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CHEM 312 Lecture 7: FissionPowerPoint Presentation

CHEM 312 Lecture 7: Fission

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CHEM 312 Lecture 7: Fission

- Readings: Modern Nuclear Chemistry, Chapter 11; Nuclear and Radiochemistry, Chapter 3
- General Overview of Fission
- Energetics
- The Probability of Fission
- Fission Product Distributions
- Total Kinetic Energy Release
- Fission Product Mass Distributions
- Fission Product Charge Distributions

- Fission in Reactors
- Delayed neutron

- Proton induced fission

Nuclear Fission

- Fission discovered by Otto Hahn and Fritz Strassman, Lisa Meitner in 1938
- Demonstrated neutron irradiation of uranium resulted in products like Ba and La
- Chemical separation of fission products

- Demonstrated neutron irradiation of uranium resulted in products like Ba and La
- For induced fission, odd N
- Addition of neutron to form even N
- Pairing energy

- In 1940 G. N. Flerov reported that 238U undergoes fission spontaneously
- half life of round 1016 y
- Several other spontaneous fission isotopes found
- Z > 90

- Partial fission half lives from nanoseconds to 2E17 years

Fission

- Can occur when enough energy is supplied by the bombarding particle for the Coulomb barrier to be surmounted
- Fast neutron
- Proton

- Spontaneous fission occurs by tunneling through barrier
- Thermal neutron induces fission from pairing of unpaired neutron, energy gain
- Nuclides with odd number of neutrons fissioned by thermal neutrons with large cross sections
- follows1/v law at low energies, sharp resonances at high energies

Energetics Calculations

- Why does 235U undergo neutron induced fission for thermal energies?
- Where does energy come from?

- Generalized energy equation
- AZ + n A+1Z + E

- For 235U
- E=(40.914+8.071)-42.441
- E=6.544 MeV

- For 238U
- E=(47.304+8.071)-50.569
- E=4.806 MeV

- For 233U
- E=(36.913+8.071)-38.141
- E=6.843 MeV

- Fission requires around 5-6 MeV

Fission Process

- Nucleus absorbs energy
- Excites and deforms
- Configuration “transition state” or “saddle point”

- Nuclear Coulomb energy decreases during deformation
- Nuclear surface energy increases

- Saddle point key condition
- rate of change of the Coulomb energy is equal to the rate of change of the nuclear surface energy
- Induces instability that drives break up of nucleus

- If the nucleus deforms beyond this point it is committed to fission
- Neck between fragments disappears
- Nucleus divides into two fragments at the “scission point.”
- two highly charged, deformed fragments in contact

- Large Coulomb repulsion accelerates fragments to 90% final kinetic energy within 10-20 s

Fission Products

- Fission yield curve varies with fissile isotope
- 2 peak areas for U and Pu thermal neutron induced fission
- Variation in light fragment peak
- Influence of neutron energy observed

235U fission yield

Fission

- Primary fission products always on neutron-excess side of stability
- high-Z elements that undergo fission have much larger neutron-proton ratios than the stable nuclides in fission product region
- primary product decays by series of successive - processes to its stable isobar

- Yields can be determined
- Independent yield: specific for a nuclide
- Cumulativeyield: yield of an isobar
- Beta decay to valley of stability

- Data for independent and cumulative yields can be found or calculated

- For reactors
- Emission of several neutrons per fission crucial for maintaining chain reaction
- “Delayed neutron” emissions important in control of nuclear reactors

Comparison of cumulative and independent yields for A=141

http://www-nds.iaea.org/sgnucdat/c2.htm

Fission Process: Delayed Neutrons

- Particles form more spherical shapes
- Converting potential energy to emission of “prompt” neutrons
- Gamma emission after neutrons
- Then decay
- Occasionally one of these decays populates a high lying excited state of a daughter that is unstable with respect to neutron emission

- “delayed” neutrons
- 0.75 % of total neutrons from fission
- 137-139I and 87-90Br as examples

Delayed Neutrons in Reactors

- Control of fission
- 0.1 msec for neutron from fission to react
- Need to have tight control
- 0.1 % increase per generation
- 1.001^100, 10 % increase in 10 msec

