Chap.18 Ex. Atom nucleid

 

Exercises

Chap.18

 

Nuclear Physics: Nucleus, Stability & Binding Energy

1. Calculate the mass defect and binding energy of iron-56 nucleus (⁵⁶Fe) with mass 55.934937 u. (Mp = 1.007276 u, Mn = 1.008665 u)

Solution:
Number of protons = 26, neutrons = 30
Theoretical mass = (26×1.007276) + (30×1.008665) = 56.449346 u
Mass defect = 56.449346 - 55.934937 = 0.514409 u
Binding energy = 0.514409 × 931.5 ≈ 479.17 MeV
Binding energy per nucleon = 479.17/56 ≈ 8.56 MeV/nucleon

2. Using the Aston curve, explain why medium-mass nuclei are the most stable.

Solution:
The Aston curve shows binding energy per nucleon peaks for medium-mass nuclei (A ≈ 50-60).
This means these nuclei have:
- The most stable configuration (minimum energy state)
- Optimal balance between strong force and Coulomb repulsion
- For heavy nuclei: Coulomb repulsion reduces stability
- For light nuclei: less benefit from short-range strong force

3. Compare the stability of these nuclei with justification: ¹²C, ⁵⁶Fe, ²³⁸U.

Solution:
Stability order: ⁵⁶Fe > ¹²C > ²³⁸U
Justification:
- ⁵⁶Fe is near the peak of Aston curve (binding energy/nucleon ≈ 8.8 MeV)
- ¹²C has binding energy/nucleon ≈ 7.7 MeV (less stable than Fe)
- ²³⁸U has binding energy/nucleon ≈ 7.6 MeV (less stable due to proton Coulomb repulsion)

4. Calculate the radius of silver-107 nucleus (A = 107) using R = R₀A^1/3 with R₀ = 1.2 fm.

Solution:
R = 1.2 × 107^1/3
107^1/3 ≈ 4.747 (since 4.747³ ≈ 107)
R ≈ 1.2 × 4.747 ≈ 5.696 fm
The Ag-107 nucleus has radius ≈ 5.7 femtometers