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Nuclear, particle & frontier orientation · Interactive lesson

Gamma decay, multipolarity and 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.

UPIV-13.06 · Lesson 6 of 9 in this unit. Position is not a completion score.
Starting model: Gamma decay, multipolarity and lifetimesWeisskopf half-life against transition energy, both on log scales, for the four lowest electric multipoles. The dashed horizontal line is one second; the dashed vertical line marks the chosen energy. At E = 0.65 MeV and A = 137, E1 gives 10^-15.0 s and E4 gives 10^0.8 s.E1E2E3E41 sWeisskopf half-life (log scale)transition energy 0.05 to 2 MeV (log scale)A = 137
Start with a picture. Change the model in the lab below.

Nuclear, particle & frontier orientation. These numbered units provide nuclear, particle and frontier orientation, not a substitute for a complete specialist course. Check your university’s required depth and prerequisites.

1

Build the model

One photon, and at least one unit of angular momentum — that floor sets everything.

The key idea

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.

The complete physical model and assumptions

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 − Jf| ≤ L ≤ Jᵢ + Jf, 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 to a hundred thousand times slower. Weisskopf is not a prediction; it is the unit in which the deviation — the real physics — gets reported.

1A 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: ER = Eγ²/2Mc², 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. Many levels can decay to more than one lower level, not only to the ground state, so what you actually fit are branching ratios between competing transitions.

2Angular 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 − Jf| ≤ L ≤ Jᵢ + Jf. Parity: πᵢ πf = +1 admits even-L electric and odd-L magnetic, while πᵢ πf = −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: 5⁻ → 2⁺. Spins give L = 3, 4, 5, 6, 7; the parity flips, so the survivors are E3, M4, E5, M6, E7. 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.

3Weisskopf 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.

4Isomers 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 = Eγ/ħc. For A = 137 and Eγ = 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.

5When 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/λγ, so a measured total rate is λₜₒₜₐₗ = λγ(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 single-photon rate is exactly zero — yet the level still empties. ⁷²Ge's 691 keV 0⁺ state converts, with T½ = 444 ns. Above 2mec² = 1.022 MeV a second channel opens, internal pair creation, and ¹⁶O's 6.05 MeV 0⁺ state uses it, emptying in 67 ps almost entirely by emitting an electron-positron pair; a second-order two-photon branch exists but carries less than 10⁻³ of the decays.

6Lifetime, 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, let the recoiling nucleus fly a measured distance to a stopper and compare Doppler-shifted with unshifted lines. From about 10 fs to 1 ps, use Doppler-shift attenuation as the recoil slows inside the stopper. Below that, 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 ER = 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 ER = Eγ²/2Mc² 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.

2

Key equations

Read the units, assumptions and limits before substituting.

Which multipoles a transition may use

|Jᵢ − Jf| ≤ L ≤ Jᵢ + Jf, 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 parity

πᵢ πf = (−1)L for ELπᵢ πf = (−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 rates

λ(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 L

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

Internal conversion

α = λₑ / λγ, λₜₒₜₐₗ = λγ(1 + α), Tₑ = Eₜᵣ − Bₑ

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

λₑ and λγ are the conversion-electron and photon partial rates, Eₜᵣ the transition energy, Bₑ the electron's atomic binding energy; α grows roughly as Z³ and with L, and falls steeply with Eₜᵣ.

Recoil energy against linewidth

ER = Eγ² / (2Mc²), Γ = ħ/τ, v₁ = cΓ/Eγ

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

M is the emitting atom's mass, τ the level's mean life, v₁ the source speed that shifts the line by one width. ⁵⁷Fe at 14.4 keV with τ = 141 ns: ER = 1.96 meV, Γ = 4.7 neV, v₁ = 0.097 mm s⁻¹.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.65 MeV
0.05 to 2 MeV; steps of 0.05.
137
20 to 240; steps of 1.

Try this. 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 point, already above the one-second line at about 6 s, climbs to about 4 years. Nothing about the nuclear force changed — only how much angular momentum the photon has to carry.

Live results

LOG₁₀ 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

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics IV modelWeisskopf half-life against transition energy, both on log scales, for the four lowest electric multipoles. The dashed horizontal line is one second; the dashed vertical line marks the chosen energy. At E = 0.65 MeV and A = 137, E1 gives 10^-15.0 s and E4 gives 10^0.8 s.E1E2E3E41 sWeisskopf half-life (log scale)transition energy 0.05 to 2 MeV (log scale)A = 137
Weisskopf half-life against transition energy, both on log scales, for the four lowest electric multipoles. The dashed horizontal line is one second; the dashed vertical line marks the chosen energy. At E = 0.65 MeV and A = 137, E1 gives 10⁻¹⁵ s and E4 gives 100.8 s.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Transition energy E: 0.65 MeV; Mass number A: 137

Baseline 0.65; choose 0.05 to 2, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

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 cannot emit a single photon, 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 = λγ(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 an 11% slip; for a heavily converted transition it is a factor of a hundred, and the Weisskopf comparison is meaningless.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

The 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

¹³⁷ᵐBa 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

The 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A nuclear level at 350 keV with Jπ = 4⁻ decays directly to the 2⁺ ground state. Which multipole assignment is right?
7

Review & continue

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.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.