WebAssign Printed Access Card for Serway/Vuille's College Physics, 11th Edition, Multi-Term
WebAssign Printed Access Card for Serway/Vuille's College Physics, 11th Edition, Multi-Term
11th Edition
ISBN: 9781337763486
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 29, Problem 14P
To determine

The binding energy per nucleon of F2656e , M2555n and C2759o .

Expert Solution & Answer
Check Mark

Answer to Problem 14P

The binding energy per nucleon of F2656e is greater than its neighbors.

Explanation of Solution

Section 1:

To determine: The binding energy per nucleon of F2656e .

Answer: The binding energy per nucleon of F2656e is 8.79MeV/nucleon .

Explanation:

Formula to calculate the binding energy per nucleon is,

BEA=[Zmp+Nmnm(F2656e)]c2A

  • mp is the mass of the proton.
  • mn is the mass of the neutron.
  • Z is the atomic number.
  • N is the neutron number.
  • m(F2656e) is the mass of F2656e .
  • c is the speed of light.
  • A is the mass number.

Substitute 26 for Z, 30 for N,1.008625 u for mn , 1.007825 u for mp , 55.934942 u for m(F2656e) , 931.5 MeV/u for c2 and 56 for A in the above equation.

BEA=[(26)(1.007825u)+(30)(1.008625u)(55.934942u)](931.5MeV/u)56=8.79MeV/nucleon

The binding energy per nucleon of F2656e is 8.79MeV/nucleon .

Section 2:

To determine: The binding energy per nucleon of M2555n .

Answer: The binding energy per nucleon of M2555n is 8.77MeV/nucleon .

Explanation:

Formula to calculate the binding energy per nucleon is,

BEA=[Zmp+Nmnm(M2555n)]c2A

  • m(M2555n) is the mass of M2555n .

Substitute 25 for Z, 30 for N,1.008625 u for mn , 1.007825 u for mp , 54.938050 u for m(M2555n) , 931.5 MeV/u for c2 and 55 for A in the above equation.

BEA=[(26)(1.007825u)+(30)(1.008625u)(55.938050u)](931.5MeV/u)55=8.77MeV/nucleon

The binding energy per nucleon of M2555n is 8.77MeV/nucleon .

Section 3:

To determine: The binding energy per nucleon of C2759o .

Answer: The binding energy per nucleon of C2759o is 8.77MeV/nucleon .

Explanation:

Formula to calculate the binding energy per nucleon is,

BEA=[Zmp+Nmnm(C2759o)]c2A

  • mp is the mass of the proton.
  • mn is the mass of the neutron.
  • Z is the atomic number.
  • N is the neutron number.
  • m(C2759o) is the mass of C2759o .
  • c is the speed of light.
  • A is the mass number.

Substitute 27 for Z, 32 for N,1.008625 u for mn , 1.007825 u for mp , 58.933200 u for m(C2759o) , 931.5 MeV/u for c2 and 59 for A in the above equation.

BEA=[(27)(1.007825u)+(32)(1.008625u)(58.933200u)](931.5MeV/u)59=8.77MeV/nucleon

The binding energy per nucleon of C2759o is 8.77MeV/nucleon .

Conclusion:

From Sections 1, 2 and 3, the binding energy per nucleon of F2656e is greater than its neighbors.

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