Gamma decay

Gamma decay is a decay mode in which an excited nucleon drops to a less energetic state by emitting a Photon. It is essentially just a State transition of a nucleon from a higher energy eigenstate to a lower one. As such, the energy of the photon equals the difference in energy between states, however it is reduced by a small nucleus recoil term. It often follows alpha or beta decay, which generally leave one or more nucleons in an excited state.

The nuclear process is

where $\ce{X}$ is a nuclide and the asterisks denotes that it is in an excited state. $\gamma$ represents a photon.

Photons emitted by gamma decay are called gamma rays. Their energy is extremely high, often in the order of hundreds of keV to several MeV, and they comprise the most energetic part of the electromagnetic spectrum. More generally, "gamma ray" refer to any photon in this range of energy. The majority of gamma rays observed on Earth are due to gamma decay, cosmic rays or artificially generated due to particle accelerator collisions.

Like all state transitions that emit a photon, gamma decay is governed by electromagnetism.

Energy conservation#

Consider some nucleus of mass $M$ at rest. It gamma decays from an initial energy eigenstate $E_{i}$ to $E_{f}$. Most of this energy is emitted through a photon, which will carry most of the energy of the decay. We'll call this $E_{\gamma}$. However, since the nucleus is not being held in place, it will suffer some recoil, so the total decay energy is split between the photon and the nuclear recoil because momentum must be conserved. Since the nucleus is quite heavy, it's fair to assume that the recoil's kinetic energy is not relativistic, so we'll call it

The conservation of energy and momentum give

The bottom simply solves to $\mathbf p_{R}=-\mathbf p_{\gamma}$. In the top equation, we can define $\Delta E\equiv E_{i}-E_{f}$ and use the photon energy $E_{\gamma}=cp_{\gamma}$¹ to state

This is a quadratic equation in $E_{\gamma}$. Using the quadratic formula gives

Now, we know that $\Delta E$ is in the MeV range whereas the nuclear rest energy $Mc^{2}$ is in the GeV range, so we can afford to consider $\Delta E/Mc^{2}$ to be quite small. Approximate the root with the binomial expansion to first order $\sqrt{ 1+x }\simeq 1+x/2$:

so

$\Delta E$ is positive by definition (since in gamma decay $E_{f}<E_{i}$), the minus case would lead to a guaranteed negative photon energy, which makes no sense. Thus, only the positive case is physically valid and we are left with

This proves that the photon carries nearly all of the decay energy. If we held on to the second term of the square root

We'd have gotten

The second term is the recoil correction:

This is tiny: usually in the order somewhere between and 1 and 100 electronvolts. Nevertheless, it's not insignificant. On the higher end, $100\text{ eV}$ recoils can displace an atom is a solid state crystal which can break things at an atomic level. This is commonly known as radiation damage.

Selection rules#

Gamma decay results in the emission of electromagnetic radiation from a nucleus. We can describe this radiation as a superposition of known radiation shapes, namely regarding the polarity of the radiation. We know that radiation from an arbitrary source can be described as a multipole expansion containing a monopole term, a dipole term, a quadrupole term, etc. Each of these has a characteristic angular distribution, which we can use as a Basis to express any form of radiation.

Electric radiation arises from the variation of the charge distribution inside the nucleus. Magnetic radiation instead arises from the variation of nuclear magnetic moment, be it orbital angular momentum or Spin. Within the nucleus, angular momentum (and by extension spin and hence parity) must be conserved. This conservation requirement forbids some transitions and allows some others: we call selection rules the set of rules we use to determine which transitions are allowed.

Consider some gamma decay transition that takes a nucleus from $J_{i}^{P_{i}}$ to $J_{f}^{P_{f}}$. From conservation of angular momentum I have

since the photon being emitted may very well carry away some of the angular momentum of the nucleus, written $\mathbf{L}_{\gamma}$. The angular momentum depends on the transition. If $L$ is the order of the multipole operator ($L=1$ is dipole, $L=2$ is quadrupole, etc.), the photon is passed $L\hbar$ angular momentum. $L$ may not be zero because a pure monopole (resting point charge) cannot emit radiation.

The values of $L$ give the set of multipole terms that describe the radiation. According to conservation of momentum, the allowed values of $L$ satisfy

For instance, if we start from $J_{i}=3/2$ and end up in $J_{f}=5/2$, the possible transitions are $L=1,2,3,4$. This means that the actual radiation being emitted is some Linear combination of these four terms (dipole, quadrupole, octopole, hexadecapole).

The parity difference instead tells us whether each pole is electric or magnetic. Electric transitions produce photons with parity given by $P_{\gamma}(\text{EL})=(-1)^{L}$ (same parity as $L$) and magnetic transitions have $P_{\gamma}(\text{ML})=(-1)^{L+1}$ (opposite of $L$)². The total parity must be conserved, so

The transition parity supplies the difference between the start and end states, so if nothing changes, you must only pick even-parity transitions ($P_{\gamma}=1$), and if the parity flips, you must only pick odd-parity transitions ($P_{\gamma}=-1$). This results in odd poles being magnetic and even ones being electric if $P_{i}=P_{f}$ and vice versa if $P_{i}\neq P_{f}$. In the previous example, if $P_{i}=P_{f}$, then our $L=1,2,3,4$ transitions become $M_{1},E_{2},M_{3},E_{4}$, but if $P_{i}\neq P_{f}$ they become $E_{1},M_{2},E_{3},M_{4}$.

So to summarize, the selection rules are

1. Gamma radiation is the linear combination of all multipole terms $L$ allowed by $\lvert J_{i}-J_{f} \rvert\leq L\leq J_{i}+J_{f}$, excluding $L=0$.
2. If parity doesn't change, odd poles are electric and even ones are magnetic. If parity changes, odd poles are magnetic and even ones are electric.

The excited state is $J^{P}_{i}=\frac{11}{2}^{-}$. The ground state is $J^{P}_{f}=\frac{3}{2}^{+}$. The angular momentum difference determines that the allowed poles are $L=4,5,6,7$. The parity flip then tells us that odds are magnetic and evens are electric. Thus, the gamma radiation produced by this decay is a linear combination of $M_{4},E_{5},M_{6}$ and $E_{7}$.

There's an edge case in the rules: if $J_{i}=J_{f}$, then $L=0$ only. But that's not permitted. In such transitions (which are observed), $L=1$.

Moreover, not all poles are made equal. The behavior of the poles can be succinctly summarized in a few statements

1. The lowest-pole radiation dominates.
2. Electric multipole emission is about 100 times more probable than the same-order magnetic.
3. Emission of order $L+1$ is $\sim 10^{-5}$ times less likely than the previous order $L$.

Points 2 and 3 can be combined to state:

and

To be clear, this are very rough order-of-magnitude estimates. The probabilities here essentially express the