University Physics IV · Nuclear Physics · 13.6

Gamma Decay, Multipolarity & Lifetimes

A nucleus that has settled its neutron and proton numbers still has spin, parity and a few hundred keV to shed. This is the toolkit for that: read the multipole off two level labels, get the rate right to within a factor of ten, and know when no photon comes out at all.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Gamma decay is the electromagnetic field taking a nucleus down between its own levels, and the subject is the multipole expansion of that field. Each term carries a definite angular momentum L and a definite parity — (−1)^L for the electric multipole EL, (−1)^(L+1) for the magnetic ML — so conserving both fixes which terms may appear: |Jᵢ − J_f| ≤ L ≤ Jᵢ + J_f, with L = 0 struck out because a real transverse photon has no monopole component. That part is exact, and it is a permission list, not a rate.

The rate comes from the golden rule applied to one nucleon moving between two shell-model orbitals, and it carries a factor (kR)²L — the nuclear radius measured against the photon's reduced wavelength, raised to twice the multipole order. At a few hundred keV in a medium nucleus that factor, with the statistical coefficients beside it, costs about five decades of half-life per extra unit of L. That one number is why the lowest permitted multipole nearly always dominates, and why a four-unit spin drop at low energy can hold a state for minutes.

The cost is honesty about the estimate: Weisskopf assumes one proton with unit radial overlap and no collective motion, so real E2 strengths in deformed nuclei run a hundred times faster and E1 strengths ten thousand times slower. Weisskopf is not a prediction; it is the unit in which the deviation — the real physics — gets reported.

Simple definition: Gamma decay is the emission of a single photon by a nucleus dropping between two of its own energy levels, with the photon's multipole order L fixed by the spin change and its electric or magnetic character fixed by the parity change.
Example: The ¹³⁷Ba isomer sits at 661.7 keV with Jπ = 11/2⁻ above a 3/2⁺ ground state: ΔJ = 4 with a parity flip, so the photon must be M4 — and that one label is why the state lives 2.55 minutes instead of a femtosecond.

Which multipoles a transition may use|Jᵢ − J_f| ≤ L ≤ Jᵢ + J_f, with L ≥ 1

Two spin labels give a short list of candidates, and nothing else in the problem can add to it.

J in units of ħ. L = 0 is struck out: a real transverse photon carries no monopole field.

Electric or magnetic, decided by parityπᵢ π_f = (−1)^L for ELπᵢ π_f = (−1)^(L+1) for ML

Parity halves the list, leaving electric and magnetic orders alternating up the ladder.

So E1, M2, E3 and M4 flip the parity; M1, E2, M3 and E4 leave it alone.

Weisskopf single-particle ratesλ(E1) = 1.0×10¹⁴ A²⁄³ E³λ(E2) = 7.3×10⁷ A⁴⁄³ E⁵λ(M4) = 4.5×10⁻⁶ A² E⁹

Defines the Weisskopf unit: a measured rate divided by this is what a level scheme is really reporting.

E in MeV, λ in s⁻¹, A the mass number; T½ = ln2/λ. Also λ(E3) = 34 A² E⁷ and λ(E4) = 1.1×10⁻⁵ A⁸⁄³ E⁹.

The price of one more unit of LkR = E_γ R / ħc, R = 1.2 A¹⁄³ fm, ħc = 197.3 MeV⋅fm

Each extra unit of L costs (kR)² and a statistical factor near 1/60 — about five decades of half-life here.

At E_γ = 0.66 MeV and A = 137: R = 6.19 fm, kR = 0.0207, (kR)² = 4.3×10⁻⁴.

Internal conversionα = λₑ / λ_γ, λₜₒₜₐₗ = λ_γ(1 + α), Tₑ = Eₜᵣ − Bₑ

Turns a measured total half-life into the gamma partial rate you can compare with Weisskopf.

α grows roughly as Z³ and with L, and falls steeply with Eₜᵣ; Bₑ is the electron's atomic binding energy.

Recoil energy against linewidthE_R = E_γ² / (2Mc²), Γ = ħ/τ, v₁ = cΓ/E_γ

E_R is 4×10⁵ linewidths, so free-atom resonance fails and only a lattice-bound nucleus absorbs.

⁵⁷Fe at 14.4 keV with τ = 141 ns: E_R = 1.96 meV, Γ = 4.7 neV, v₁ = 0.097 mm s⁻¹.