- 0.1 msec for neutron from fission to react
- Delayed neutrons useful in control
- Longer than 0.1 msec
- 0.65 % of neutrons delayed from 235U
- 0.26 % f

- Power proportional to number of available neutrons
- Should be kept constant under changing conditions
- Control elements and burnable poisons

- k=1 (multiplication factor)
- Ratio of fissions from one generation to the next
- k>1 at startup

- or 233U and 0.21 % for 239Pu

- Ratio of fissions from one generation to the next

- Should be kept constant under changing conditions

Energetics

- Any nucleus of A> 100 into two nuclei of approximately equal size is exoergic.
- Why fission at A>230

- Separation of a heavy nucleus into two positively charged fragments is hindered byCoulomb barrier
- Treat fission as barrier penetration
- Barrier height is difference between
- Coulomb energy between the two fragments when they are just touching
- the energy released in the fission process

- Barrier height is difference between

- Treat fission as barrier penetration
- Near uranium both these quantities have values close to 200 MeV

Energetics

- Generalized Coulomb barrier equation
- Compare with Q value for fission

- Determination of total kinetic energy
- Equation deviates at heavy actinides (Md, Fm)

- Consider fission of 238U
- Assume symmetric
- 238U119Pd + 119Pd + Q
- Z=46, A=119
- Vc=462*1.440/(1.8(1191/3)2)=175 MeV
- Q=47.3087-(2*-71.6203) = 190.54 MeV

- Z=46, A=119
- asymmetric fission
- 238U91Br + 147La + Q
- Z=35, A=91
- Z=57, A=147
- Vc=(35)(57)*1.44/(1.8*(911/3+1471/3))=164 MeV
- Q=47.3087-(-61.5083+-66.8484) = 175.66 MeV

- Realistic case needs to consider shell effects

Fission Isomers

- Some isomeric states in heavy nuclei decay by spontaneous fission with very short half lives
- Nano- to microseconds
- De-excite by fission process rather than photon emission

- Fissioning isomers are states in these second potential wells
- Also called shape isomers
- Exists because nuclear shape different from that of the ground state
- Proton distribution results in nucleus unstable to fission

- Around 30 fission isomers are known
- from U to Bk

- Can be induced by neutrons, protons, deuterons, and aparticles
- Can also result from decay

Fission Isomers: Double-humped fission barrier

- At lower mass numbers, the second barrier is rate-determining, whereas at larger A, inner barrier is rate determining
- Symmetric shapes are the most stable at two potential minima and the first saddle, but some asymmetry lowers second saddle

Proton induced fission

- Energetics impact fragment distribution
- excitation energy of the fissioning system increases
- Influence of ground state shell structure of fragments would decrease
- Fission mass distributions shows increase in symmetric fission

Topic Review

- Mechanisms of fission
- What is occurring in the nucleus during fission

- Understand the types of fission
- Particle induced
- Spontaneous

- Energetics of fission
- Q value and coulomb barrier

- The Probability of Fission
- Cumulative and specific yields

- Fission Product Distributions
- Total Kinetic Energy Release
- Fission Product Mass Distributions

Questions

- Compare energy values for the symmetric and asymmetric fission of 242Am.
- What is the difference between prompt and delayed neutrons in fission.
- What is the difference between induced and spontaneous fission.
- What influences fission product distribution?
- Compare the Coulomb barrier and Q values for the fission of Pb, Th, Pu, and Cm.
- Describe what occurs in the nucleus during fission.
- Compare the energy from the addition of a neutron to 242Am and 241Am. Which isotope is likely to fission from an additional neutron.

