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I am trying to understand gamma radiation and trying to figure out how to calculate radiative width. Is the radiative width how far the atom can be from another one and the probability of it then emitting a photon? How could I then calculate the radiative width for different states of an atom?

rob
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2 Answers2

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Radiative width is a terminology used in electromagnetic decays attributed to spectral lines .

For atomic states this is seen in the width of the lines and it has to do with the specific molecules in their specific locations as intermolecular fields change the boundary conditions for the wavefunction describing the state and thus change the widths.

For nuclear decays the gamma ray spectrum width are inferred from the probabilistic nature of quantum mechanics, the Heisenberg uncertainty principle, from the measured lifetimes. The widths vary with the nucleus under study and also the lattices containing the nuclei. The lifetimes are measured and recorded in tables, and from them the width of the energy level can be calculated. Lifetime to widths is discussed here.

The other answer has covered elementary particle decays.

anna v
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The "radiative width" seems to be a shorthand way to refer to the decay width --- that is, the Heisenberg uncertainty in the energy $\Gamma = \hbar / \Delta t$ for a bound state with finite mean life $\Delta t$ --- due to a decay involving photons.

Some (most?) particles can decay via several different modes. For example, the $K^+$ meson (up plus anti-strange quark) decays

  • 64% of the time to $\mu^+\nu_\mu$
  • 21% of the time to $\pi^+\pi^0$
  • 6% of the time to $\pi^+\pi^+\pi^0$
  • 5% of the time to $\pi^0 e^+\nu_e$
  • 0.6% of the time to $\mu^+ \nu_\mu \gamma$

I left some modes out. The fractions at the front are technically "partial widths" $\Gamma_i / \Gamma_\text{total}$. Decay widths are great because all the decay modes have the same decay width, so you can use the spread in reconstructed kaon masses in $K\to\pi\pi$ decays, plus knowledge of how many kaons you started with and how many two-pion events you caught, to determine that 79% of charged kaon decays are something other than $\pi\pi$.

The first decay on my abbreviated list is a purely leptonic process; the next two, with only pions, are purely hadronic processes; the fourth is a "semileptonic" process; the last is radiative. So you might encounter a sentence like

... the radiative decay width of the charged kaon is about 0.6% of its total ...

For atomic transitions essentially all excited states decay by photon emission, so I'd expect to hear about "decay width" rather than "radiative decay width." For nuclear excited states you might distinguish the radiative decay width of a state --- that is, its probability for emitting a photon --- from the decay width for nucleon emission, or beta or alpha decay, or fission.

rob
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