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Lecture 12: Review of Structure & Decay

• Highlights from Lectures 1-10 of PHYS7501 • This doesn’t necessarily cover everything PHYS7501 Greatest Hits that will be asked on the midterm, but it encompasses most of the main take-aways

VOLUME I

Lecture 12: Ohio University PHYS7501, Fall 2017, Z. Meisel ([email protected]) The definition of u being based on How big is a nucleus? 12C means it is a valuable tool for high-precision mass measurements, e.g. C.Scheidenberger et al., Nuc. Phys. A 2002 Phenomenological estimates: • A nucleus’s mass is roughly: M(Z,A) = A*amu • 1amu = atomic mass unit = u = 931.494MeV/c2 ≈1.66x10-24g • The amu is defined such that M(12C) ≡ 12u 1/3 • A nucleus’s (charge) radius is roughly: R(Z,A) = (1.2fm)*A -15 Since A cancels in the ρ expression, • fm = femtometer (a.k.a. fermi) = 10 m the nuclear density is independent of the nuclear size, much as a liquid’s • The radius of a is often referred to as r0=1.2fm density is independent of the size of • For the RMS radius, multiply by 3/5 the drop. Partly inspired by this property, • Therefore, an estimate for the nuclear density is: some basic nuclear calculations are based on this liquid drop analogy. • = = = 0.14 / ( . ) (G. Gamow, Proc. Roy. Soc. A 1929,1930) 𝑀𝑀 𝑀𝑀 1𝑢𝑢 𝐴𝐴 3 4 3 4 3 . × • For𝜌𝜌 fun,𝑉𝑉 in3 𝜋𝜋terms𝑅𝑅 3of𝜋𝜋 mass1 2𝑓𝑓𝑓𝑓-density:𝐴𝐴 ≈ =𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑓𝑓𝑓𝑓 2.3 × ( . −24) −25 1 66 10 𝑔𝑔 𝐴𝐴 10 𝑔𝑔 4 14 3 …which doesn’t sound like much, but this is 2x10 3g/cm (the Great Pyramid3 of Giza is only ~1012 grams) ρ 3𝜋𝜋 1 2𝑓𝑓𝑓𝑓 𝐴𝐴 ≈ 𝑓𝑓𝑓𝑓 2 Nuclear Transmutation • Rules for converting one (or ) to another (or others) • Charge conservation: = (q from + positrons + electrons) • Baryon conservation: = (A from + protons) 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 • Lepton number conservation∑ 𝑞𝑞 : [ ∑ 𝑞𝑞 ] = [ ] 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 * Transmutation likelihoods∑ 𝐴𝐴 are impacted∑ 𝐴𝐴 by energetics and spin/parity selection rules 𝑁𝑁𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 − 𝑁𝑁𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎−𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑁𝑁𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 − 𝑁𝑁𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎−𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 The first intentionally made in the Two Types: laboratory was 14N(α,p) in 1919. (E. Rutherford, Nature 1935). • Reactions The first measured was α decay from uranium. (H. Becquerel, Comptes Rendus 1896). • Multiple reactants create one or more products • Notation: A+b  c+D is written as A(b,c)D, where M(b)

• BE(Z,A) = Volume - Surface - Coulomb - Asymmetry ± Pairing • One mathematical parameterization* (of many!): *from B. Martin, Nuclear and Particle Physics (2009) • , = , , + ( ) •Volume: Nucleons cohesively bind, so: = 𝐵𝐵𝐵𝐵 𝑍𝑍 𝐴𝐴 𝑎𝑎𝑣𝑣𝑓𝑓𝑣𝑣 𝐴𝐴 − 𝑎𝑎𝑠𝑠𝑓𝑓𝑠𝑠 𝐴𝐴 − 𝑎𝑎𝑐𝑐𝑓𝑓𝑐𝑐 𝑍𝑍 𝐴𝐴 − 𝑎𝑎𝑎𝑎𝑓𝑓𝑎𝑎 𝑍𝑍 𝐴𝐴 𝑖𝑖𝑎𝑎𝑝𝑝𝑓𝑓𝑝𝑝 𝐴𝐴 •Surface: Since radius goes as and𝒗𝒗 surface area goes as , = 1 𝒇𝒇 𝑨𝑨 𝑨𝑨 ( )𝟐𝟐 ⁄3 2, = �𝟑𝟑 •Coulomb: Energy for a charged sphere goes as and , so 𝒔𝒔 𝑅𝑅 ∝ 𝐴𝐴 𝑞𝑞2 1 𝑆𝑆𝑆𝑆 ∝ 𝑅𝑅 𝒇𝒇 𝑨𝑨 𝑨𝑨 ⁄3 𝒁𝒁 𝒁𝒁−𝟏𝟏 𝑅𝑅 𝟏𝟏 𝒄𝒄 �𝟑𝟑 •Asymmetry: Z=N favored (want Z=A/2) but lesser problem𝑅𝑅 ∝ 𝐴𝐴for large𝒇𝒇 A, 𝒁𝒁so𝑨𝑨 , 𝑨𝑨 = 𝑨𝑨 𝟐𝟐 𝒁𝒁−𝟐𝟐 -1 •Pairing: Favor spin-0 nucleon pairs & disfavor unpaired nucleons, empirically𝒂𝒂 == 𝒇𝒇 𝒁𝒁 𝑨𝑨 𝑨𝑨 ) •Even-Z, Even-N: = + −𝟏𝟏 𝒑𝒑𝒑𝒑 •Odd-Z, Odd-N: = 𝒇𝒇𝒇𝒇 𝑨𝑨 𝑨𝑨𝑨𝑨 •Even-Odd: = 𝒊𝒊 𝟏𝟏 • are fit to data 𝒊𝒊 −𝟏𝟏 𝒊𝒊 𝟎𝟎 A mnemonic for remembering 𝑖𝑖 𝑎𝑎 SEMF contributions is “VSCAP”. 4 Nuclear Mass Differences