A level scheme is a spin, a parity, and an energy

A gamma spectrum is a list of photon energies; a level scheme is the model that explains them. Excited states sit anywhere from tens of keV to several MeV above the ground state — 14.4 keV in ⁵⁷Fe, 661.7 keV in ¹³⁷Ba, 4.44 MeV in ¹²C — and each carries a spin and a parity, written Jπ. Gamma decay moves the nucleus between two of them and changes nothing else: Z and A are untouched, so no new element appears. The transition energy is very slightly above the photon energy, because the nucleus recoils: E_R = E_γ²/2Mc², which for a 1 MeV photon from A = 100 is 1²/(2 × 93 150) MeV = 5.4 eV. Against a germanium detector's 2 keV resolution that is invisible; against a Mössbauer linewidth of a few neV it is enormous, and the same formula decides the last section below. Most levels also feed the level immediately beneath rather than jumping to the ground state, so what you actually fit are branching ratios between competing transitions.

Angular momentum and parity name the multipole

Expand the radiation field in multipoles. The EL term carries L units of angular momentum with parity (−1)^L; the ML term carries L units with parity (−1)^(L+1). Conservation then gives two hard rules. Spins: |Jᵢ − J_f| ≤ L ≤ Jᵢ + J_f. Parity: πᵢ π_f = +1 admits even-L electric and odd-L magnetic, while πᵢ π_f = −1 admits odd-L electric and even-L magnetic. And L ≥ 1 always, because a free photon's field has no monopole term — that exclusion is not a selection rule about nuclei, it is a statement about light. Work one: 4⁻ → 2⁺. Spins give L = 2, 3, 4, 5, 6; the parity flips, so the survivors are M2, E3, M4, E5, M6. The lowest order dominates, but the next one up is of the opposite character and can mix in — an M1 transition almost always carries some E2, and the mixing ratio δ is measured from angular correlations, never assumed.

Weisskopf sets the scale; the deviation is the result

To get a rate you need a matrix element, and Weisskopf's estimate supplies the crudest defensible one: a single proton moving between two shell-model orbitals inside a uniform sphere of radius R = 1.2A¹⁄³ fm, with unit radial overlap and no collective motion. With E in MeV and λ in s⁻¹ the results are λ(E1) = 1.0×10¹⁴ A²⁄³E³, λ(E2) = 7.3×10⁷ A⁴⁄³E⁵, λ(E3) = 34 A²E⁷ and λ(E4) = 1.1×10⁻⁵ A⁸⁄³E⁹, with magnetic partners λ(M1) = 5.6×10¹³ E³, λ(M2) = 3.5×10⁷ A²⁄³E⁵, λ(M3) = 16 A⁴⁄³E⁷ and λ(M4) = 4.5×10⁻⁶ A²E⁹. Every one of them goes as E^(2L+1). Nobody expects these to be right, and they are not: divide a measured rate by the estimate and you have the strength in Weisskopf units. E2 transitions in deformed rare earths reach 100 W.u., because whole nuclei rather than single protons are moving. E1 transitions typically land near 10⁻⁵ W.u., most of the dipole strength having been swallowed by the giant resonance at 15 MeV. M1 sits near 10⁻² W.u. Those deviations are the physics; the estimate is only the yardstick.

Isomers are the multipole penalty made visible

Take the ratio of two Weisskopf rates one unit of L apart and it is dominated by (kR)², with k = E_γ/ħc. For A = 137 and E_γ = 0.66 MeV, R = 6.19 fm and kR = 0.0207, so (kR)² = 4.3×10⁻⁴; the statistical coefficients contribute roughly another factor of 60, giving about five decades of half-life per extra unit. That is the whole mechanism of nuclear isomerism. ¹³⁷ᵐBa is the case to memorise: an 11/2⁻ state 661.7 keV above a 3/2⁺ ground state, hence M4, living 2.55 minutes where an E1 of the same energy would be gone in 10⁻¹⁵ s. The spins are no accident. Just below the magic numbers 50, 82 and 126 a high-j intruder orbital — g9/2, h11/2, i13/2 — sits beside low-j neighbours, so nuclei there routinely offer a level pair differing by four units of J. Those are the islands of isomerism. ⁹⁹ᵐTc, a 1/2⁻ p1/2 state above a 9/2⁺ g9/2 ground state, is the same physics, and its six-hour half-life is why it is in a hospital rather than a laboratory.

When no photon comes out at all

The nuclear multipole field does not only make photons. Where a bound atomic electron's wavefunction overlaps the nucleus, the field can couple to it directly and eject it with kinetic energy Tₑ = Eₜᵣ − Bₑ. This is internal conversion, a first-order process in its own right. Its strength is booked as α = λₑ/λ_γ, so a measured total rate is λₜₒₜₐₗ = λ_γ(1 + α); α grows roughly as Z³ and with L, falls steeply with transition energy, and reaches 10² and beyond for low-energy high-multipole transitions in heavy atoms. The extreme case is 0⁺ → 0⁺. The spins permit only L = 0, and there is no L = 0 photon, so the γ rate is exactly zero — yet the level still empties. ⁷²Ge's 691 keV 0⁺ state converts, with T½ = 444 ns. Above 2m_ec² = 1.022 MeV a second channel opens, internal pair creation, and ¹⁶O's 6.05 MeV 0⁺ state uses it, emptying in 67 ps by emitting an electron-positron pair and no photon whatever.

