2- Calculate the mass defect of this nucleus in atomic mass units then in kilograms.
Solution:
Theoretical mass = (92 × 1.00728 u) + (143 × 1.00866 u) = 92.66976 u + 144.23838 u = 236.90814 u
Mass defect = 236.90814 u - 234.99332 u = 1.91482 u
In kg: 1.91482 u × 1.66054×10⁻²⁷ kg/u ≈ 3.179×10⁻²⁷ kg
3- Calculate, in joules then in MeV, the binding energy of this nucleus.
Given: 1 eV = 1.6022×10⁻¹⁹ J; c = 3×10⁸ m/s; 1u = 931.5 MeV/c²
Solution:
In MeV: E = Δm × 931.5 = 1.91482 u × 931.5 MeV/u ≈ 1783.66 MeV
In joules: E = Δm × c² = (3.179×10⁻²⁷ kg) × (3×10⁸ m/s)² ≈ 2.861×10⁻¹⁰ J
Or via conversion: 1783.66×10⁶ eV × 1.6022×10⁻¹⁹ J/eV ≈ 2.857×10⁻¹⁰ J
4- Calculate the binding energy per nucleon of this nucleus.
Solution:
Binding energy per nucleon = 1783.66 MeV / 235 nucleons ≈ 7.59 MeV/nucleon
5- Compare the stability of uranium-235 nucleus to that of radium-226 nucleus whose binding energy is 7.66 MeV per nucleon.
Solution:
Uranium-235 has binding energy per nucleon of 7.59 MeV
Radium-226 has binding energy per nucleon of 7.66 MeV
Conclusion: The radium-226 nucleus is slightly more stable than uranium-235 because its binding energy per nucleon is higher.
💞 Questions // Group B
Study of Gold Radionuclide
6) a. Calculate the mass of a gold atom Au.
Solution:
Molar mass = 198 g/mol = 0.198 kg/mol
Mass of one atom = Molar mass / NA
= 0.198 kg/mol / 6.022×10²³ mol⁻¹ ≈ 3.288×10⁻²⁵ kg
Or in atomic units:
Nucleus mass = 197.925 u
Electrons mass = 79 × 5.50×10⁻⁴ u ≈ 0.04345 u
Atom mass ≈ 197.925 u + 0.04345 u ≈ 197.968 u
6) b. Compare the mass of gold atom Au to its nucleus mass.
Solution:
Nucleus mass = 197.925 u
Atom mass ≈ 197.968 u
Difference ≈ 0.043 u (only 0.022% of total mass)
Conclusion: The atom mass is practically equal to its nucleus mass,
electrons contribute very little to the total mass.
7) The average radius of gold atom is r = 1.6×10⁻¹⁰ m. The average nucleon radius is r₀ = 1.2×10⁻¹⁵ m. Compare the density of gold atom to its nucleus density. Conclude about matter distribution in the atom.
Solution: For nucleus:
Radius R = r₀ × A^(1/3) = 1.2×10⁻¹⁵ × 197^(1/3) ≈ 6.98×10⁻¹⁵ m
Volume V = (4/3)πR³ ≈ 1.43×10⁻⁴² m³
Density ρ = m/V ≈ 197.925×1.66×10⁻²⁷ kg / 1.43×10⁻⁴² m³ ≈ 2.3×10¹⁷ kg/m³
For atom:
Volume V = (4/3)πr³ ≈ 1.72×10⁻²⁹ m³
Density ρ = m/V ≈ 3.288×10⁻²⁵ kg / 1.72×10⁻²⁹ m³ ≈ 1.9×10⁴ kg/m³
Conclusion:
Matter is extremely concentrated in the nucleus,
which contains almost all the mass in a very small volume,
the rest of the atom being essentially empty space.