• The energy released in a nuclear reaction is the “Q-value” • = , ( , ), • For example, = + 𝑄𝑄 ∑𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑀𝑀𝑀𝑀 (𝑍𝑍, 𝐴𝐴) − ∑𝑝𝑝𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑀𝑀𝑀𝑀 𝑍𝑍 𝐴𝐴 • 68 69 = 5468. + (7. 69) ( 46. ) 𝑄𝑄 𝑆𝑆𝑆𝑆 𝑝𝑝 𝛾𝛾 𝐵𝐵𝐵𝐵 𝑀𝑀𝑀𝑀 𝑆𝑆𝑆𝑆 𝑀𝑀𝑀𝑀 𝑝𝑝 − 𝑀𝑀𝑀𝑀 𝐵𝐵𝐵𝐵 • = 0. • Considering the case above, we calculated− 189𝑀𝑀 the energy𝑀𝑀𝑀𝑀 released288𝑀𝑀 by𝑀𝑀𝑀𝑀 adding− − one260𝑀𝑀 𝑀𝑀𝑀𝑀 to 68Se, which corresponds to the energy it takes− 641𝑀𝑀 to remove𝑀𝑀𝑀𝑀 one proton from 69Br, a.k.a. the “proton ”, Sp • Similarly, can calculate the energy to remove 1- Sn, two-protons S2p, or two-neutrons S2n • , = 1, + ( , ) • , = , 1 + ( , ) 𝑆𝑆𝑝𝑝 𝑍𝑍 𝑁𝑁 𝑀𝑀𝑀𝑀 𝑍𝑍 − 𝑁𝑁 𝑀𝑀𝑀𝑀 𝑝𝑝 − 𝑀𝑀𝑀𝑀 𝑍𝑍 𝑁𝑁 • , = 2, + 2 ( , ) 𝑆𝑆𝑛𝑛 𝑍𝑍 𝑁𝑁 𝑀𝑀𝑀𝑀 𝑍𝑍 𝑁𝑁 − 𝑀𝑀𝑀𝑀 𝑛𝑛 − 𝑀𝑀𝑀𝑀 𝑍𝑍 𝑁𝑁 • , = , 2 + 2 ( , ) 𝑆𝑆2𝑝𝑝 𝑍𝑍 𝑁𝑁 𝑀𝑀𝑀𝑀 𝑍𝑍 − 𝑁𝑁 ∗ 𝑀𝑀𝑀𝑀 𝑝𝑝 − 𝑀𝑀𝑀𝑀 𝑍𝑍 𝑁𝑁 2𝑛𝑛 𝑆𝑆 𝑍𝑍 𝑁𝑁 𝑀𝑀𝑀𝑀 𝑍𝑍 𝑁𝑁 − ∗ 𝑀𝑀𝑀𝑀 𝑛𝑛 − 𝑀𝑀𝑀𝑀 𝑍𝑍 𝑁𝑁 5 Nuclear charge distribution Review: R. Hofstadter, Rev. Mod. Phys. (1956)

• Comparing measured scatt( ) to calculations w/ various F (q ) reveals a Fermi-like distribution (R. Woods & D. Saxon, Phys. Rev. (1954))

125MeV e- on Be σ θ 125MeV e- on Au R. Hofstadter, H. Fechter, & J. McIntyre, Phys. Rev. (1953) Measured ρch examples

B.A. Brown, Lecture Notes in Physics, 2005.

6 In general, a moment is a distance multiplied by a physical quantity. For distributions you integrate the quantity’s Electric & Magnetic Moments distribution with respect to distance.

• The nuclear magnetic moments describe the distribution of electric currents in the nucleus • Causes nuclei to align along an external magnetic field, which can be exploited using NMR, MRI, etc.

• The nuclear electric moments describe the distribution of electric charges in the nucleus • Used as a measure of the nuclear shape

7 First stab at the potential, : The Harmonic Oscillator • Based on some evidence (and logic) that nuclei aren’t perfectly constant in density, Heisenberg (Z. Phys. 1935) posited that a parabolic potential could be assumed, conveniently allowing the adoption of the𝑉𝑉 harmonic oscillator solutions (one of the few analytically solved systems!) i.e. only odd or even • This provides evenly spaced energy levels , with = + . functions are allowed • The corresponding angular momenta are = 1, 3, … 1 0. for each oscillator shell 2 • The number of particles per angular momentum𝑛𝑛 is𝐸𝐸 2𝑛𝑛( +𝑛𝑛 1) forℎ υ + 1 projections & 2 spins • So, the number of particles per level is: 𝑙𝑙 𝑛𝑛 − 𝑛𝑛 − ≥ 2𝑙𝑙 2𝑙𝑙

Could the HO potential still be useful for some cases?

…can get the job done for nuclei (e.g. H. Guo et al. PRC 2017) 8 Loveland, Morrissey, & Seaborg, Modern Nuclear Chemistry (2006) …but need to be careful, because can impact results (B.Kay et al arXiv 2017) Move to an empirical potential: Woods-Saxon Cumulative • Since the nuclear interaction is short-range, a natural improvement would be to adopt a central potential mimicking the empirical density distribution • This is basically a square well with soft edges, 92 as described by the Woods-Saxon potential: 58 R. Woods & D. Saxon, Phys. Rev. (1954) • Using the Woods-Saxon 40 is a good idea because of commitment to 20 22 MeV protons on Pt reality… but we’re no wiser as to the origin of the magic numbers 8

2 Was this step completely useless? No! It broke the degeneracy in ℓ Loveland, Morrissey, & Seaborg, Modern Nuclear Chemistry (2006) 9 The missing link: the spin-orbit interaction LS

• Due to desperation or genius (or both) Maria Göppert-Mayer [Phys. Rev. February 1949] (and nearly simultaneously Haxel, Jensen, & Suess [Phys. Rev. April 1949]) posited that nucleon spin and orbital angular momentum interacted strongly, making j the good quantum number for a nucleon: = + • Prior to this approach, ⃗ angular momentum was coupled as is typically done𝚥𝚥⃗ for𝑙𝑙 atoms,𝑠𝑠⃗

where = + , = , and = A. Kastberg, Lecture Notes Physique Atomique • This is “LS coupling” jj 𝐽𝐽⃗ 𝐿𝐿 𝑆𝑆⃗ 𝐿𝐿 ∑𝑛𝑛𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑙𝑙⃗ 𝑆𝑆⃗ ∑𝑛𝑛𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑠𝑠⃗ • Positing that the spin-orbit interaction is stronger than spin-spin or orbit-orbit means that instead, = and = + • This is “jj coupling” 𝐽𝐽⃗ ∑𝑛𝑛𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝚥𝚥⃗ 𝚥𝚥⃗ 𝑙𝑙⃗ 𝑠𝑠⃗

10 Filling the shells Loveland, Morrissey, & Seaborg, Modern Nuclear Chemistry (2006) • We can construct a nucleus using our “shell model”: • A nucleon will go in the lowest-energy level which isn’t already filled, i.e. • the largest angular momentum, • for the lowest orbital angular momentum, • for the lowest oscillator shell, n 𝑗𝑗 • + 1 protons or neutrons are allowed𝑙𝑙 per level • Each level is referred to by its 2𝑗𝑗• by the # for the oscillator shell (convention either starts with 0 or𝑛𝑛𝑛𝑛 1)𝑛𝑛 • 𝑛𝑛by spectroscopic notation (s=0,p=1,d=2,f=3,…) • by the half-integer corresponding to the spin 𝑙𝑙 • For 𝑗𝑗example: 7Li (Z=3, N=4)

11 Isomers on the Nuclear Chart L. van Dommelen, Quantum Mechanics for Engineers (2012) Even-Z, Odd N Odd-Z, Even N Odd-Z, Odd N

Special cases exist (mostly for higher-A nuclides) where even-even nuclei have isomers (e.g. M. Müller-Veggian et al., Z.Phys.A (1979)) 12 Collective Model

• There are compelling reasons to think that our nucleus isn’t a rigid sphere • The liquid drop model gives a pretty successful description of some nuclear properties. …can’t liquids slosh around? • Many nuclei have non-zero electric quadrupole moments (charge distributions) …this means there’s a non-spherical shape. …can’t non-spherical things rotate? • Then, we expect nuclei to be able to be excited rotationally & vibrationally • We should (and do) see the signature in the nuclear excited states

