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Fundamentals of Nuclear Engineering
Module 1: Atomic and Nuclear Physics
Dr. John H. Bickel
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Objectives:1. Explain key concepts of energy release from nuclear
vs. chemical reactions.2. Explain range of units used for: energy (eV), distances
in SI units3. Explain differences between Coulomb and nuclear
force4. Explain why α, β, γ, sf decay reactions proceed5. Explain types of nuclear reactions, conversion of mass
to energy, and magnitude of energy release 6. Explain Periodic Table and Isotope Table7. Explain how to get information from On-line Isotope
Table data bases.
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Atomic Physics
• Binding of electrons to nucleus• Electromagnetic forces• Interactions between outer electronic shells of atoms and molecules
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An Atom:• Atom composed of:• Z Positively charged (+) protons at center• Z Negatively charged (-) electrons surrounding nucleus in
shells• Protons-electrons bound together by Coulomb force• Binding energy of atomic electrons is: energy required to
separate electron from atom• Binding energy typically expressed in electron Volts (eV)• 1 eV is energy required to move 1 electron against
electrostatic field of 1 Volt.• 1 eV = 1.52 x 10-22 BTU = 4.45 x 10-26 kW-hrs
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Comparing attraction vs. repulsion
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Smaller distance unit makes sense for atoms
• MKS units (meters, kilograms, seconds) are useful for “outside world”
• Separation distances involving atoms, molecules require significantly smaller metric
• 1 Angstrom (Å)= 10-10 meters = 10-8 cm.
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Atoms can be arranged in Periodic Tablebased upon: Z and electron shell structure
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Chemical reaction energies relatively small
• Chemical reactions involve outer electronic shellsouter electronic shells
• Transition to final energy state with lower internal energylower internal energyis done via giving off heat (thermal radiation)
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Example chemical reaction energies
• H H+ + e- -13.58 eV (ground state B.E.)
• Na Na+ + e- -5.1 eV (from outer most shell)
• Molecule* Molecule 1-6 eV (electron shell transition)
• O2 + C … CO2 4.45 eV (Carbon combustion)
• All of these in range of: 11--15 15 eVeV
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Nuclear Physics
• Binding of neutrons, protons, sub-atomic particles• Electromagnetic forces, nuclear forces• Interactions between neutrons, protons, sub-atomic particles• Conversions of matter-energy within nucleus
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Basic Relativity Theory:• Einstein (1904) explained
matter and energy were inter-related
• Speed of light “c” is not relative to any inertial frame of reference
• c=3x108 meters/sec.• All objects have rest mass “mo”• As objects speed up:
m = mo/(1-v2/c2)1/2
• ∆E = ∆mc2 …. net changes in “energy” are related to net changes in “mass”
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Physics Data Tables Frequently Describe Components of Nucleus via Rest Mass Energy
• Example: neutron mn = 1.675x10-27kg
• In terms of rest energy:
• mnc2 = (1.675x10-27kg)(3x108 m./sec)2
= 1.5075x10-8 Joule
• To convert from Joules to eV: x 6.241x1018 eV /Joule
• mnc2 = 9.39x109 eV or ~939 MeV
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Physics data tables ALSO frequently describe components of nucleus in AMU
• Atomic Mass Unit, or AMU is 1/12 mass of C12
• 1 AMU = 1.6604x10-27 kg.
• Energy equivalent of 1 AMU is:
• (AMU) x c2 = (1. 6604x10-27kg)(3x108 m./sec)2
= 931 MeV
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Basic building blocks of atoms:
938.8 MeV1.0078 AMUHydrogen
0.511 MeV5.49x10-4 AMUElectron
939.6 MeV1.00867 AMUNeutron
938.3 MeV1.00728 AMUProton
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A Nucleus:• Z Positively charged (+) protons• (A-Z) neutrons• A is atomic number – sum of: neutrons + protons• A ≥ Z• For given Z, multiple possible isotopes depending on
number of neutrons• Some isotopes stable, others decay away rapidly• Periodic Table of Elements has 104-118 stable elements
(depends on what one calls “stable”)• Isotope Table contains >3100 entries, ~265 stable• Stable isotopes tend to be composed of mixture of
protons and neutrons.
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Forces Holding Nucleus Together are Large
• Consider Coulomb force of repulsion within nucleus of He atom
• For He Z = 2 • 2 positively (+) charged
protons• Dimension of nucleus on
order of ~10-15 m.• Binding energy has to be >>
106 eV just to hold protons together
• Source of binding energy must be other than electromagnetic
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Simplified Representation of Nuclear Force:
“Potential well” Uo• Uo ~ MeV• Ro ~ 2x10-15m• Coulomb type
repulsion for: r > Ro
• This simple model can actually explain many issues of nuclear physics.
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Qualitative Observations of Nuclear Force:
• Nucleus combination of: strong, short ranged attractive forces + relatively weaker long range Coulomb-type repulsive forces.
• Pushing 2 atoms close enough to engage short range nuclear forcecan form stable nucleus provided Coulomb barrier can be overcome
• Spreading large nucleus >2x10-13cm beyond range of nuclear force allows longer range Coulomb force to split nucleus
• As with chemical reactions: nuclear reactions proceed to lower internal energy end state.