💞 Questions // Group C
Nuclear Physics - Nucleus, Stability & Binding Energy
8. 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
9. 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
10. 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)
11. Calculate the radius of silver-107 nucleus (A = 107) using the formula R = R₀A1/3 with R₀ = 1.2 fm.
Solution:
R = 1.2 × 1071/3
1071/3 ≈ 4.747 (since 4.747³ ≈ 107)
R ≈ 1.2 × 4.747 ≈ 5.696 fm
The Ag-107 nucleus has radius ≈ 5.7 femtometers
12. Why are nuclei with magic numbers of protons/neutrons exceptionally stable?
Solution:
Magic numbers (2, 8, 20, 28, 50, 82, 126) correspond to filled nuclear shells:
- Analogous to stable electron configurations
- Minimum energy configuration
- More difficult to add/remove nucleons
- Examples: ⁴He (2p+2n), ¹⁶O (8p+8n), ⁴⁸Ca (20p+28n)
- "Doubly magic" nuclei (magic p and n) are extremely stable
Exercice 1 : Étude du radionucléide Au
On donne :
- Masse molaire de Au : 198 g/mol
- Nombre d'Avogadro : 6,022×10²³ mol⁻¹
- Masse de l'électron : 5,50×10⁻⁴ u
- Célérité de la lumière : c = 3×10⁸ m/s
- 1 u = 931,5 MeV/c² = 1,66×10⁻²⁷ kg
- 1 eV = 1,6×10⁻¹⁹ J
- Masse du noyau Au : 197,925 u
- Masse du noyau Hg : 197,923 u
- Masse du proton : mp = 1,00728 u
- Masse du neutron : mn = 1,00866 u
A- Comparaison de la masse volumique du noyau d'or et de celle de l'atome d'or
1) a. Calculer la masse d'un atome d'or Au.
Solution:
Masse molaire = 198 g/mol = 0,198 kg/mol
Masse d'un atome = Masse molaire / NA
= 0,198 kg/mol / 6,022×10²³ mol⁻¹ ≈ 3,288×10⁻²⁵ kg
Ou en unités atomiques :
Masse noyau = 197,925 u
Masse électrons = 79 × 5,50×10⁻⁴ u ≈ 0,04345 u
Masse atome ≈ 197,925 u + 0,04345 u ≈ 197,968 u
1) b. Comparer la masse de l'atome d'or Au à celle de son noyau.
Solution:
Masse noyau = 197,925 u
Masse atome ≈ 197,968 u
Différence ≈ 0,043 u (soit seulement 0,022% de la masse totale)
Conclusion : La masse de l'atome est pratiquement égale à celle de son noyau,
les électrons contribuent très peu à la masse totale.
2) Le rayon moyen d'un atome d'or est r = 1,6×10⁻¹⁰ m. Le rayon moyen d'un nucléon est r₀ = 1,2×10⁻¹⁵ m. Comparer la masse volumique de l'atome d'or à celle de son noyau. Conclure à propos de la répartition de la matière dans l'atome.
Solution: Pour le noyau :
Rayon R = r₀ × A^(1/3) = 1,2×10⁻¹⁵ × 197^(1/3) ≈ 6,98×10⁻¹⁵ m
Volume V = (4/3)πR³ ≈ 1,43×10⁻⁴² m³
Masse volumique ρ = m/V ≈ 197,925×1,66×10⁻²⁷ kg / 1,43×10⁻⁴² m³ ≈ 2,3×10¹⁷ kg/m³
Pour l'atome :
Volume V = (4/3)πr³ ≈ 1,72×10⁻²⁹ m³
Masse volumique ρ = m/V ≈ 3,288×10⁻²⁵ kg / 1,72×10⁻²⁹ m³ ≈ 1,9×10⁴ kg/m³
Conclusion :
La matière est extrêmement concentrée dans le noyau,
qui contient presque toute la masse dans un volume très petit,
le reste de l'atome étant essentiellement du vide.