• The relative energetics of rotation vs vibration can be inferred from geometry • The rotational frequency should go as (because and ) 1 𝑟𝑟 2 𝐿𝐿 2 𝜔𝜔 ∝ 𝑅𝑅 • The vibrational𝐼𝐼 ≡ 𝜔𝜔 frequency𝐼𝐼 ∝ 𝑀𝑀 should𝑅𝑅 go as 1 (because it’s like an oscillator) 2 𝜔𝜔𝑣𝑣 ∝ ∆𝑅𝑅 • So 13

𝜔𝜔𝑟𝑟 ≪ 𝜔𝜔𝑣𝑣 Predicted regions of deformation P. Möller et al. ADNDT (2016) B. Harvey, Introduction to and Chemistry (1962)

b 4 = , 3 5 𝜋𝜋 𝑎𝑎 − 𝑏𝑏 2 2 0 14 Loveland, Morrissey, Seaborg, Modern Nuclear Chemistry (2006) 𝛽𝛽 ≡ 𝛼𝛼 𝑅𝑅𝑠𝑠𝑠𝑠푠 Rotational bands: sequences of excited states

( ) • = , so for a given , ( + 1) 2 ћ 𝑗𝑗 𝑗𝑗+1 • Note𝑟𝑟𝑟𝑟𝑟𝑟 that parity needs to be maintained because rotation is 𝐸𝐸 2𝐼𝐼 𝐼𝐼 ∆𝐸𝐸 ∝ 𝑗𝑗 𝑗𝑗 symmetric upon reflection and so 0 ground-states can only have j=0,2,4,… (because = (+ 1) ) • Without observing the decay scheme, picking𝐽𝐽 -out associated 𝜋𝜋 − rotational states could be pretty difficult B. Harvey, Introduction to Nuclear Physics and Chemistry (1962) • Experimentally, coincidence measurements allow schemes to be mapped

158Er

L. van Dommelen, Quantum Mechanics for Engineers (2012) T.Dinoko et al. EPJ Web Conf. (2013) 15 Vibrational energy levels • Just as the quantum harmonic oscillator eigenvalues are quantized, so too will the energy levels for different quanta (phonons) of a vibrational mode. • Similarly, the energy levels have an even spacing, = ( + ) 1 • Even𝑛𝑛 -even nuclides2 have 0+ ground states, and thus, for𝐸𝐸 a =𝑛𝑛 2 vibration,ћ𝜔𝜔 = 2 excitations will maintain the symmetry of the wave-function (i.e. λ= 1 excitations𝑛𝑛 would violate parity) Loveland, Morrissey, Seaborg, Modern Nuclear Chemistry (2006) • Therefore, the 1st vibrational state will be 2+ 𝑛𝑛 • We can excite an independent quadrupole vibration by adding a second phonon • The second phonon will build excitations on the first, coupling to either 0+,2+, or 4+ • Employing a nuclear potential instead winds up breaking the degeneracy for states associated with a given number of phonons

16 Nilsson model: single-particle level splitting

• Consider the options for our nucleon’s orbit around the nucleus 𝑗𝑗 • Orbits with the same principle quantum number will have the same radius

• Notice that the orbit with the smaller projection of ( ) R.Casten, Nuclear Structure from a Simple Perspective (1990) sticks closer to the bulk of the nucleus during its orbit 1 • Since the is attractive, 𝑗𝑗 𝐾𝐾 the orbit will be more bound (i.e. lower energy) than the orbit

• The opposite1 would be true if the nucleus in our picture was oblate,2 squishing𝐾𝐾 out toward the orbit 𝐾𝐾 K K • Therefore, for prolate nuclei,2 lower single-particle levels will be more bound (lower𝐾𝐾-energy), whereas larger states will be more𝐾𝐾 bound for oblate nuclei

𝐾𝐾

Loveland, Morrissey, Seaborg, Modern Nuclear Chemistry (2006) 17 B. Harvey, Introduction to Nuclear Physics and Chemistry (1962) Nilsson Model: Example • Consider 25Al, for which we expect 0.2, like 27Al • There are 13 protons and 12 neutrons, so the unpaired proton will be responsible𝛽𝛽2 ≈ for • Filling the single-particle levels, 𝜋𝜋 • We place two protons in the 1s1/2 level, which isn’t 𝐽𝐽shown • Then two more in 1/2-, two more in 3/2-, two more in 1/2-, two more in the 1/2+, two more in the 3/2+ • And the last one winds up in the 5/2+ level • So, we predict . = • For the first excited𝜋𝜋 sate,5 + it seems likely the𝐽𝐽𝑔𝑔 𝑠𝑠 proton⁄2 will hop up to the nearby 1/2+ level • Agrees with data • Since 25Al is deformed, we should see rotational bands with states that have (integer)+ and ( + 1) spacing 𝑗𝑗 ∝ 𝑗𝑗 𝑗𝑗 18 Connection to thermodynamics

= + + • Now we can solve for our entropy: ( ) 𝑑𝑑𝑑𝑑 𝑎𝑎 ∗ 𝐸𝐸∗ • Since a zero-temperature system has𝑆𝑆 zero𝐸𝐸 entropy,∫ 𝑇𝑇 𝐸𝐸 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐2≈ ∫ 𝑘𝑘𝐵𝐵 𝑑𝑑𝑑𝑑 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 • Recall from the microscopic picture, = ln( ) ∗ 𝑆𝑆 𝐸𝐸 ≈ 𝑘𝑘𝐵𝐵 𝑎𝑎𝐸𝐸 • So, the number of accessible configurations𝐵𝐵 (a.k.a. nuclear states) for our system is exp 2 𝑆𝑆 𝑘𝑘 𝑔𝑔 • The density of states∗ is going to be proportional to the total number of states 𝑔𝑔 ≈ 𝑎𝑎𝐸𝐸 • So, the state density = exp 2 , where is a constant • A more careful treatment∗ using partition functions∗ and other statistical mechanics tools 𝜌𝜌 𝐸𝐸 𝐶𝐶 𝑎𝑎𝐸𝐸 𝐶𝐶 H. Bethe, Phys. Rev. (1936) yields: = / / exp 2 𝜋𝜋 ∗ ∗ 1 4 ∗5 4 238 = 1/ = 3 × 10 • Going back𝜌𝜌 𝐸𝐸 to our12 𝑎𝑎estimate𝐸𝐸 for U𝑎𝑎 and𝐸𝐸 using and , we get • In practice, and are usually fit to data ∗ 4 −1 𝑎𝑎 𝑑𝑑 𝐸𝐸 𝑆𝑆𝑛𝑛 𝜌𝜌 ≈ 𝑀𝑀𝑀𝑀𝑀𝑀 • in particular isn’t so relevant, since we can normalize to the region at low excitation 𝐶𝐶 𝑎𝑎 = energy where individual levels can be counted and ideally also∗ to 19 𝐶𝐶 𝜌𝜌 𝐸𝐸 ∗ 𝜌𝜌 𝐸𝐸 𝑆𝑆𝑛𝑛 Experimental results confirm the exponential behavior of ( )

One challenge in comparing to counts of Techniques which are sensitive to the∗ discrete states is knowing if your integrated number of levels can𝜌𝜌 overcome𝐸𝐸 measurement missed any levels this …though with assistance from models

