FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
Question
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Chapter 27, Problem 43QAP
To determine

(a)

The binding energy of per nucleon of H2

Expert Solution
Check Mark

Explanation of Solution

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of H2=2.01355321270u

  H12+Energyp++n0

Mass defect

  Δm=(1.007276467u+1.008664917u)2.01355321270=0.023881713u

Converting this into energy by below equation

  E=mc2

  E=0.023881713u×931.494061MeVu=2.224567MeV

The H2 nucleus has four nucleons (one neutron and one proton), therefore the binding energy per nucleon is

  E2=2.224567MeV2=1.1122835MeV/nucleon

Conclusion:

The binding energy of per nucleon of H2 is 1.1122835MeV/nucleon.

To determine

(b)

The binding energy of per nucleon of H4e

Expert Solution
Check Mark

Explanation of Solution

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of H4e=4.002602u

  H24e+Energy2p++2n0

Mass defect

  Δm=(2×1.008665+2×1.007825)4.002602=0.030378u

Converting this into energy by below equation

  E=mc2

  E=0.030378u×931.494061MeVu=28.2992659MeV

The H4e nucleus has four nucleons (two neutrons and one protons), therefore the binding energy per nucleon is

  E4=28.2992659MeV4=7.0742MeV/nucleon

Conclusion:

The binding energy of per nucleon of H4e is 7.0742MeV/nucleon.

To determine

(c)

The binding energy of per nucleon of L6i

Expert Solution
Check Mark

Explanation of Solution

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of L6i=6.01348 u

  L36i+Energy3p++4n0

Mass defect

  Δm=(3×1.008664917+3×1.007276467)6.01348=0.034344152u

Converting this into energy by below equation

  E=mc2

  E=0.034344152u×931.494061MeVu=31.99137362MeV

The L6i nucleus has 6 nucleons therefore the binding energy per nucleon is

  E6=31.99137362MeV6=5.331MeV/nucleon

Conclusion:

The binding energy of per nucleon of L6i is 5.331MeV/nucleon.

To determine

(d)

The binding energy of per nucleon of C12

Expert Solution
Check Mark

Explanation of Solution

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of C12=12.000 u

  C612+Energy6p++6n0

Mass defect

  Δm=(6×1.008664917+6×1.007276467)12=0.095648304u

Converting this into energy by below equation

  E=mc2

  E=0.095648304u×931.494061MeVu=89.0958MeV

The C12 nucleus has 12 nucleons therefore the binding energy per nucleon is

  E12=89.0958MeV12=7.6802MeV/nucleon

Conclusion:

The binding energy of per nucleon of C12 is 7.6802MeV/nucleon.

To determine

(e)

The binding energy of per nucleon of F56e

Expert Solution
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Explanation of Solution

Calculation:

 

There are 26 protons and 30 neutrons in the stable iron isotope (Iron / Fe-56). Isotopes are atoms of the same element with the same number of atoms (the same number of protons), but with a different mass. The atomic number of all iron isotopes is 26. However, their masses are different.

Natural iron (Fe) consists of 4 isotopes: 5.845% of radioactive 54Fe, 91.754% of stable 56Fe, 2.119% of stable 57Fe and 0.282% of stable 58Fe. 56Fe of which is the most common because it is the most common endpoint of fusion chains in extremely massive stars.

The mass of Iron/Fe-56 M(F2656e)=55.9349375u

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

The weight of an electron (it has a weight of about 1/1836 that of the proton):

  Me=5.4857990943×104u

  F2656e+Energy26p++30n0

  26M(H1)+30mn=56.46339748u

  Δm=(56.46339748u55.9349375u)=0.52845998u

Converting this into energy by below equation

  E=mc2

  E=0.52845998u×931.494061MeVu=492.26MeV

The F56e nucleus has 56 nucleons therefore the binding energy per nucleon is

  E56=492.26MeV56=8.79031MeV/nucleon

Conclusion:

The binding energy of per nucleon of F56e is 8.79031MeV/nucleon.

To determine

(f)

The binding energy of per nucleon of S90r

Expert Solution
Check Mark

Explanation of Solution

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of S90r=89.908 u

  S90r+Energy38p++52n0

Mass defect

  Δm=(52×1.008664917+38×1.007276467)89.908=1.377170663u

Converting this into energy by below equation

  E=mc2

  E=1.377170663u×931.494061MeVu=1282.826294MeV

The S90r nucleus has 90 nucleons therefore the binding energy per nucleon is

  E90=1282.826294MeV90=8.69594MeV/nucleon

Conclusion:

The binding energy of per nucleon of S90r is 8.69594MeV/nucleon.