Exercise: Study of Gold Radionuclide
Given:
- Molar mass of Au: 198 g/mol
- Avogadro's number: 6.022×10²³ mol⁻¹
- Electron mass: 5.50×10⁻⁴ u
- Speed of light: c = 3×10⁸ m/s
- 1 u = 931.5 MeV/c² = 1.66×10⁻²⁷ kg
- 1 eV = 1.6×10⁻¹⁹ J
- Gold nucleus mass: 197.925 u
- Mercury nucleus mass: 197.923 u
- Proton mass: mp = 1.00728 u
- Neutron mass: mn = 1.00866 u
A- Comparison of gold nucleus density and gold atom density
1) a. Calculate the mass of a gold atom Au.
Solution:
Molar mass = 198 g/mol = 0.198 kg/mol
Mass of one atom = Molar mass / NA
= 0.198 kg/mol / 6.022×10²³ mol⁻¹ ≈ 3.288×10⁻²⁵ kg
Or in atomic units:
Nucleus mass = 197.925 u
Electrons mass = 79 × 5.50×10⁻⁴ u ≈ 0.04345 u
Atom mass ≈ 197.925 u + 0.04345 u ≈ 197.968 u
1) b. Compare the mass of gold atom Au to its nucleus mass.
Solution:
Nucleus mass = 197.925 u
Atom mass ≈ 197.968 u
Difference ≈ 0.043 u (only 0.022% of total mass)
Conclusion: The atom mass is practically equal to its nucleus mass,
electrons contribute very little to the total mass.
2) The average radius of gold atom is r = 1.6×10⁻¹⁰ m. The average nucleon radius is r₀ = 1.2×10⁻¹⁵ m. Compare the density of gold atom to its nucleus density. Conclude about matter distribution in the atom.
Solution: For nucleus:
Radius R = r₀ × A^(1/3) = 1.2×10⁻¹⁵ × 197^(1/3) ≈ 6.98×10⁻¹⁵ m
Volume V = (4/3)πR³ ≈ 1.43×10⁻⁴² m³
Density ρ = m/V ≈ 197.925×1.66×10⁻²⁷ kg / 1.43×10⁻⁴² m³ ≈ 2.3×10¹⁷ kg/m³
For atom:
Volume V = (4/3)πr³ ≈ 1.72×10⁻²⁹ m³
Density ρ = m/V ≈ 3.288×10⁻²⁵ kg / 1.72×10⁻²⁹ m³ ≈ 1.9×10⁴ kg/m³
Conclusion:
Matter is extremely concentrated in the nucleus,
which contains almost all the mass in a very small volume,
the rest of the atom being essentially empty space.
Exercise 2:
Cobalt-60 Nuclear Excited State
A cobalt-60 nucleus 6027Co in an excited state emits γ radiation with energy 1.33 MeV when returning to its ground state.
Given:
• Ground state mass = 59.93382 u
• γ energy = 1.33 MeV
• 1 u = 931.5 MeV/c²
Find: The mass of the nucleus in the excited state.
Solution:
Step 1: Convert γ energy to atomic mass units (u)
1.33 MeV ÷ 931.5 MeV/c²/u = 0.0014278 u
Step 2: Calculate excited state mass
Massexcited = Massground + Eγ/c²
= 59.93382 u + 0.0014278 u
= 59.9352478 u
Final Answer:
The mass of the cobalt-60 nucleus in the excited state is 59.93525 u (rounded to 6 decimal places).
Exercise 3:
Calculating Number of Nuclei
A sample contains 5 mg of radium-226 (22688Ra) where each nucleus has a mass of 225.9770 u.
Given:
• Sample mass = 5 mg = 5×10-6 kg
• Mass per nucleus = 225.9770 u
• 1 u = 1.66×10-27 kg
Calculate: The number of nuclei in this sample.
Solution:
Step 1: Convert nuclear mass to kg
Mass per nucleus = 225.9770 u × 1.66×10-27 kg/u
= 3.7512182×10-25 kg
Step 2: Calculate number of nuclei
Number of nuclei = Total mass / Mass per nucleus
= 5×10-6 kg / 3.7512182×10-25 kg
≈ 1.33×1019 nuclei
Final Answer:
The 5 mg sample of Ra-226 contains approximately 1.33×1019 nuclei.
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