T. Ericson, Nuc. Phys. (1959)

M. Guttormsen et al. Euro.Phys.J.A (2015)

20 and the spin distribution IAEA, RIPL-2 Handbook • Having2 found = , now we need to estimate 2 𝐼𝐼𝑘𝑘𝐵𝐵𝑇𝑇 the moment of inertia2 𝜎𝜎 𝜎𝜎 ћ • Since we assumed a spherical nucleus earlier to justify the degeneracy of in ,𝐼𝐼 we’ll double-down and use for a rigid sphere: =∗ 𝐸𝐸 2 𝑀𝑀2 𝐼𝐼 • Using this and the5 previously derived formula for the 𝐼𝐼 𝑀𝑀𝑅𝑅 nuclear temperature / , and the standard estimate 1 ∗ 𝑘𝑘𝐵𝐵 Implied spin distributions for the nuclear radius𝑇𝑇 ≈= 𝐸𝐸 : 𝑎𝑎 = 2 1 2 �3 ∗ ⁄3 2 2𝑀𝑀𝑟𝑟0 𝐴𝐴 𝐸𝐸 2 • Using = 931𝑅𝑅 .5𝑟𝑟0𝐴𝐴 𝜎𝜎A, 5ћ 𝑎𝑎 𝑀𝑀𝑀𝑀𝑀𝑀 2 𝑢𝑢 , 1. 𝑐𝑐: 0.014 𝑀𝑀 𝐴𝐴𝑚𝑚 ≈ ⁄ 5 ∗ 2 ⁄3 𝐸𝐸 • ћIt𝑐𝑐 turns≈ 197𝑀𝑀 out, 𝑀𝑀fits𝑀𝑀 𝑓𝑓𝑓𝑓 to neutron𝑟𝑟0 ≈ 2𝑓𝑓𝑓𝑓resonances𝜎𝜎 ≈ yield 𝐴𝐴 𝑎𝑎

𝐴𝐴 −1 • Therefore,8 0.014 21 𝑎𝑎 ≈ ⁄ 𝑀𝑀𝑀𝑀𝑀𝑀 7 2 ⁄6 ∗ 𝜎𝜎 ≈ 𝐴𝐴 𝐸𝐸 Decay Equations for One Nuclear Species • Solving for ( ) from the seemingly innocuous equation = ( ), is actually pretty hard, unless one employs the Laplace transform, which turns our𝑑𝑑𝑑𝑑 differential equation into an algebraic one𝑁𝑁 𝑡𝑡 𝑑𝑑𝑑𝑑 −λ𝑁𝑁 𝑡𝑡 • For the LHS, we assume ( ) to be an exponential function (empirically a safe bet) and use the derivative property of Laplace transform: 𝑁𝑁 𝑡𝑡 • (0) 𝑑𝑑𝑑𝑑 For the RHS, simply swap-in for time: ( ) • 𝑑𝑑𝑑𝑑 → 𝑠𝑠𝑠𝑠 𝑠𝑠 − 𝑁𝑁 • Therefore, ( ) (0) = ( ) 𝑠𝑠 −λ𝑁𝑁 𝑡𝑡 → −λ𝑁𝑁 𝑠𝑠 ( ) = • Which is re𝑠𝑠𝑠𝑠-written𝑠𝑠 − as:𝑁𝑁 −𝜆𝜆(𝑁𝑁 𝑠𝑠) 𝑁𝑁 0 • Using one of several different𝑁𝑁 𝑠𝑠 methods𝑠𝑠+λ (See E.g. D.Pressyanov, Am.J.Phys. 2002, or a Math Methods book), the inverse Laplace transform can be employed, yielding the familiar relation: • = (0) , the number of nuclei existing at time • Since the Activity (decays/second)−𝜆𝜆𝑡𝑡 = , 𝑁𝑁 𝑡𝑡 𝑁𝑁 𝑒𝑒 𝑡𝑡 • = (0) 22 −λ𝑡𝑡 𝐴𝐴 𝜆𝜆𝑁𝑁 𝐴𝐴 𝑡𝑡 𝐴𝐴 𝑒𝑒 Common Descriptors for Decay

•Aside from the decay constant , other more intuitive quantities are often used •It is common to state the time at which half of the nuclei in a radioactive sample will have λ undergone decay, a.k.a. the half-life: ½ ( ) •By definition: ½ = … so, = ½ ( ) 𝑡𝑡 𝑁𝑁 𝑡𝑡 1 1 −λ𝑡𝑡 ( ) ½ •Re-write as 2 =𝑁𝑁 0 , 2which makes2 𝑒𝑒 it apparent that ½ = ln 2 An alternative piece ofλ𝑡𝑡 trivia is the mean lifetime for the nuclei in the sample, • 𝑒𝑒 𝑡𝑡 𝜆𝜆 ( )/ •By definition: = = = = ∞ ∞ ( )/ ( )/ 2 𝜏𝜏 ∫0 𝑡𝑡𝑡𝑡 𝑡𝑡 𝑑𝑑𝑑𝑑 ∫0 𝑡𝑡𝑡𝑡 𝑡𝑡 𝑑𝑑𝑑𝑑 𝑁𝑁 0 λ 1 = ln(2∞) 0. •Therefore ½ 𝜏𝜏 ∫0 𝑁𝑁 𝑡𝑡 𝑑𝑑𝑑𝑑 𝑁𝑁 0 λ 𝑁𝑁 0 λ λ •We can re-state the lifetime in terms of the equivalent energy width using the Heisenberg uncertainty principle:𝑡𝑡 𝜏𝜏 ≈ 693𝜏𝜏 • . × • Taking the mean lifetime as the time uncertainty, = ∆𝐸𝐸 � ∆𝑡𝑡 ≥ ћ ( −22 ) ћ 6 582 10 𝑀𝑀𝑀𝑀𝑀𝑀 23 ∆𝐸𝐸 𝜏𝜏 ≈ 𝜏𝜏 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Secular Equilibrium & Radioactivity from Nuclear Reactions • If a radioactive nucleus is made in a nuclear reaction (e.g. in a star, from cosmic rays, or using an accelerator), the constant replenishment of the daughter is similar to the case of a very long-lived parent • In particular this is true of the reaction rate is much slower than the daughter decay …which is pretty much always the case • Starting with 0 = 0 and recasting “decay” of 1 2 as a reaction, = 0 = (0), = 0 + 0 = 2 becomes 12 1 1 1 λ2 𝑁𝑁 −λ1𝑡𝑡 −λ2𝑡𝑡 −λ2𝑡𝑡→ 𝑅𝑅 λ2 λ 𝑁𝑁 −λ1𝑡𝑡 𝐴𝐴 −λ2𝑡𝑡 • 𝐴𝐴Since2 𝑡𝑡 λ2−0λ, 1 𝐴𝐴 1 𝑒𝑒 −1𝑒𝑒 𝐴𝐴, 2 𝑒𝑒 𝐴𝐴2 𝑡𝑡 λ2−λ1 𝑅𝑅12 𝑒𝑒 − 𝑒𝑒 which is known as the “activation equation”−λ2𝑡𝑡 1 2 12 • Linearλ growth≈ of𝐴𝐴 the𝑡𝑡 daughter≈ 𝑅𝑅 at− short𝑒𝑒 times, A. Bielajew, Intro. to Special but have diminishing returns as time increases Relativity, Quantum Mechanics, and Nuclear Physics For Nuclear Engineers • If you’re making a custom on-site, (2014) this lets you know how long to bother performing the production reaction • This relation allows nuclear reaction cross sections to be measured using the daughter decay, and it lets you know how long to perform the reaction 24 α energy from Qα • When an α is emitted, it will share some energy with the heavy recoil, so isn’t quite equal to • We just need to employ conservation of momentum and energy 𝐾𝐾𝐾𝐾𝛼𝛼 𝑄𝑄𝛼𝛼 • = + • Conveniently = 0, so the daughter and α will move in opposite directions 𝑝𝑝and⃗𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑�= 𝑝𝑝⃗𝛼𝛼 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 • 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑�+ = 𝛼𝛼 + = + = + 2 𝑝𝑝 −2𝑝𝑝 2 2 2 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑푑𝑑𝑑𝑑𝑑𝑑𝑑 𝛼𝛼 𝛼𝛼 𝛼𝛼 𝛼𝛼 𝑝𝑝 𝑝𝑝 𝑝𝑝 𝐴𝐴 Spectrum from M.Mroz, K. Brandenburg, A. Mamum, & A. Pun • 2𝐴𝐴𝑝𝑝𝑝𝑝=𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝛼𝛼 2𝐴𝐴𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑푑=𝑑𝑑𝑑𝑑𝑑𝑑 2𝐴𝐴𝛼𝛼 2𝐴𝐴𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑푑𝑑𝑑𝑑𝑑𝑑𝑑 2𝐴𝐴𝛼𝛼 𝐴𝐴𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑푑𝑑𝑑𝑑𝑑𝑑𝑑 𝛼𝛼 𝛼𝛼 𝑄𝑄 226Ra decay𝐾𝐾𝐾𝐾 𝐾𝐾𝐾𝐾 𝐴𝐴𝛼𝛼+𝐴𝐴𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑푑𝑑𝑑𝑑𝑑𝑑𝑑 𝐴𝐴𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 sequence • 𝑄𝑄𝛼𝛼 = 𝐴𝐴𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑푑𝑑𝑑𝑑𝑑𝑑𝑑 𝐾𝐾𝐾𝐾𝛼𝛼 𝐴𝐴𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑푑𝑑𝑑𝑑𝑑𝑑𝑑 𝐾𝐾𝐾𝐾𝛼𝛼 𝐴𝐴𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑푑𝑑𝑑𝑑𝑑𝑑𝑑 • So it’s𝛼𝛼 a pretty𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 small𝛼𝛼 effect (though𝐾𝐾𝐾𝐾 not𝐴𝐴 so for β-delayed𝑄𝑄 particle emission in lighter nuclei)

