University Physics V · Radioactive Decay · 15.5
Gamma Decay, Multipoles & Isomers
Given two levels with their spins and parities, work out which photon can carry the difference, how fast it goes, and when the honest answer is that no single photon can. That is what turns a level scheme into a set of half-lives, and what makes a six-hour isomer out of a nuclear excitation.
Build the model
Connect the measurement to the mechanism.
Gamma decay is a first-order golden-rule transition in which a nucleus hands its surplus angular momentum and parity to one photon, so the model is built in a basis where the photon's angular momentum is diagonal. Expanding the quantised vector potential in vector spherical harmonics rather than plane waves does exactly that: every mode carries a total angular momentum L ≥ 1, a projection μ, and a definite parity — electric EL with (−1)L, magnetic ML with (−1)(L+1). The nuclear side is then a spherical tensor operator of rank L; in the long-wavelength limit M(EL, μ) = ∫ρ(r) rL YLμ* d³r, with a current-and-magnetisation partner for ML.
Wigner–Eckart splits every matrix element of it into a Clebsch–Gordan coefficient holding all the m-dependence and one reduced matrix element holding all the nuclear structure, so what survives is two selection rules and a single number B(σL). The cost sits in the same formula: the operator brings rL and the phase space brings k(2L+1), so with kR ≈ 0.02 for a 660 keV photon in a mid-heavy nucleus, each extra unit of L divides the rate by roughly 10⁵. That is why a level whose lowest allowed multipole is M4 lives minutes rather than picoseconds — and why the long-wavelength truncation, the Weisskopf normalisation and the competing conversion-electron channel must all be named before a rate is quoted.
- Simple definition
- A gamma transition is labelled by the angular momentum L the emitted photon carries and by the family of the operator that emits it — electric EL of parity (−1)L or magnetic ML of parity (−1)(L+1) — with the two level spins fixing which L are possible and the two parities fixing the family.
- Example
- Ba-137m sits at 661.66 keV with Jᵖ = 11/2⁻ above a 3/2⁺ ground state: the triangle rule gives 4 ≤ L ≤ 7 and the parity flips, so the lowest route is M4 — and that L = 4 is why the level lives 2.55 minutes instead of picoseconds.
Lists the photons that can exist. L = 0 is missing, so a 0 → 0 transition has no single-photon route at all.
J in units of ħ; L is the photon's total angular momentum and μ = mf − mᵢ
Halves the list: a parity change leaves E1, M2, E3, M4; no change leaves M1, E2, M3, E4.
π = ±1 read off the level scheme; the product is +1 when the parity is unchanged
Collapses every m onto one number, so a rate needs only B(σL) = |⟨Jf‖M(σL)‖Jᵢ⟩|²/(2Jᵢ+1).
the CG coefficient is dimensionless; the reduced element carries e⋅fmᴸ for EL, μN⋅fm(L−1) for ML
The (Eγ/ħc)(2L+1) factor is the whole reason a high-L transition at low energy is slow.
(2L+1)!! = 3, 15, 105, 945 for L = 1 to 4; ħc = 197.33 MeV fm; λ in s⁻¹
1 W.u. = λmeasured/λW. Collective E2s reach 100 W.u., M1s run near 10⁻², E1s 10⁻⁴ or less.
Eγ in MeV, R = 1.2A¹⁄³ fm; 0.307A(−2/3) is [μN/(e⋅fm)]² = [ħc/2mpc²fm]² times geometry
Turns a measured half-life into the photon partial rate before any comparison, and Γ is the width Mössbauer absorption resolves.
ħ = 6.582×10⁻¹⁶ eV s; α = λₑ/λγ is dimensionless and grows with Z³ and with L
Where the label EL or ML comes from
The coupling is the nuclear current against the quantised field, H′ = −(1/c)∫ j⋅A d³r, and the golden rule needs a final photon state. Expanding A in plane waves hides angular momentum; expanding it in vector spherical harmonics diagonalises J² and parity instead, so every mode is stamped with a total angular momentum L, a projection μ, and a parity: electric EL with (−1)L, magnetic ML with (−1)(L+1). Two structural facts fall out. There is no L = 0 photon — the field is transverse and spin-1, and the L = 0 electric moment is the total charge, which cannot change. And in the long-wavelength regime kR ≪ 1, where kR = EγR/ħc ≈ 0.02 for a 660 keV photon in a mid-heavy nucleus, only the leading term of each expansion survives, leaving M(EL, μ) = ∫ρ(r) rL YLμ* d³r and its convection-plus-magnetisation partner for ML. Everything downstream is a matrix element of one of those two operators.