Pop Quiz

- Provide calculations showing why 239Pu can be fissioned by thermal neutron but not 240Pu.
- Compare the Q value and Coulomb energy (Vc) from the fission of 239Pu resulting in 138Ba and 101Sr. Is this energetically favored?
- Provide comments on blog
- Bring to class on 17 October

CHEM 312 Lecture 8: Nuclear Force, Structure and Models

- Readings:
- Nuclear and Radiochemistry: Chapter 10 (Nuclear Models)
- Modern Nuclear Chemistry: Chapter 5 (Nuclear Forces) and Chapter 6 (Nuclear Structure)

- Characterization of strong force
- Charge Independence
- Introduce isospin

- Nuclear Potentials
- Simple Shell Model (Focus of lecture)
- Nucleus as a Fermi Gas

Nuclear Force

- For structure, reactions and decay of nuclei
- electromagnetic, strong and weak interactions are utilized

- Fundamental forces exhibit exchange character
- operate through virtual exchange of particles that act as force carriers

Charge Independent Force

- Strong force not effected by charge
- np, nn, pp interactions the same
- Electromagnetic force for charge

- np, nn, pp interactions the same
- Strong force examined by
- Nucleon-nucleon scattering
- Mirror nuclei
- Isobars with number of p in one nuclei equals number of n in other
- Similar energy for net nuclear binding energy
- Normalize influence of Coulomb Energy

- Shows proton and neutron two states of same particle

- Isospin is conserved in processes involving the strong interaction
- Isospin forms basis for selection rules for nuclear reactions and nuclear decay processes
- Property of nucleon
- Analogy to angular momentum
- T=1/2 for a nucleon
- +1/2 for proton, -1/2 for neutron

Shell Model

- Model nucleus as a spherical rigid container
- square-well potential
- potential energy assumed to be zero when particle is inside walls
- Particle does not escape

- square-well potential
- Harmonic oscillator potential
- parabolic shape
- steep sides that continue upwards
- useful only for the low-lying energy levels
- equally spaced energy levels
- Potential does not “saturate”
- not suitable for large nuclei

- Change from harmonic oscillator to square well lowers potential energy near edge of nucleus
- Enhances stability of states near edge of nucleus
- States with largest angular momentum most stabilized

Shell Model

- Shell filling
- States defined by n and l
- 1s, 1p, 1d, …
- Compare with electrons

- 1s, 1p, 1d, …
- States with same 2n+l degenerate with same parity (compose level)
- 2s = 2*2+0=4
- 1d = 2*1+2 =4
- 1g=2*1+4=6
- 2d=2*2+2=6
- 3s=2*3+0=6

- States defined by n and l
- Spin-Orbit Interaction
- Addition of spin orbit term causes energy level separation according to total angular momentum (j=ℓ+s)
- For p, l=1
- s=±1/2
- j= 1+1/2=3/2 and 1-1/2=1/2
- split into fourfold degenerate 1p3/2 and twofold degenerate 1p1/2 states

- For g, l=4, j=7/2 and 9/2

- For p, l=1
- states with parallel coupling and larger total angular momentum values are favored
- closed shells 28, 50, 82, and 126
- splitting of the 1f, 1g, 1h, and 1i

- Addition of spin orbit term causes energy level separation according to total angular momentum (j=ℓ+s)
- Each principal quantum number level is a shell of orbitals
- Energy gap between shell the same

In odd A nucleus of all but one of nucleons considered to have their angular momenta paired off

forming even-even core

single odd nucleon moves essentially independently in this core

net angular momentum of entire nucleus determined by quantum state of single odd nucleon

Spin of spin of state, parity based on orbital angular momentum

Even (s, d, g, i,….)

Odd (p, f, h,….)

Configuration Interaction

For nuclides with unpaired nucleons number half way between magic numbers nuclei single-particle model is oversimplification

Contribution from other nucleons in potential well, limitation of model

Odd-Odd Nuclei

one odd proton and one odd neutron each producing effect on nuclear moments

No universal rule can be given to predict resultant ground state

Level Order

applied independently to neutrons and protons

proton levels increasingly higher than neutron levels as Z increases

Coulomb repulsion effect

order given within each shell essentially schematic and may not represent exact order of filling

Ground States of Nuclei

filled shells spherically symmetric and have no spin or orbital angular momentum and no magnetic moment

ground states of all even-even nuclei have zero spin and even parity

increased binding energy of nucleons

Filling ShellsFilling Example

- Consider isotope 7Li
- 3 protons and 4 neutrons
- 2 protons in 1s1/2, 1 proton in 1p3/2
- 2 neutrons in 1s1/2, 2 neutrons in 1p3/2