• Conveniently, α sources typically have several Eα from the , and so they provide several energy calibration points

25 Geiger-Nuttall relation Geiger & Nuttall, Philisoph. Mag. (1912)

• In an early effort to characterize α-decay, Geiger & Nuttall (H. Geiger & J.M. Nuttall, Philisoph. Mag. (1911, 1912)) compared the range of α particles in a material vs t½ of the α-source and found a linear relationship in log-log space • In modern terms, using instead of range, ½ we get the Geiger-Nuttall relation: log ½ = + 𝛼𝛼 𝑄𝑄 − • Obviously the energy somehow impacts10 ½ 𝛼𝛼 𝑡𝑡 𝑎𝑎 𝑏𝑏𝑏𝑏𝑄𝑄 Loveland, Morrissey, & Seaborg, Modern Nuclear Chemistry (2006) …incredibly strongly 𝛼𝛼 𝑡𝑡 • For ~ × 2 increase in , nearly 20 orders of magnitude decrease in ½!! 𝑄𝑄𝛼𝛼 𝑡𝑡 What does this imply about useful α sources? There’s a relatively limited range of Eα available. • Large Eα sources aren’t active for long enough, • while low Eα sources require huge amounts to

have an appreciable activity (A=λN). 26 Tunneling through a thick barrier

• = , where = 2 1 is pretty ugly 2 2 −2𝐺𝐺 𝑒𝑒 2𝜇𝜇𝑐𝑐 −1 𝑅𝑅 𝑅𝑅 𝑅𝑅 • 𝑇𝑇Conveniently,𝑒𝑒 for most2𝐺𝐺 casesћ𝑐𝑐 𝑍𝑍 𝛼𝛼𝑍𝑍𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑�, 𝑄𝑄𝛼𝛼 𝑐𝑐𝑐𝑐𝑐𝑐 𝑏𝑏 − 𝑏𝑏 − 𝑏𝑏 B.A. Brown, Lecture Notes in Nuclear Structure Physics (2005) so the Gamow factor 𝑏𝑏2≫ 𝑅𝑅 2 𝜋𝜋 𝑒𝑒 2𝜇𝜇𝑐𝑐 238U • Since the decay half-life2𝐺𝐺 will≈ 2beћ𝑐𝑐 inversely𝑍𝑍𝛼𝛼𝑍𝑍𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑� proportional𝑄𝑄𝛼𝛼 to / the tunneling probability, ½ • You may notice this is what Geiger2𝐺𝐺 & Nuttall𝑍𝑍 𝑄𝑄 told𝛼𝛼 us all along ½ 𝑡𝑡 ∝ 𝑒𝑒 ∝ 𝑒𝑒 log ½ = + − 10 𝛼𝛼 Aside: 𝑡𝑡 𝑎𝑎 𝑏𝑏𝑏𝑏𝑄𝑄 Gamow realized this formalism would work just as well for a charged-particle tunneling in (i.e. for ). 𝑅𝑅 𝑏𝑏 For nuclear fusion, 2G is often written instead as 2πη, where η is the Sommerfeld parameter. In , T=P=exp(-2πη) is multiplied by the Maxwell-Boltzmann distribution to get the Gamow window. 27 Why is it α particles that are being emitted?

• So far we’ve been smugly pleased with ourselves about our ability to describe α decay …but why α decay? Why not proton decay, or 3He decay, or 12C decay? • The short answer is Q-values, Coulomb barriers, and clustering probabilities • Q-value: The cluster decay must be energetically favorable • Coulomb barrier: Higher-Z particles will have a larger barrier to tunnel through • Clustering probability: It’s less likely for more nucleons to congregate within a nucleus

28 β decay spectrum, spin conservation, and the neutrino R. Evans, The (1955) • Early experiments investigating the “β ray” showed that it was not emitted with a singular energy, like the “α ray”, but rather in a continuum of energies • Though the maximum energy is equal to the decay Q-value • Furthermore, the reaction + doesn’t conserve spin! • = = = … so 0 + − 1 1 1 𝑛𝑛 → 𝑝𝑝 𝑒𝑒 rd • To remedy this issue,2 Pauli proposed the involvement2 of a 3 𝐽𝐽𝑛𝑛 𝐽𝐽𝑝𝑝 𝐽𝐽𝑒𝑒 ≤ 𝐽𝐽𝑝𝑝 𝐽𝐽𝑒𝑒 ≤ ≠ hypothetical particle, the neutrino Like a proper old-timey physicist, he made this proposal not in a paper, but in a letter to physicist • Given the above considerations, it was postulated that Lisa Meitner is a spin-½ particle (“fermion”) thatν it is massless* and electrically neutral (of course this isn’t quite true, but true enough for our purposes) ν • In one of his last works before switching to primarily performing experimental work, Fermi postulated (E. Fermi, Z. Phys. 1934) that nucleons could act as sources & sinks of electrons and neutrinos, in analogy to charged particles acting as sources and sinks of photons in quantum electrodynamics (the only successful theory of interactions between quantum particles at that point) For what it’s worth, Nature rejected Fermi’s paper for being “too remote from physical reality” 29 β decay phase space factor & the β energy spectrum • Considering the β decay rate for an electron momentum within + ,