• Lower internal energy state reached by emission of radiations (α,β,γ, ….others) and reduction in mass
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Expressing Nuclear Reactions:• Nuclear reactions also written: A + B C + D with some associated ∆Q
• When ∆Q < 0 energy input required for reaction to proceed (typically kinetic energy)
• When ∆Q > 0 rest mass energy is converted to kinetic-mass energy and/or radiation-mass energy
• Nuclear reaction “short-hand” shows Z, A conserved
• 2 ways nuclear reactions commonly written:
• 90Th232 + 0n1 90Th233 + γ 90Th232(n,γ)90Th233
• 90Th233 91Pa233 + -1β0
90Th233 (*,β)91Pa233
• 5B11 +1H1
4Be8 + 2He4 5B11(p,α)4Be8
• 5B10 + 0n1 3Li7 + 2He4
5B10(n,α)3Li7
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Energy Release From Nuclear Reactions:
• Energy yield can be obtained from relationship: ∆Q = ∆mc2.
• For: A + B C + D total energy must be conserved.• ∆Q =[(mA + mB)-(mC +mD)] c2
• Example: thermal neutron capture 0n1 + 1p1 1H2 + γ
• Kinetic energies of n, p are negligible.• ∆m =[(mn + mp)-(md)]
= [(1.00728AMU + 1.00867AMU)-2.01355AMU] = 0.00240AMU
• ∆Q = 0.00240AMU x 931MeV/AMU = 2.24 MeV• Thus reaction proceeds → yielding: 2.24 MeV γ-ray
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Nuclear Fission Reactions
• Neutron induced fission involves: neutron capture and splitting nucleus into 2 fission products, 2-3 neutrons.
• Common fissionable isotopes include: 92U235 94Pu239
• Spontaneous fission also possible in heavy elements
• Isotopes capable of spontaneous fission: 995Am24198Cf252
• Hundreds of possible fission product combinations are possible• Average fission product mass is ~118 AMU
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Energy release from fission reactions:• One of >100 possible fission reactions:
• 92U235 + 0n1 54Xe140 + 38Sr94 +20n1
• ∆m =[(mU + mn)-(mXe + mSr + 2mn)] = [(235.0439AMU +1.00728AMU) –(139.92164AMU + 93.915361AMU +2x1.00728AMU)]
= 0.1996AMU
• ∆E = 0.1996AMU x 931MeV/AMU = 186 MeV
• This reaction proceeds, resulting in various radiations (γ,α,β) being emitted as well as recoil kinetic energy of fission fragments.
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Nuclear Forces Characterized by Plotting: Binding Energy / Nucleon:
• For: 1< A <58, adding nucleons increases Binding Energy, stability with increasing A
• These isotopes favor nuclear fusion reactions
• For: A > 58, binding energy decreases with increasing A
• Large A: favor nuclear fission reactions
Source: http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html#c1
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Isotopes Can Be Arranged by Z,n in Table Based On Number of Protons and Neutrons
• Source: http://ie.lbl.gov/schematics.html
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Radioactive Decay Modes in Isotope Table:• α+ decay is emission of totally ionized Helium atom nucleus, or: 2He4
Example: 92U23890Th234 + 2He4
• β- decay is emission of “electron-like” particle from nucleus. Example: 90Th234
91Pa234 + β-
• β- decay is actually decay of a neutron via reaction: n p + β- + ν ( ν: “neutrino”)
• β+ decay is “positron” emission – a positively charged “electron-like” particle from nucleus
• β+ decay is actually decay of a proton via reaction: p n + β+ + ν
• K-capture is capture of K-shell electron by nucleus
• sf is “spontaneous fission” Example: 98Cf238 sf
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Specific Decay Trends Found Based on Location
• Gray region is stable • For higher Z values, more neutrons needed to be stable• Region of Z > 57 (Lanthanum) increasing trend to α+ decay• Isotopes < Gray region tend to β- decay• Isotopes > Gray region tend to β+ decay, or K-capture (electron capture)
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Radioactive Decay is: Chain of Separate α, β, Decays
• Decay of natural Uranium92U238
• 92U23890Th234 + 2He4
• 90Th234 91Pa234 + -1β0
• 91Pa234 92U234 + -1β0
• 92U23490Th230 + 2He4
• Alternate decay chains can exist• Alternate chains have different decay
rates• More on this in Lecture 2• Source: http://hyperphysics.phy-
astr.gsu.edu/hbase/nucene/nucbin.html#c1
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Summary• Chemical reactions involve exchanges of electrons, ions
yield ∆Q values: 1 – 15 eV
• Chemical properties characterized by Periodic Table of Periodic Table of ElementsElements
• Nuclear reactions involve protons, neutrons, α, β particles yield ∆Q values: typically 100 keV – 200 MeV
• Nuclear properties characterized by Table of IsotopesTable of Isotopes
• Isotope position on Table of Isotopes shows tendency for specific decay modes (α, β, sf, n)
• Unstable isotopes decay tend to decay to more stable isotopes