Wigner–Eckart: one reduced element, many m values
M(σL, μ) is a spherical tensor of rank L, so ⟨Jf mf|M(σL, μ)|Jᵢ mᵢ⟩ = ⟨Jᵢ mᵢ L μ|Jf mf⟩⟨Jf‖M(σL)‖Jᵢ⟩/√(2Jf+1). Three consequences arrive at once. The Clebsch–Gordan coefficient vanishes unless μ = mf − mᵢ and |Jᵢ − Jf| ≤ L ≤ Jᵢ + Jf, so the selection rule is geometry, not dynamics — no nuclear model is involved in it. The (2Jᵢ+1)(2L+1)(2Jf+1) matrix elements collapse onto one number, so a rate summed over mf and averaged over mᵢ depends only on B(σL) = |⟨Jf‖M(σL)‖Jᵢ⟩|²/(2Jᵢ+1); this is why a rate never depends on how the source is oriented. And the m-dependence that was factored out is not lost — it is precisely what a γ–γ angular correlation, or the angular distribution from an aligned source, measures.
Reading the multipole off two levels
Subtract the spins, multiply the parities, then take the smallest surviving L. Ba-137: the 661.66 keV level is 11/2⁻ and the ground state 3/2⁺, so 4 ≤ L ≤ 7; the parity product is −1, which lets EL live only at odd L and ML only at even L, leaving M4, E5, M6, E7 — the answer is M4. Fe-57: 3/2⁻ → 1/2⁻ at 14.4 keV gives 1 ≤ L ≤ 2 with no parity change, so M1 and E2 both survive; the transition is M1 with an E2 admixture near 10⁻⁷. Mixing is the normal case whenever L and L+1 are both allowed, and the mixing ratio δ is measured, not predicted. The clean case is an even-even ground-state band: 2⁺ → 0⁺ forces L = 2 exactly and no parity change makes it pure E2, which is why B(E2) values are the standard currency of collective structure.
The (kR)² penalty per unit of L, and how an isomer forms
Both factors in the rate punish large L. The operator carries rL integrated over a nucleus of radius R = 1.2A¹⁄³ fm, and the photon phase space carries k(2L+1). Collecting them in the Weisskopf estimate, neighbouring orders stand in the ratio λ(L+1)/λ(L) ≈ (kR)²⋅L(L+2)(L+3)²/[(L+1)²(L+4)²(2L+3)²] — in practice (kR)² divided by something close to (2L+3)². For the 661.66 keV transition in A = 137, kR = 0.0207, so (kR)² = 4.3×10⁻⁴ and one extra unit of L costs about 10⁻⁵. Run the ladder at that energy: E1 near 10⁻¹⁵ s, M2 near 10⁻⁸ s, E3 near 2×10⁻⁵ s, M4 near 450 s. That is isomerism in one line. A level whose lowest allowed multipole is M4 — and an isomer is usually low in energy too, which shrinks kR further — simply has no fast way to shed its angular momentum.
Weisskopf units are a diagnostic, not a prediction
λW is what you get by assuming a single proton hops between two shell-model orbitals with a uniform-density radial integral and every geometric factor set to unity: BW(EL) = (1/4π)[3/(L+3)]²R(2L)e², with the magnetic estimate at the same L fixed at 0.307A(−2/3) times the electric one, since μN/(e⋅fm) = ħc/(2mpc²⋅fm) = 0.105. Nothing there describes a real nucleus, which is exactly why the ratio λmeasured/λW — one Weisskopf unit — is useful as a diagnostic. Collective E2 transitions in deformed nuclei run 100–300 W.u. because many nucleons move coherently; M1 transitions typically run 10⁻² to 10⁻³ W.u.; E1 transitions are hindered to 10⁻⁴ or below by isospin and by the centre-of-mass constraint; the M4 in Ba-137m comes out near 2.6 W.u. Enhancement and hindrance are the structure the single-particle picture threw away.
When no photon can leave: E0, conversion, and pairs
Two things stop a photon. First, no allowed L: for 0 → 0 both |Jᵢ − Jf| and Jᵢ + Jf are zero, so the triangle admits only L = 0, which no photon has. Such a level empties by E0 internal conversion, whose matrix element ⟨f|Σ rₖ²|i⟩ is weighted by the electron density inside the nuclear volume, or by internal pair creation once the energy exceeds 2mec² = 1.022 MeV. Second, competition: internal conversion runs alongside every γ, so a measured half-life gives λₜₒₜₐₗ = λγ(1 + α), and only λγ may be set against a Weisskopf estimate. The coefficient α grows with Z³, with L, and with falling transition energy. For the 661.66 keV M4 in Ba-137 it is 0.112, which is exactly why 94.7% of Cs-137 decays feed the isomer while only 94.7/1.112 = 85.1% of them yield the 661.66 keV photon a detector sees.
Change one variable at a time
Make the relationship visible.
Set |ΔJ| = 4 with a parity change: the lowest route is L = 4 magnetic, and at 660 keV its bar stands near 102.5 s — that is Ba-137m. Drag the energy down to 100 keV and the same bar climbs 7.4 decades into centuries; drop |ΔJ| to 1 and the marker jumps twenty-two decades down the ladder.
LOWEST ALLOWED L4 ħ
log₁₀(T½/s) AT L = 1-15.0
log₁₀(T½/s) AT L = 42.7
kR = E R / ħc0.0207
Live interpretationLOWEST ALLOWED L: 4 ħ. log₁₀(T½/s) AT L = 1: −15.0. log₁₀(T½/s) AT L = 4: 2.7. kR = E R / ħc: 0.0207
Catch the common trap
Explain before calculating.