( ) = ( ) 𝑒𝑒 𝑒𝑒 2𝜋𝜋 2 𝑝𝑝 𝑑𝑑𝑝𝑝 ′ • The matrix elementћ is just some number, so the functional form is from ( ) λ 𝑝𝑝𝑒𝑒 𝑑𝑑𝑑𝑑𝑒𝑒 Ψ𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐻𝐻 Ψ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝜌𝜌 𝐸𝐸 • Therefore, we expect = 2 𝜌𝜌 𝐸𝐸 = 2 2 2 2 2 𝑝𝑝𝑒𝑒 2 λ 𝑝𝑝𝑒𝑒 𝑑𝑑𝑑𝑑𝑒𝑒 ∝ 𝑄𝑄 − 𝐾𝐾𝐾𝐾𝑝𝑝 𝑝𝑝𝑒𝑒 𝑄𝑄 − 𝐾𝐾𝐾𝐾𝑒𝑒 𝑚𝑚𝑒𝑒𝐾𝐾𝐾𝐾𝑒𝑒 𝑄𝑄 − 2𝑚𝑚𝑒𝑒 𝑝𝑝𝑒𝑒

k Wu & Albert, Phys. Rev (1949)

Not too bad, but what effect are we forgetting that will cause positrons and electrons to behave differently? Coulomb repulsion! 30 Decay selection rules and “forbidden” decays = 2 ( ) 2𝜋𝜋 ∗ ′ 𝑓𝑓 ( ) 𝜆𝜆 ћ � 𝛹𝛹𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝐻𝐻 𝛹𝛹𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑑𝑑𝜏𝜏 𝜌𝜌 𝐸𝐸 • As was just alluded to, ignoring higher-order terms of the 𝑟𝑟 Taylor expansion omits the possibility for angular momentum transfer −𝑖𝑖 𝑝𝑝⃗ν+𝑝𝑝⃗𝑒𝑒 �ћ • If angular momentum transfer is to occur, higher-order terms𝑒𝑒 need to be included and it will no longer be the case that is independent of 2 • In fact, for these cases >𝑀𝑀1𝑓𝑓𝑓𝑓and/or = , the𝑝𝑝 leading𝑒𝑒 -order overlap = 0 and so a higher-order term will be necessary 2 ∆𝐽𝐽 ∆𝜋𝜋 𝑦𝑦𝑦𝑦𝑦𝑦 𝑀𝑀𝑓𝑓𝑓𝑓 • The order that’s required will correspond to the angular momentum transfer of the decay • This combined with whether or not parity is changed is referred to as how “forbidden” a transition is…even though it’s just a hindrance ∆𝐽𝐽 • 0 0 " " • 0+ 1+ = 0 1 = − → 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 • + = 0+ 1, = " " • =− 2, 𝑜𝑜𝑜𝑜=∆𝐽𝐽 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎 ∆𝜋𝜋 𝑛𝑛𝑛𝑛 → "𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎� ∆𝐽𝐽 𝑜𝑜𝑜𝑜 ∆𝜋𝜋 𝑦𝑦𝑦𝑦𝑦𝑦 → 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 • For ∆a𝐽𝐽 given∆ transition𝜋𝜋 𝑛𝑛𝑛𝑛 → " 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠type, 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓�will typically be within an order of magnitude of some value 31 𝑓𝑓𝑓𝑓 for = 2 ( ) 2𝜋𝜋 ∗ ′ 𝜆𝜆 ћ � 𝛹𝛹𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝐻𝐻 𝛹𝛹𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑑𝑑𝜏𝜏 𝜌𝜌 𝐸𝐸𝑓𝑓 • Rather than a nucleon undergoing transmutation by its lonesome, instead e--capture can occur λ• This is either due to a capture of a low-lying (usually the “K-shell”) electron or due to the electron Fermi energy in an electron-degenerate environment being high enough to overcome the electron-capture Q-value • The decay constant for electron-capture decay is a bit different than for β decay, because the final state only consists of a nucleon and a neutrino … i.e. = 0 and = , = (0) 𝑒𝑒 𝑓𝑓 𝑑𝑑 𝑘𝑘 ν • The decay constant is then: 2 𝐾𝐾, 𝐾𝐾 Ψ ψ 𝜑𝜑 where (0) is the wave-function 𝐺𝐺for𝐹𝐹 the inner2 -2most atomic2 electron (the one in the “K-shell”) 3 3 3 𝑓𝑓𝑓𝑓 ν 𝐾𝐾 𝜆𝜆 2𝜋𝜋 ћ 𝑐𝑐 𝑀𝑀 𝑇𝑇 𝜑𝜑 / • You may𝜑𝜑 𝐾𝐾recall from your Quantum class, 0 = 2 3 2 1 𝑍𝑍𝑚𝑚𝑒𝑒𝑒𝑒 2 • As such, the ratio of electron-capture to β𝜑𝜑+ 𝐾𝐾decay for𝜋𝜋 a nucleus4𝜋𝜋𝜀𝜀0ћ goes as λ𝐾𝐾 (of course, > 2 is a requirement for β+ decay to be possible in the first place) 3 λβ+ ∝ 𝑍𝑍 Since EC decay𝑄𝑄𝛽𝛽 only 𝑚𝑚emits𝑒𝑒 a neutrino, which will be almost impossible for us to detect, how do you figure EC decay is usually detected? X-ray and Auger electron emission due to atomic electrons filling the vacated orbital 32 γ decay basics

• γ decay is a de-excitation from an excited bound state to a lower energy state, preceded by some decay or reaction • [Just to be clear] Z & A are unchanged • γ ray energies can span anywhere from several keV to several MeV • γ decay lifetimes are typically extremely short ( femtoseconds) [with the exception of isomeric states] 𝜏𝜏 ≲

INEL γ Spectrum Catalog

33 γ decay types • Parity and angular momentum are conserved during γ decay • Photons carry some integer angular momentum with a minimum = 1, where is referred to by the multipole 2 • = 1: , = 2 , 𝑙𝑙 𝑙𝑙 𝑙𝑙 • A photon’s parity depends not only on , but also on the decay type 𝑙𝑙 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 ⋯ • A photon decay corresponds to shift in the nucleus’s charge and matter distribution 𝑙𝑙 •Shift in the charge distribution = change in electric field = Electric •Shift in the current distribution [i.e. orbitals of protons]= change in magnetic field = Magnetic • The selection rules corresponding to a particular decay type are:

34 How does 0 0 happen? = 1 • Since for +a photon, +de-excitation by photon emission isn’t possible • Instead𝑚𝑚𝑚𝑚𝑚𝑚 the process of internal conversion can happen, whereby𝑙𝑙 a nucleus interacts→ electromagnetically with an orbital electron and de-excites by ejecting that orbital electron • This process operates in competition with γ decay for any transition, not just 0+→0+

