Fundamentals of Nuclear EngineeringFundamentals of Nuclear Engineering Module 1: Atomic and Nuclear...

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1 Fundamentals of Nuclear Engineering Module 1: Atomic and Nuclear Physics Dr. John H. Bickel

Transcript of Fundamentals of Nuclear EngineeringFundamentals of Nuclear Engineering Module 1: Atomic and Nuclear...

Page 1: Fundamentals of Nuclear EngineeringFundamentals of Nuclear Engineering Module 1: Atomic and Nuclear Physics Dr. John H. Bickel 2 3 Objectives: 1. Explain key concepts of energy release

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Fundamentals of Nuclear Engineering

Module 1: Atomic and Nuclear Physics

Dr. John H. Bickel

Page 2: Fundamentals of Nuclear EngineeringFundamentals of Nuclear Engineering Module 1: Atomic and Nuclear Physics Dr. John H. Bickel 2 3 Objectives: 1. Explain key concepts of energy release

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Page 3: Fundamentals of Nuclear EngineeringFundamentals of Nuclear Engineering Module 1: Atomic and Nuclear Physics Dr. John H. Bickel 2 3 Objectives: 1. Explain key concepts of energy release

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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

Page 12: Fundamentals of Nuclear EngineeringFundamentals of Nuclear Engineering Module 1: Atomic and Nuclear Physics Dr. John H. Bickel 2 3 Objectives: 1. Explain key concepts of energy release

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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