A 4⁻ level decays to a 2⁺ level in the same nucleus. Which multipole carries essentially the whole photon rate?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe 661.66 keV level of Ba-137 has Jᵖ = 11/2⁻ and the ground state is 3/2⁺. List every multipole the two selection rules allow, and name the one that carries the rate.
- Triangle rule first: |11/2 − 3/2| = 4 and 11/2 + 3/2 = 7, so 4 ≤ L ≤ 7. The condition L ≥ 1 is already satisfied, so nothing is lost there.
- Parity next: πᵢ πf = (−1)(+1) = −1, a change. An EL needs (−1)L = −1, so odd L only: E5 and E7. An ML needs (−1)(L+1) = −1, so even L only: M4 and M6.
- The allowed list is therefore M4, E5, M6, E7 — and the lowest of them is a magnetic multipole, not an electric one.
- Rank them: with R = 1.2 × 137¹⁄³ = 6.19 fm and k = 0.66166/197.33 fm⁻¹, kR = 0.0207, so each extra unit of L costs roughly (kR)²/(2L+3)² ≈ 10⁻⁵. M4 is the transition to better than a part in 10⁴.
AnswerM4. The allowed set is M4, E5, M6 and E7, but each extra unit of L costs about 10⁻⁵, so M4 carries the rate to better than a part in 10⁴.
MediumEstimate the M4 half-life of that 661.66 keV transition from the Weisskopf single-particle formula, then compare it with the measured 2.552 min, given a total internal-conversion coefficient α = 0.112.
- Anchor on the electric estimate at the same L: λW(E4) = 1.1×10⁻⁵ A⁸⁄³ Eγ⁹ s⁻¹. With A = 137, A⁸⁄³ = 4.988×10⁵, so λW(E4) = 5.487 Eγ⁹ s⁻¹ for Eγ in MeV.
- Convert to magnetic at the same L with λW(M4) = 0.307 A(−2/3) λW(E4). Here 0.307/26.575 = 0.011552, so λW(M4) = 0.06338 Eγ⁹ s⁻¹.
- Insert the energy: Eγ⁹ = 0.66166⁹ = 2.431×10⁻², giving λW(M4) = 1.541×10⁻³ s⁻¹ and T½(W) = ln2/λ = 450 s = 7.50 min.
- The measured 2.552 min = 153.1 s is the total rate, λₜₒₜₐₗ = ln2/153.1 = 4.527×10⁻³ s⁻¹, and conversion electrons take part of it: λγ = λₜₒₜₐₗ/(1 + α) = 4.527×10⁻³/1.112 = 4.071×10⁻³ s⁻¹.
- Compare like with like: λγ/λW = 4.071×10⁻³/1.541×10⁻³ = 2.6, so the transition runs at 2.6 Weisskopf units.
AnswerT½(W) = 450 s = 7.5 min against a measured 2.55 min. In photon-partial terms the transition is 2.6 W.u. — the same order of magnitude, which is all a single-particle estimate ever claims.
HardThe 14.413 keV first excited state of Fe-57 is 3/2⁻ above a 1/2⁻ ground state and has T½ = 98.3 ns. Assign the multipolarity, find the natural linewidth Γ, and compare it with the free-atom recoil energy to say why Mössbauer absorption needs a lattice.
- Selection rules: |3/2 − 1/2| = 1 ≤ L ≤ 2 with no parity change, so ML takes odd L (M1) and EL takes even L (E2). Both are allowed, and the Weisskopf ratio λW(E2)/λW(M1) = 1.1×10⁻⁷ at this energy — the same order as (kR)² = (3.37×10⁻⁴)² — so the transition is M1 to a part in 10⁷.
- Mean life: τ = T½/ln2 = 98.3/0.6931 = 141.8 ns.
- Width: Γ = ħ/τ = 6.582×10⁻¹⁶ eV s ÷ 1.418×10⁻⁷ s = 4.64×10⁻⁹ eV, so Γ/Eγ = 4.64×10⁻⁹/14413 = 3.22×10⁻¹³. One full width corresponds to a Doppler shift of v = cΓ/Eγ = 0.097 mm s⁻¹, which is why a Mössbauer drive scans in millimetres per second.
- Free-atom recoil: ER = Eγ²/(2Mc²) with Mc² = 56.935 u × 931.494 MeV/u = 5.304×10¹⁰ eV, giving ER = (1.4413×10⁴)²/(1.0607×10¹¹) = 1.96×10⁻³ eV.
- Compare: ER/Γ = 1.96×10⁻³/4.64×10⁻⁹ = 4.2×10⁵. Emitter and absorber are each displaced by ER, so free atoms miss resonance by about 8×10⁵ linewidths; only the recoil-free fraction, where the lattice takes the momentum and Mc² becomes the crystal's mass, overlaps at all.
AnswerM1, with an E2 admixture near 10⁻⁷. Γ = ħ/τ = 4.6×10⁻⁹ eV, a fractional width of 3.2×10⁻¹³, while the free-atom recoil of 1.96×10⁻³ eV is 4.2×10⁵ linewidths — hence recoilless emission.