• The energy of the emitted electron is: = , , where is the decay transition energy, and , is the− electron binding energy 𝐼𝐼𝐼𝐼 𝑥𝑥𝑥𝑥 𝐵𝐵𝐵𝐵 𝑒𝑒 𝐸𝐸 𝐸𝐸 −−𝐸𝐸 𝑥𝑥𝑥𝑥 𝐵𝐵𝐵𝐵 𝑒𝑒 𝐸𝐸 𝐸𝐸 hyperphysics

Ahmad & Wagner, Nucl. Instrum. Meth. (1974)

2 A similar, but different phenomenon is Internal Pair Conversion, where a photon with Eγ>2mec interacts with 35 the coulomb field of the nucleus to create an e+-e- pair. See e.g. A. Wuosmaa et al. Phys. Rev. C Rapid Comm. 1998 Weisskopf (a.k.a. single-particle) estimates for

2𝑙𝑙 +1 • , , , = 𝐸𝐸𝛾𝛾 𝛾𝛾 ( , , , ), 8𝜋𝜋 𝑙𝑙𝛾𝛾+1 �ћ𝑐𝑐 𝛾𝛾 𝑖𝑖 𝑓𝑓 2 𝛾𝛾 𝑖𝑖 𝑓𝑓 • Reducedλ 𝑙𝑙 𝐽𝐽 𝜋𝜋 → transition𝐽𝐽 𝜋𝜋 𝑙𝑙𝛾𝛾 probabilities2𝑙𝑙𝛾𝛾+1 ‼ ћ assuming𝐵𝐵 𝑙𝑙 𝐽𝐽 the𝜋𝜋 → initial𝐽𝐽 𝜋𝜋 to final state transitionλ is due to a single nucleon re-orienting itself within a nucleus of uniform density with = / are: / 1 3 • . . , = ( ) 0 2 𝛾𝛾 The units for B change𝑅𝑅 with𝑟𝑟 𝐴𝐴l ! 1 3 2𝑙𝑙 2𝑙𝑙𝛾𝛾 3 2 2𝑙𝑙𝛾𝛾 γ 𝑠𝑠 𝑝𝑝 𝛾𝛾 4𝜋𝜋 𝑙𝑙𝛾𝛾+3 0( )/ • 𝐵𝐵 . . 𝐸𝐸 ,𝑙𝑙 = 𝑟𝑟 𝐴𝐴 𝑒𝑒 (𝑓𝑓𝑓𝑓) , where the nuclear magneton = 2 𝛾𝛾 10 3 2𝑙𝑙 −2 2 2 2𝑙𝑙𝛾𝛾−2 𝑒𝑒ћ • Note:𝐵𝐵𝑠𝑠 𝑝𝑝There𝑀𝑀 𝑙𝑙𝛾𝛾 is a steep𝜋𝜋 𝑙𝑙𝛾𝛾+3 dependency𝑟𝑟0 𝜇𝜇𝑛𝑛 of𝑓𝑓𝑓𝑓 on , so only one multipole of 𝜇𝜇a𝑛𝑛 decay2𝑚𝑚𝑝𝑝 𝑐𝑐type will matter • The equations above are still a huge pain L. van Dommelen, Quantum Mechanics for Engineers (2012) to work with and more noble souls haveλ 𝑙𝑙 worked-out the decay constant for various situations. • Using in MeV, in for a nucleus with is −given1 by: 𝛾𝛾 • Weisskopf𝑄𝑄 estimatesλ are𝑠𝑠 generally within an order of magnitude of the real answer, so γ decay constants𝐴𝐴 are often quoted as the ratio to this estimate in “Weisskopf Units” [w.u.] 36 Weisskopf (a.k.a. single-particle) estimates for ½

½( ) for Electric Transitions (from Weisskopf) ½( ) for Magnetic Transitions (from Moszkowski)

𝑡𝑡 𝐸𝐸𝛾𝛾 𝑡𝑡 𝐸𝐸𝛾𝛾 𝑡𝑡

L. van Dommelen, Quantum Mechanics for Engineers (2012)

Note that E transitions of a given multipole and Eγ are ~100X faster than M transitions with the same Eγ,ℓ

Now we see how it is that low-energy high-spin states exist as isomeric states. 37 Internal conversion coefficient, α • Don’t forget about our old friend, internal conversion, which competes with γ decay • Competition between the two is described by the internal conversion coefficient = , λ𝐼𝐼𝐼𝐼 so = + = (1 + ) 𝛼𝛼 λ𝛾𝛾 • depends𝐼𝐼𝐼𝐼 on theγ densityγ of electrons near the nucleus, and so some friendly atomic physicists haveλ doneλ theλ dirtyλ work of𝛼𝛼 calculating the following approximate formulas: 𝛼𝛼 These rely on the Born • = 5 ; = 3 , approximation, so 3 . . 2 𝑙𝑙+ ⁄2 3 . . 2 𝑙𝑙+ ⁄2 𝑍𝑍 𝑙𝑙 4 2𝑚𝑚𝑒𝑒𝑐𝑐 𝑍𝑍 4 2𝑚𝑚𝑒𝑒𝑐𝑐 Z<<137 ought to apply where 3 , is the transition Q-value, 3 𝛼𝛼 𝐸𝐸𝐸𝐸 𝑛𝑛. . 𝑙𝑙+1 𝛼𝛼𝑓𝑓 𝑠𝑠 𝑄𝑄 𝛼𝛼 𝑀𝑀𝑙𝑙 𝑛𝑛 𝛼𝛼𝑓𝑓 𝑠𝑠 𝑄𝑄 and is the principal1 quantum number of the orbital electron being ejected 𝛼𝛼𝑓𝑓 𝑠𝑠 ≈ 137 𝑄𝑄 • The atomic orbitals , , , , , correspond to = 1, 2, 3, 4, 5, 𝑛𝑛 • Clearly this process is favored for high- nuclei, …but also for < 1. = 0 transitions 𝐾𝐾 𝐿𝐿 𝑀𝑀 𝑁𝑁 𝑂𝑂 ⋯ 𝑛𝑛 ⋯ • For 0 0 transitions, = 3.8 / / , with Q in MeV and λ in s-1 𝑍𝑍 𝑄𝑄 022𝑀𝑀𝑀𝑀𝑀𝑀 𝑙𝑙 + + 3 4 3 1 2 → λ𝐸𝐸� � 𝑍𝑍 𝐴𝐴 𝑄𝑄 38 Arfken, Klema, & McGowan, Phys. Rev. Lett. (1952) γ angular correlations, general case • Generally speaking, ( ) for any γ-γ coincidence is defined by a sum of Legendre polynomials: • = 𝑊𝑊 (𝜃𝜃cos ) • i.e. =𝑖𝑖=1𝑙𝑙 + cos + cos ( )+ cos ( ), 𝑖𝑖=0 2𝑖𝑖 2𝑖𝑖 𝑊𝑊where𝜃𝜃 the∑ normalization𝑎𝑎 𝑃𝑃 2 is𝜃𝜃 such that 4 90° = 1 2𝑙𝑙 2 4 2𝑙𝑙 • The coefficients𝑊𝑊 𝜃𝜃 are𝑎𝑎 fit to𝜃𝜃 data 𝑎𝑎and the 𝜃𝜃results⋯ 𝑎𝑎 are 𝜃𝜃 106Pd checked against the expected results for particular𝑊𝑊 𝑎𝑎𝑖𝑖 combinations of , , , , R.Evans, The Atomic Nucleus (1955) • For common cases, 𝑖𝑖 𝑖𝑖 𝑓𝑓 1 2 pre-tabulated values𝐽𝐽 𝐽𝐽 𝐽𝐽 𝑙𝑙 𝑙𝑙 are available to compare to

