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A Look at Nuclear Science and Technology Larry Foulke
2.2 Atomic and Nuclear Physics – The Einstein Connection Binding Energy and Nuclear Glue
The Nucleus
Protons
Neutrons
• Protons each carry a positive charge, so how is it possible to hold so many protons together in such a small volume?
• Answer: The (Strong) Nuclear Force Image Source: See Note 1
(Strong) Nuclear Force Coulomb Forces p p repulsive
p
n
no effect
p
e attractive
Nuclear Force p p attractive
p
n
p
e no effect
• (Strong) Nuclear Force – Force between nucleons (protons and neutrons). – Extremely attractive force at ranges up to 2×10-15 m (≈ 2 nucleon diameters). – Force drops to zero beyond 2×10-15 m. – Force acts equally on protons and neutrons. 3
Nuclear Stability Lithium-7 Nucleus n p n
p
n p
n
• Nuclear Stability – The attractive nuclear force and repulsive coulomb force exactly balance in a stable nucleus. – Any change in the nuclear composition will change the balance of forces.
Nuclear Force (Attractive) Coulomb Force (Repulsive)
– Lots of potential energy in the nucleus. 4
Nuclear Binding Energy Due to the structure of the nucleus and the balance of forces, nucleons bound in a nucleus are more stable (have a lower energy) than free nucleons.
E=mc2 • When bound, each nucleon turns a small fraction of its mass into energy, which is typically radiated from the nucleus. • This binding energy must be added to the nucleus to remove (unbind) a nucleon. Image Source: See Note 2
5
Nuclear Binding Energy Mass Defect = Σ Massconstituents - Massbound nucleus – Constituents may be individual nucleons or two (or more) nuclei. Example: 6Li + 6Li = 12C
• Mass Energy Equivalence:
E=m c2
• Binding Energy:
Binding Energy =
[Σ Mass
constituents
- Massbound nucleus
]c
2
6
Montessori Muddle by Montessori Muddle is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License.
Nuclear Binding Energy The binding
p
p
2 protons = 2 x 1.007276 u = 2.014552 u
n
n
2 neutrons = 2 x 1.008665 u = 2.017330 u
Total mass of individual particles = 4.031882 u
p Total mass of bound nucleus = 4.001475 u n n p Mass loss = 0.030407 u
energy can be seen as a mass defect between the weight of the nucleus and the individual (unbound) weights of its constituent nucleons.
u = abbreviation for E=mc2 – Lots of amu Energy Released from mass defect worth energy for a very 28.3 MeV small mass defect (0.030407amy)(931.5MeV/amu)=28.3 Mev
Conversion Factors • Mass 1 amu = 1 u = 1.661× 10-27 kg • Energy 1 eV = 1.602 × 10-19 Joule 1 Joule = 1 kg × m2 / s2 • Velocity (Speed of Light) c = 2.998 × 108 m / s • Mass to Energy Conversion (E=mc2) 1 amu = 931.50 MeV 8
Binding Energy Tips • Nuclei are more stable than free nucleons. • The binding energy provides a measure of how tightly bound a nucleus is. • Nuclei with larger binding energies require more energy to break apart. • Binding Energy Per Nucleon – Gives a measure of the forces acting on each nucleon in the nucleus. – Amount of energy (average) required to rip a single nucleon out of the nucleus. 9
Binding Energy Per Nucleon
Image Source: See Note 3
10
Decreasing Nuclear Strength
Decreasing Nuclear Strength
Binding Energy Per Nucleon
Most Tightly Bound (Nickel-62) Image Source: See Note 3
11
Binding Energy Example Change in binding energy per nucleon
118 2 × 8.5 × 118 = 2006 MeV
236 7.5 × 236 = 1770 MeV
Tighter bound nuclei = Lower energy state
BE118 > BE235
Reaction: 2×118 → 236, Add 236 MeV (endothermic) Reaction: 236 → 2×118, Remove 236 MeV (exothermic)
Image Source: See Note 3
12
Binding Energy Example
Change in binding energy per nucleon
D-T Fusion
Image Source: See Note 3
4 (Helium)4 × 7 = 28 MeV
BE4 > BE2 + BE3 Reaction: 2 + 3 → 4 (9+2)-(28)=-17 Remove 17 MeV (exothermic)
3 (Tritium) 3 × 3 Reaction: 4 → 2 + 3 = 9 MeV (28)-(9+2)=17 2 (Deuterium) 2 × 1 Add 17 MeV (endothermic) = 2 MeV 13
Image Source Notes 1. 2. 3.
Creative Commons: http://en.wikipedia.org/wiki/File:Helium_atom_QM.svg Public domain: http://en.wikipedia.org/wiki/File:Einsteinformal_portrait-35.jpg Background graph in public domain. Source: https://commons.wikimedia.org/wiki/File:Binding_energ y_curve_-_common_isotopes.svg; Overlay reprinted with permission from the American Nuclear Society. Nuclear Engineering – Theory and Technology of Commercial Nuclear Power by Ronald Allen Knief, 2nd Edition. Copyright 2008 by the American Nuclear Society, La Grange Park, Illinois. Figure 2-1.