- 3 protons and 4 neutrons
- spin and angular momentum based on unpaired proton
- spin should be 3/2
- nuclear parity should be negative
- parity of a p-state (odd l value, l=1)

- Excited state for 7Li?
- Proton from 1p3/2 to 1p1/2
- Breaking paired nucleons requires significant energy, neutrons remain paired

- Bound excited state corresponds to promotion of proton
- 1p1/2 corresponds to 1/2-

- Proton from 1p3/2 to 1p1/2

Filling Levels Not always so straight forward high spin does not appear as ground Deformation impacts level filling

- consider 13C
- 7th neutron is unpaired
- p ½ state
- ½-

- 51V unpaired nucleon is 23rd proton, f 7/2
- 7/2-

- examine 137Ba
- 81st neutron is unpaired, h 11/2
- spin 11/2-
- measured as 3/2+

Configurations with both odd proton and odd neutron have coupling rules to determine spin

Integer spin value

Determine spin based on Nordheim number N

Nordheimnumber N (=j1+j2+ l1+ l2) is even, then I=j1-j2

if N is odd, I=j1j2

Parity from sum of l states

Even positive parity

Odd negative parity

prediction for configurations in which there is combination of particles and holes is I=j1+j2-1

Examples on following page

Shell Filling: Spin and parity for odd-odd nucleiShell Model Example

- Consider 38Cl
- 17 protons (unpaired p in 1d3/2)
- l=2 (d state) and j=3/2

- 21 neutrons (unpaired n in 1f7/2)
- l=3 (f state) and j=7/2
- N= 2+3/2+3+7/2 = 10
- Even; I=j1-j2
- Spin = 7/2-3/2=2
- Parity from l (3+2)=5 (odd), negative parity

- 2-

- 17 protons (unpaired p in 1d3/2)
- Consider 26Al (13 each p and n)
- Hole in 1d5/2, each j = 5/2, each l=2
- N=5/2+5/2+2+2=9
- N=odd; I=j1j2
- I = 0 or 5 (5 actual value)
- Parity 2+2=4, even, +
- 5+

Particle Model: Collective Motion in Nuclei

- Effects of interactions not included in shell-model description
- pairing force
- lack of spherically symmetric potential

- Nonspherical Potential
- intrinsic state
- most stable distribution of nucleons among available single-particle states

- since energy require for deformation is finite, nuclei oscillate about their equilibrium shapes
- Deformities 150 <A<190 and A<220
- vibrational levels

- Deformities 150 <A<190 and A<220
- nuclei with stable nonspherical shape have distinguishable orientations in space
- rotational levels
- polarization of even-even core by motion of odd nucleon

- intrinsic state
- Splitting of levels in shell model
- Shell model for spherical nuclei

- Deformation parameter e2

Prolate: polar axis greater than equatorial diameter

DR=major-minor axis

ProlateDR is positive

Oblate DR is negative

Oblate: polar axis shorter than diameter of equatorial circle

Nilsson Diagram

- 50<Z<82
- 127I
- 53rd proton is unpaired
- 7/2+ from shell model

- measured as 5/2+

- 53rd proton is unpaired
- Deformation parameter should show 5/2, even l
- Oblate nuclei

Fermi Gas Model

Potential energy well derived from the Fermi gas model. The highest filled energy levels reach up to the Fermi level of approximately 28 MeV. The nucleons are bound by approximately 8 MeV.

Review and Questions

- What is a nuclear potential
- What are the concepts behind the following:
- Shell model
- Fermi model

- How do nuclear shapes relate to quadrupole moments
- Utilize Nilsson diagrams to correlate spin and nuclear deformation

Pop Quiz

- Using the shell model determine the spin and parity of the following
- 19O
- 99Tc
- 156Tb
- 90Nb
- 242Am
- 4He

- Compare your results with the actual data. Which isotopes maybe non-spherical based on the results?
- Post comments on the blog
- E-mail answers or bring to class

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