39 Steps of fission 1. A nucleus becomes deformed either due to an external perturbation that brings in energy or an internal cluster rattling around within the potential well 2. The energy is absorbed as a collective excitation that manifests itself as a drastic shape change, elongating the nucleus into a peanut shape 3. The separation of the two lobes of the peanut becomes great enough that the two repel each other, splitting apart at the scission point 4. The coulomb repulsion accelerates the two fragments apart 5. The two fragments are each highly excited and de-excite initially via , followed by γ emission (meaning prompt neutrons will be emitted along the direction of the fragments) 6. The neutron-rich fragments will then β decay back to stability, possibly emitting delayed neutrons via β-delayed neutron emission

Why do you figure there is A massive particle is better suited to remove the large angular momentum present in high-lying excited states. neutron emission at first and At lower excitation energies, a particle has to tunnel out of the Loveland, Morrissey, & Seaborg, finally only γ emission? nucleus, while the γ doesn’t. Modern Nuclear Chemistry (2006) 40 Energetics of shape change: Liquid drop picture • For the deformed shape, which is an ellipsoid in this picture, the nuclear radius can be parameterized as an expansion in terms of Legendre polynomials [which for axial symmetry will only keep the = 2 term] • = [1 + (cos )] B.R. Martin, Nuclear and Particle Physics (2009) 𝑙𝑙 • is the quadrupole distortion parameter, which is related to the quadrupole deformation by 𝑅𝑅 𝜃𝜃 𝑅𝑅0 𝛼𝛼2𝑃𝑃2 𝜃𝜃 = = (1 + ) = 𝛼𝛼2 , and the ellipsoid axes by: , 5 1 4 • It turns𝛼𝛼2 out, 𝜋𝜋expanding𝛽𝛽2 the Coulomb and surface𝑎𝑎 𝑅𝑅 energy0 𝛼𝛼 terms2 𝑏𝑏 as a𝑅𝑅 power0 1+𝛼𝛼 2series in yields / • 1 and 1 + 2 2/ 𝛼𝛼 ′ 𝑍𝑍 1 2 ′ 2 3 2 2 1 3 5 5 / • Meaning𝑐𝑐 the𝑐𝑐 𝐴𝐴 energy cost2 for deformation𝑠𝑠 is 𝑠𝑠 = 2 2 𝐸𝐸 ≈ 𝑎𝑎 − 𝛼𝛼 𝐸𝐸 ≈ 𝑎𝑎 𝐴𝐴 2 𝛼𝛼 2/ 𝛼𝛼2 2 3 𝑍𝑍 1 3 • So, when the non-deformed Coulomb energy∆ 𝐸𝐸is twice5 the𝑎𝑎𝑠𝑠 non𝐴𝐴 -deformed− 𝑎𝑎𝑐𝑐 𝐴𝐴 surface energy or greater, there is zero energetic cost (or even an energetic gain) to deform (and ultimately fission!)

• The fissionability parameter = = is a measure of fission favorability 2 𝑍𝑍2 𝐸𝐸𝑐𝑐 𝑎𝑎𝑐𝑐 𝑍𝑍 �𝐴𝐴 41 𝑍𝑍2 2𝐸𝐸𝑠𝑠 2𝑎𝑎𝑠𝑠 𝐴𝐴 𝑥𝑥 ≡ �𝐴𝐴 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Loveland, Morrissey, & Seaborg, Modern Nuclear Chemistry (2006) rate • Spontaneous fission is akin to α decay (or proton or cluster decay, for that matter), where a barrier is “assaulted” at some rate and there is probability for tunneling through the barrier • Here, the difference is that the potential is not from the nucleus, but rather the potential energy surface for the landscape of possible shapes. The nucleus itself is tunneling through the barrier. ( ) • ½ = , where is the assault frequency and is the tunneling probability ln 2 • 𝑡𝑡 corresponds𝑓𝑓𝑓𝑓 to the𝑓𝑓 rate at which the nuclear shape𝑃𝑃 is changing, i.e. the frequency of surface oscillations, which is ~10 (Hill & Wheeler, Phys. Rev. 1953) • For𝑓𝑓 the simplest case of a one-humped barrier, 20 −1 approximated as a inverted parabola, 𝑠𝑠 (due to the liquid-drop +single particle potential for deformation, See Lec. 4), the tunneling probability is (Hill & Wheeler, Phys. Rev. 1953) = / Extremely sensitive 1 to predictions of Ef • 𝑃𝑃So, ½1+exp102𝜋𝜋𝐸𝐸𝑓𝑓expћ𝜔𝜔 ( ) 42 −21 𝑡𝑡 ≈ 2𝜋𝜋𝐸𝐸𝑓𝑓 ⁄ ћ𝜔𝜔 Spontaneous fission rate • The height of the is related to the fissionability parameter (recall ~48 means fission immediately) • So without resorting to fancy calculations of , 𝑥𝑥 the half-life for spontaneous fission can be ball-parked 𝑓𝑓 • But this method is extremely rough 𝐸𝐸

• Loveland, Morrissey, & Seaborg, Modern Nuclear Chemistry (2006) ….using shell-corrected masses gives some improvement: Vandenbosch & Huizenga, (1973) Hill & Wheeler, Phys. Rev. 1953

Even-Odd Even-Even

43 Fission fragment mass distribution K. Flynn et al. PRC Vol 5 1972 • The culprit for the asymmetric mass distribution are the = 50, = 82 shell closures, which favors nuclei in this range for one of the fission fragments.𝑍𝑍 𝑁𝑁 • The other fragment has the remainder of most of the rest of the nucleons

44 Fission fragments: kinetic energy • The total kinetic energy of fission products will roughly be the coulomb repulsion energy of the two main fission fragments, • = …where r =1.8fm is used instead of 1.2 because of the strong deformation at scission . / / 0 𝑍𝑍1𝑍𝑍2𝛼𝛼ћc 1 3 1 3 • E.g.𝑇𝑇 𝑇𝑇𝑇𝑇240Pu 1 8 𝐴𝐴1 +𝐴𝐴2 • / 2.55 . If = 50, = 82, then = 94 50 = 44, = 240 132 44 = 64 ( )( ) • = / ℎ𝑖𝑖𝑖𝑖�/ ℎ𝑖𝑖𝑖𝑖� 𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙 𝐴𝐴 𝑍𝑍 ≈ . 𝑍𝑍 𝑁𝑁 𝑍𝑍 − 𝑁𝑁 − − 50 44 𝛼𝛼ћc Schunk & Robledo, Rep. Prog. Phys. (2016) 1 3 1 3 𝑇𝑇𝑇𝑇𝑇𝑇 1 8 132 +108 ≈ 178𝑀𝑀𝑀𝑀𝑀𝑀 240Pu

45