Lifetime, linewidth, and the recoilless line

A level's lifetime and its energy width are one measurement: Γτ = ħ, with ħ = 6.582×10⁻¹⁶ eV s. The timescale picks the technique. Above about a nanosecond, time the delayed coincidence between the feeding photon and the decaying one. Between roughly 1 ps and 1 ns, use Doppler-shift attenuation as the recoiling nucleus slows in a stopper. Below 10⁻¹⁵ s, abandon clocks and measure Γ directly — 1 fs corresponds to 0.66 eV. The sharp-line end becomes a tool. ⁵⁷Fe's 14.4 keV level has τ = 141 ns, so Γ = 4.7 neV, a fractional width of 3.2×10⁻¹³. But a free ⁵⁷Fe atom recoils with E_R = 1.96 meV, some 4×10⁵ linewidths, so emission and absorption lines miss each other entirely. Lock the nucleus into a crystal and a sizeable fraction of events leave the lattice in its original vibrational state; M in E_R = E_γ²/2Mc² becomes the mass of the whole crystal and the recoil vanishes. Pound and Rebka then used that line to measure a gravitational shift of 2.5×10⁻¹⁵ over 22.5 m.

Change one variable at a time

Make the relationship visible.

Interactive model

Transition energy E0.65 MeV

Mass number A137

Hold A at 137 and pull the energy down to 0.10 MeV: E1 still empties the level in 10⁻¹³ s, while the E4 curve climbs past the one-second line into years. Nothing about the nuclear force changed — only how much angular momentum the photon has to carry.

LOG₁₀ T½ FOR E1-15.0 log₁₀ s

LOG₁₀ T½ FOR E40.8 log₁₀ s

E1 TO E4 SPAN15.8 decades

COST OF ONE MORE L5.27 decades

Live interpretationLOG₁₀ T½ FOR E1: −15.0 log₁₀ s. LOG₁₀ T½ FOR E4: 0.8 log₁₀ s. E1 TO E4 SPAN: 15.8 decades. COST OF ONE MORE L: 5.27 decades

Catch the common trap

Explain before calculating.

Misconception check

Internal conversion is not the nucleus emitting a photon that its own K-shell electron then absorbs. There is no photon in the intermediate state; the nuclear multipole field couples directly to the electron's wavefunction where it overlaps the nucleus, in one first-order step. Two facts kill that picture. A 0⁺ → 0⁺ transition emits no photon at all, since there is no L = 0 radiation field, and yet it converts perfectly well — there would be nothing to absorb. And α is not capped by any absorption probability; it passes 100 for low-energy, high-L transitions in heavy atoms. The damage is in the bookkeeping: a measured half-life gives λₜₒₜₐₗ = λ_γ(1 + α), so for ¹³⁷ᵐBa with α = 0.110 the gamma partial half-life is 170 s, not the 153 s on the chart. Here that is a 10% slip; for a heavily converted transition it is a factor of a hundred, and the Weisskopf comparison is meaningless.

A nuclear level at 350 keV with Jπ = 4⁻ decays directly to the 2⁺ ground state. Which multipole assignment is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe 140.5 keV level of ⁹⁹Tc has Jπ = 7/2⁺ and decays to the 9/2⁺ ground state. List the multipoles the transition is allowed to use, name the one that carries it, and estimate its Weisskopf half-life. The measured half-life of the level is 0.19 ns.

Spins: |7/2 − 9/2| = 1 and 7/2 + 9/2 = 8, so L may run from 1 to 8. L = 0 is excluded whatever the spins say, because a real photon has no monopole component.
Parity: both states are positive, so πᵢ π_f = +1. Electric multipoles then need (−1)^L = +1, i.e. even L, and magnetic multipoles need (−1)^(L+1) = +1, i.e. odd L. The allowed set is M1, E2, M3, E4, M5, E6, M7, E8.
Each extra unit of L costs (kR)² and its statistical factor. Here R = 1.2 × 99¹⁄³ = 5.55 fm and k = 0.1405/197.3 = 7.12×10⁻⁴ fm⁻¹, so (kR)² = 1.6×10⁻⁵. The lowest order wins by a huge margin: M1 carries the transition, with a small E2 admixture.
Weisskopf for M1 carries no A dependence: λ = 5.6×10¹³ E³ = 5.6×10¹³ × (0.1405)³ = 5.6×10¹³ × 2.77×10⁻³ = 1.55×10¹¹ s⁻¹, so T½ = ln2/λ = 4.5 ps.
Against the measured 0.19 ns, the real transition is about 43 times slower — a strength of 0.02 W.u. That is unremarkable for M1, which typically lands near 10⁻² W.u.