2.2 Atomic and Nuclear Physics – The Einstein Connection Binding Energy and Nuclear Glue
The Nucleus
Protons
Neutrons
• Protons each carry a positive charge, so how is it possible to hold so many protons together in such a small volume?
• Answer: The (Strong) Nuclear Force Image Source: See Note 1
(Strong) Nuclear Force Coulomb Forces p p repulsive
p
n
no effect
p
e attractive
Nuclear Force p p attractive
p
n
p
e no effect
• (Strong) Nuclear Force – Force between nucleons (protons and neutrons). – Extremely attractive force at ranges up to 2×10-15 m (≈ 2 nucleon diameters). – Force drops to zero beyond 2×10-15 m. – Force acts equally on protons and neutrons. 3
Nuclear Stability Lithium-7 Nucleus n p n
p
n p
n
• Nuclear Stability – The attractive nuclear force and repulsive coulomb force exactly balance in a stable nucleus. – Any change in the nuclear composition will change the balance of forces.
Nuclear Force (Attractive) Coulomb Force (Repulsive)
– Lots of potential energy in the nucleus. 4
Nuclear Binding Energy Due to the structure of the nucleus and the balance of forces, nucleons bound in a nucleus are more stable (have a lower energy) than free nucleons.
E=mc2 • When bound, each nucleon turns a small fraction of its mass into energy, which is typically radiated from the nucleus. • This binding energy must be added to the nucleus to remove (unbind) a nucleon. Image Source: See Note 2
5
Nuclear Binding Energy Mass Defect = Σ Massconstituents - Massbound nucleus – Constituents may be individual nucleons or two (or more) nuclei. Example: 6Li + 6Li = 12C
• Mass Energy Equivalence:
E=m c2
• Binding Energy:
Binding Energy =
[Σ Mass
constituents
- Massbound nucleus
]c
2
6
Montessori Muddle by Montessori Muddle is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License.
Nuclear Binding Energy The binding
p
p
2 protons = 2 x 1.007276 u = 2.014552 u
n
n
2 neutrons = 2 x 1.008665 u = 2.017330 u
Total mass of individual particles = 4.031882 u
p Total mass of bound nucleus = 4.001475 u n n p Mass loss = 0.030407 u
energy can be seen as a mass defect between the weight of the nucleus and the individual (unbound) weights of its constituent nucleons.
u = abbreviation for E=mc2 – Lots of amu Energy Released from mass defect worth energy for a very 28.3 MeV small mass defect (0.030407amy)(931.5MeV/amu)=28.3 Mev
Conversion Factors • Mass 1 amu = 1 u = 1.661× 10-27 kg • Energy 1 eV = 1.602 × 10-19 Joule 1 Joule = 1 kg × m2 / s2 • Velocity (Speed of Light) c = 2.998 × 108 m / s • Mass to Energy Conversion (E=mc2) 1 amu = 931.50 MeV 8
Binding Energy Tips • Nuclei are more stable than free nucleons. • The binding energy provides a measure of how tightly bound a nucleus is. • Nuclei with larger binding energies require more energy to break apart. • Binding Energy Per Nucleon – Gives a measure of the forces acting on each nucleon in the nucleus. – Amount of energy (average) required to rip a single nucleon out of the nucleus. 9
Binding Energy Per Nucleon
Image Source: See Note 3
10
Decreasing Nuclear Strength
Decreasing Nuclear Strength
Binding Energy Per Nucleon
Most Tightly Bound (Nickel-62) Image Source: See Note 3
11
Binding Energy Example Change in binding energy per nucleon
118 2 × 8.5 × 118 = 2006 MeV
236 7.5 × 236 = 1770 MeV
Tighter bound nuclei = Lower energy state
BE118 > BE235
Reaction: 2×118 → 236, Add 236 MeV (endothermic) Reaction: 236 → 2×118, Remove 236 MeV (exothermic)
Image Source: See Note 3
12
Binding Energy Example
Change in binding energy per nucleon
D-T Fusion
Image Source: See Note 3
4 (Helium)4 × 7 = 28 MeV
BE4 > BE2 + BE3 Reaction: 2 + 3 → 4 (9+2)-(28)=-17 Remove 17 MeV (exothermic)
3 (Tritium) 3 × 3 Reaction: 4 → 2 + 3 = 9 MeV (28)-(9+2)=17 2 (Deuterium) 2 × 1 Add 17 MeV (endothermic) = 2 MeV 13
Image Source Notes 1. 2. 3.
Creative Commons: http://en.wikipedia.org/wiki/File:Helium_atom_QM.svg Public domain: http://en.wikipedia.org/wiki/File:Einsteinformal_portrait-35.jpg Background graph in public domain. Source: https://commons.wikimedia.org/wiki/File:Binding_energ y_curve_-_common_isotopes.svg; Overlay reprinted with permission from the American Nuclear Society. Nuclear Engineering – Theory and Technology of Commercial Nuclear Power by Ronald Allen Knief, 2nd Edition. Copyright 2008 by the American Nuclear Society, La Grange Park, Illinois. Figure 2-1.
