Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 44, Problem 16P

(a)

To determine

The difference in binding energy per nucleon for nuclei N1123a and M1223g.

(a)

Expert Solution
Check Mark

Answer to Problem 16P

The difference in binding energy per nucleon for nuclei N1123a and M1223g is 0.210 MeV.

Explanation of Solution

Write the expression to find the binding energy of a nucleus.

  Eb=[ZM(H)+NmnM(XZA)](931.494MeV/u)

Here, Eb is the binding energy, M(H) is the atomic mass of hydrogen atom, mn is the mass of neutron, M(XZA) is the atomic mass an atom of isotope XZA, Z is the number of electrons, N is the number of neutrons.

Re-write the expression to find the binding energy of a nucleus for N1123a.

  Eb(N1123a)=[11M(H)+12mnM(N1123a)](931.494MeV/u)

Substitute 1.007825 u for M(H), 1.008665 u for mn, 22.989769 u for M(N1123a) to find the binding energy of N1123a.

  Eb(N1123a)=[11(1.007825 u)+12(1.008665 u)22.989769 u]                                                       ×(931.494 MeV/u)=186.565 MeV

Divide binding energy with mass number to find the difference in binding energy per nucleon for N1123a.

  EbA=186.565 MeV23=8.11 MeV

Re-write the expression to find the binding energy of a nucleus for M1223g.

  Eb(M1223g)=[12M(H)+11mnM(M1223g)](931.494MeV/u)

Substitute 1.007825 u for M(H), 1.008665 u for mn, 22.994124 u for M(M1223g) to find the binding energy of M1223g.

  Eb=Eb(M1223g)=[12M(H)+11mnM(M1223g)](931.494MeV/u)=[12(1.007825 u)+11(1.008665 u)22.994124 u]×(931.494MeV/u)=181.726 MeV

Divide binding energy with mass number to find the difference in binding energy per nucleon for M1223g.

  EbA=181.726 MeV23=7.90 MeV

Write the expression to find the difference in binding energy per nucleon for nuclei N1123a and M1223g.

  ΔEbA=Eb(N1123a)Eb(M1223g)A

Here, ΔEbA is the difference in binding energy per nucleon.

Conclusion:

Substitute 8.11 MeV for Eb(N1123a)A and 7.90 MeV for Eb(M1223g)A to find the difference in binding energy per nucleon for nuclei N1123a and M1223g.

  ΔEbA=8.11 MeV7.90 MeV=0.210 MeV

Therefore, the difference in binding energy per nucleon for nuclei N1123a and M1223g is 0.210 MeV.

(b)

To determine

The findings from the difference in binding energy per nucleon for the nuclei N1123a and M1223g

(b)

Expert Solution
Check Mark

Answer to Problem 16P

The N1123a is more stable nucleus than M1223g.

Explanation of Solution

Write the expression to find the difference in binding energy per nucleon for nuclei N1123a and M1223g.

  ΔEbA=Eb(N1123a)Eb(M1223g)A

Here, ΔEbA is the difference in binding energy per nucleon.

Substitute 8.11 MeV for Eb(N1123a)A and 7.90 MeV for Eb(M1223g)A to find the difference in binding energy per nucleon for nuclei N1123a and M1223g.

  ΔEbA=8.11 MeV7.90 MeV=0.210 MeV

Conclusion:

Therefore, the difference in binding energy per nucleon for nuclei N1123a and M1223g is 0.210 MeV. It indicates that the binding energy per nucleon is higher in N1123a by 0.210 MeV.

Thus, In N1123a the proton repulsion will be less compared to M1223g. therefore, the N1123a is more stable nucleus than M1223g.

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Chapter 44 Solutions

Physics for Scientists and Engineers With Modern Physics

Ch. 44 - Prob. 8OQCh. 44 - Prob. 9OQCh. 44 - Prob. 10OQCh. 44 - Prob. 11OQCh. 44 - Prob. 12OQCh. 44 - Prob. 13OQCh. 44 - Prob. 1CQCh. 44 - Prob. 2CQCh. 44 - Prob. 3CQCh. 44 - Prob. 4CQCh. 44 - Prob. 5CQCh. 44 - Prob. 6CQCh. 44 - Prob. 7CQCh. 44 - Prob. 8CQCh. 44 - Prob. 9CQCh. 44 - Prob. 10CQCh. 44 - Prob. 11CQCh. 44 - Prob. 12CQCh. 44 - Prob. 13CQCh. 44 - Prob. 14CQCh. 44 - Prob. 15CQCh. 44 - Prob. 16CQCh. 44 - Prob. 17CQCh. 44 - Prob. 1PCh. 44 - Prob. 2PCh. 44 - Prob. 3PCh. 44 - Prob. 4PCh. 44 - Prob. 5PCh. 44 - Prob. 6PCh. 44 - Prob. 7PCh. 44 - Prob. 8PCh. 44 - Prob. 9PCh. 44 - Prob. 10PCh. 44 - Prob. 11PCh. 44 - Prob. 12PCh. 44 - Prob. 13PCh. 44 - Prob. 14PCh. 44 - Prob. 15PCh. 44 - Prob. 16PCh. 44 - Prob. 17PCh. 44 - Prob. 18PCh. 44 - Prob. 19PCh. 44 - Prob. 20PCh. 44 - Prob. 21PCh. 44 - Prob. 22PCh. 44 - Prob. 23PCh. 44 - Prob. 24PCh. 44 - Prob. 25PCh. 44 - Prob. 26PCh. 44 - Prob. 27PCh. 44 - Prob. 28PCh. 44 - Prob. 29PCh. 44 - Prob. 31PCh. 44 - Prob. 32PCh. 44 - Prob. 33PCh. 44 - Prob. 34PCh. 44 - Prob. 35PCh. 44 - Prob. 36PCh. 44 - Prob. 37PCh. 44 - Prob. 38PCh. 44 - Prob. 39PCh. 44 - Prob. 40PCh. 44 - Prob. 41PCh. 44 - Prob. 42PCh. 44 - Prob. 43PCh. 44 - Prob. 44PCh. 44 - Prob. 45PCh. 44 - Prob. 46PCh. 44 - Prob. 47PCh. 44 - Prob. 48PCh. 44 - Prob. 49PCh. 44 - Prob. 50PCh. 44 - Prob. 51PCh. 44 - Prob. 52PCh. 44 - Prob. 53PCh. 44 - Prob. 54APCh. 44 - Prob. 55APCh. 44 - Prob. 56APCh. 44 - Prob. 57APCh. 44 - Prob. 58APCh. 44 - Prob. 59APCh. 44 - Prob. 60APCh. 44 - Prob. 61APCh. 44 - Prob. 62APCh. 44 - Prob. 63APCh. 44 - Prob. 64APCh. 44 - Prob. 65APCh. 44 - Prob. 66APCh. 44 - Prob. 67APCh. 44 - Prob. 68APCh. 44 - Prob. 69APCh. 44 - Prob. 70APCh. 44 - Prob. 71APCh. 44 - Prob. 72APCh. 44 - As part of his discovery of the neutron in 1932,...Ch. 44 - Prob. 74APCh. 44 - Prob. 75APCh. 44 - Prob. 76APCh. 44 - Prob. 77CPCh. 44 - Prob. 78CP
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