AnswerAllowed: M1, E2, M3, E4 and upward; M1 carries it. Weisskopf gives T½ = 4.5 ps against the measured 0.19 ns, so about 0.02 Weisskopf units — ordinary M1 hindrance, not a puzzle.

Medium¹³⁷ᵐBa is the 661.7 keV isomer fed by ¹³⁷Cs. It has Jπ = 11/2⁻, the ground state is 3/2⁺, the measured half-life is 2.552 min, and the total internal conversion coefficient is α = 0.110. Assign the multipolarity, compare the gamma partial rate with the Weisskopf estimate, and say how much slower it is than an E1 of the same energy.

Spins: |11/2 − 3/2| = 4 up to 11/2 + 3/2 = 7, so L = 4, 5, 6, 7. The parity flips, so electric multipoles need odd L and magnetic need even L: the allowed set is M4, E5, M6, E7, and the lowest order, M4, carries it.
Strip the conversion out first. λₜₒₜₐₗ = ln2/153.1 s = 4.53×10⁻³ s⁻¹, so λ_γ = λₜₒₜₐₗ/(1 + α) = 4.53×10⁻³/1.110 = 4.08×10⁻³ s⁻¹ — a gamma partial half-life of 170 s, not 153 s.
Weisskopf M4: λ = 4.5×10⁻⁶ A²E⁹ = 4.5×10⁻⁶ × 1.877×10⁴ × 2.43×10⁻² = 2.05×10⁻³ s⁻¹, i.e. T½ = 338 s.
The measured rate is 4.08×10⁻³/2.05×10⁻³ = 2.0 times the estimate, so 2.0 W.u. For an M4 that is the estimate doing its job.
Now the same energy as E1: λ = 1.0×10¹⁴ × 137²⁄³ × 0.6617³ = 1.0×10¹⁴ × 26.6 × 0.290 = 7.7×10¹⁴ s⁻¹, T½ = 9.0×10⁻¹⁶ s. Three extra units of L stretch a femtosecond into minutes — a factor of 1.9×10¹⁷.

AnswerM4. λ_γ = 4.08×10⁻³ s⁻¹, a gamma partial half-life of 170 s, against a Weisskopf 2.05×10⁻³ s⁻¹ — 2.0 W.u. An E1 of the same energy would run 1.9×10¹⁷ times faster.

HardThe 14.413 keV first excited state of ⁵⁷Fe has mean life τ = 141 ns, and the ⁵⁷Fe atom has rest energy Mc² = 53.0 GeV. Find the natural linewidth Γ, the recoil energy of a free atom emitting that photon, and the source speed that shifts the line by one linewidth. Then say what makes resonant absorption possible in a solid.

Linewidth from the lifetime: Γ = ħ/τ = 6.582×10⁻¹⁶ eV s ÷ 1.41×10⁻⁷ s = 4.7×10⁻⁹ eV. Against 14 413 eV that is a fractional width Γ/E_γ = 3.2×10⁻¹³.
Recoil: the atom takes momentum p = E_γ/c, so E_R = p²/2M = E_γ²/2Mc² = (1.4413×10⁴ eV)² ÷ (2 × 5.30×10¹⁰ eV) = 2.077×10⁸/1.060×10¹¹ = 1.96×10⁻³ eV.
Compare them: E_R/Γ = 1.96×10⁻³/4.7×10⁻⁹ = 4.2×10⁵. The emitted photon falls short of the transition energy by E_R and the absorber needs an extra E_R, so the two lines miss by 2E_R — some 8×10⁵ linewidths. Free-atom resonance fluorescence is dead.
Doppler scale: v = cΓ/E_γ = 2.998×10⁸ m s⁻¹ × 3.2×10⁻¹³ = 9.7×10⁻⁵ m s⁻¹ = 0.097 mm s⁻¹. A source on a millimetres-per-second drive therefore sweeps the whole line.
In a crystal a fraction f of emissions leave the lattice in its original vibrational state. For those, M in E_R = E_γ²/2Mc² is the mass of the entire crystal, some 10²⁰ times larger, and E_R drops far below Γ. That recoil-free fraction is what makes the resonance, and the technique, exist.

AnswerΓ = 4.7×10⁻⁹ eV and E_R = 1.96×10⁻³ eV, a shift of 4.2×10⁵ linewidths; one linewidth corresponds to 0.097 mm s⁻¹. Only lattice-bound emission, with the crystal taking the recoil, restores the overlap.