To determine

(g)

The binding energy of per nucleon of I129

Expert Solution
Check Mark

Explanation of Solution

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of I129=89.908 u

  I129+Energy53p++76n0

Mass defect

  Δm=(76×1.008664917+53×1.007276467)129.913=0.11186443u

Converting this into energy by below equation

  E=mc2

  E=0.13186443u×931.494061MeVu=121.1994MeV

The I129 nucleus has 129 nucleons therefore the binding energy per nucleon is

  E129=121.1994MeV129=8.43602MeV/nucleon

Conclusion:

The binding energy of per nucleon of U235 is 8.43602MeV/nucleon.

To determine

(H)

The binding energy of per nucleon of U235

Expert Solution
Check Mark

Explanation of Solution

Mass of a proton Mp=1.672621777×1027kg=1.007276467u

Mass of Hydrogen atom M(H1)=1.007825u

Mass of a neutron Mn=1.674927351×1027kg=1.008664917u

Atomic mass of U235=235.04393

  I129+Energy53p++76n0

Mass defect

  Δm=(143×1.008664917+92×1.007276467)235.04393=1.864588095u

Converting this into energy by below equation

  E=mc2

  E=1.864588095u×931.494061MeVu=1736.852736703803795MeV

The U235 nucleus has 235 nucleons therefore the binding energy per nucleon is

  E235=1736.852736703803795MeV235=7.39086MeV/nucleon

Conclusion:

The binding energy of per nucleon of U235 is 7.39086MeV/nucleon.

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Chapter 27 Solutions

FlipIt for College Physics (Algebra Version - Six Months Access)

Ch. 27 - Prob. 11QAPCh. 27 - Prob. 12QAPCh. 27 - Prob. 13QAPCh. 27 - Prob. 14QAPCh. 27 - Prob. 15QAPCh. 27 - Prob. 16QAPCh. 27 - Prob. 17QAPCh. 27 - Prob. 18QAPCh. 27 - Prob. 19QAPCh. 27 - Prob. 20QAPCh. 27 - Prob. 21QAPCh. 27 - Prob. 22QAPCh. 27 - Prob. 23QAPCh. 27 - Prob. 24QAPCh. 27 - Prob. 25QAPCh. 27 - Prob. 26QAPCh. 27 - Prob. 27QAPCh. 27 - Prob. 28QAPCh. 27 - Prob. 29QAPCh. 27 - Prob. 30QAPCh. 27 - Prob. 31QAPCh. 27 - Prob. 32QAPCh. 27 - Prob. 33QAPCh. 27 - Prob. 34QAPCh. 27 - Prob. 35QAPCh. 27 - Prob. 36QAPCh. 27 - Prob. 37QAPCh. 27 - Prob. 38QAPCh. 27 - Prob. 39QAPCh. 27 - Prob. 40QAPCh. 27 - Prob. 41QAPCh. 27 - Prob. 42QAPCh. 27 - Prob. 43QAPCh. 27 - Prob. 44QAPCh. 27 - Prob. 45QAPCh. 27 - Prob. 46QAPCh. 27 - Prob. 47QAPCh. 27 - Prob. 48QAPCh. 27 - Prob. 49QAPCh. 27 - Prob. 50QAPCh. 27 - Prob. 51QAPCh. 27 - Prob. 52QAPCh. 27 - Prob. 53QAPCh. 27 - Prob. 54QAPCh. 27 - Prob. 55QAPCh. 27 - Prob. 56QAPCh. 27 - Prob. 57QAPCh. 27 - Prob. 58QAPCh. 27 - Prob. 59QAPCh. 27 - Prob. 60QAPCh. 27 - Prob. 61QAPCh. 27 - Prob. 62QAPCh. 27 - Prob. 63QAPCh. 27 - Prob. 64QAPCh. 27 - Prob. 65QAPCh. 27 - Prob. 66QAPCh. 27 - Prob. 67QAPCh. 27 - Prob. 68QAPCh. 27 - Prob. 69QAPCh. 27 - Prob. 70QAPCh. 27 - Prob. 71QAPCh. 27 - Prob. 72QAPCh. 27 - Prob. 73QAPCh. 27 - Prob. 74QAPCh. 27 - Prob. 75QAPCh. 27 - Prob. 76QAPCh. 27 - Prob. 77QAPCh. 27 - Prob. 78QAPCh. 27 - Prob. 79QAPCh. 27 - Prob. 80QAPCh. 27 - Prob. 81QAPCh. 27 - Prob. 82QAPCh. 27 - Prob. 83QAPCh. 27 - Prob. 84QAPCh. 27 - Prob. 85QAPCh. 27 - Prob. 86QAPCh. 27 - Prob. 87QAPCh. 27 - Prob. 88QAPCh. 27 - Prob. 89QAPCh. 27 - Prob. 90QAPCh. 27 - Prob. 91QAPCh. 27 - Prob. 92QAPCh. 27 - Prob. 93QAPCh. 27 - Prob. 94QAPCh. 27 - Prob. 95QAPCh. 27 - Prob. 96QAPCh. 27 - Prob. 97QAP
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