amu for given mass,

Δm=(4.002603  amu)(4.032980 amu)=0.03092 amu

The value of mass defect is converted into Kg by using the relation represented as follows:.

1 Kg=6.0221418×1026 amu(0.03092 amu×1 Kg6.022×1026 amu)=5.134×1029Kg

The nuclear binding energy is calculated by the relation as follows:

ΔE=(Δm)c2

Substitute 5.134×1029Kg for Δm and 3.00×108 m/s for c in the equationas follows:

ΔE=(5.134×1029Kg)(3.00×108 m/s)2=46.2×1013 Kg.m2/s2=4.62×1012 J 

The magnitude of binding energy is considered.

Now, the nuclear binding energy per nucleon is calculated as follows:

Nuclear binding energytotal number of nucleons

The number of nucleons is defined as the sum of protons and neutrons.

Thus, 4 nucleons are present in nucleus 24He.

Substitute 4.62×1012 J  for nuclear binding energy and 4 nucleons for the total number of nucleons.

(4.62×1012 J   4 nucleons)=1.15×1012 J/nucleon

Therefore, nuclear binding energy per nucleon is 1.15×1012 J/nucleon.

b) 74184W (183.950928  amu)

In the given nucleus, 74184W, the number of protons and neutrons are 74 and 110, respectively.

The mass of 74 11H atoms is as follows:

Mass of 74 11=74×mass of hydrogen=(74×1.007825 amu)=74.57905 amu

Mass of 110 neutrons =110×mass of neutron=(110×1.008665 amu)=110.95315 amu

The predicted mass of 74184W is calculated as follows:

Predicted mass = Mass of 74 11H+Mass of 110 neutrons=(74.57905 amu+110.95315 amu)=185.5323 amu

The mass defect is calculated by using the relation as follows:

Δm= Given mass Predicted mass

Substitute 185.5323 amu for predicted mass and 183.950928  amu for given mass,

Δm=(183.950928  amu)(185.5323 amu)=1.5814 amu.

The value of massdefect is converted into Kg by using the relation:

1 Kg=6.0221418×1026 amu(1.5814 amu×1 Kg6.022×1026 amu)=2.626×1027Kg

The nuclear binding energy is calculated by the relation as follows:

ΔE=(Δm)c2

Substitute 2.626×1027Kg for Δm and 3.00×108 m/s for c in the above expression,

ΔE=(2.626×1027Kg)(3.00×108 m/s)2=2.36×1010 Kg.m2/s2=2.36×1010 J 

The magnitude of binding energy is considered.

Now, the nuclear binding energy per nucleon is calculated as follows:

Nuclear binding energytotal number of nucleons

The number of nucleons is defined as the sum of protons and neutrons.

Thus, 184 nucleons are present in nucleus 74184W.

Substitute 2.36×1010 J  for nuclear binding energy and 184 nucleons for thetotal number of nucleons,

(2.36×1010 J    184 nucleons)=1.28×1012 J/nucleon

Therefore, the nuclear binding energy per nucleon is 1.28×1012 J/nucleon.

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Chapter 20 Solutions

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