Why an excited nucleus emits gamma rays

Problem

Why does an excited nucleus, after an $\alpha$ or $\beta$ decay, tend to emit $\gamma$ photons? Compare this with the atomic case, where electronic transitions produce visible or ultraviolet photons.

On the left, nuclear de-excitation (a $\gamma$ photon, MeV energies); on the right, an atomic electronic transition (a visible/UV photon, eV energies).

Solution

After an $\alpha$ or $\beta$ decay, the daughter nucleus is almost never found directly in its lowest-energy state: it remains in an excited state. The nucleus’s energy levels are quantised, exactly like those of the electrons in an atom, but they are much more widely spaced in energy.

Dropping from the excited state to the ground state, the nucleus sheds the excess energy by emitting a photon, whose energy equals the difference between the two levels.

The crucial difference is the energy scale. Atomic electronic transitions involve energies of the order of an electronvolt (eV), and the emitted photons fall in the visible or ultraviolet range. Nuclear transitions, by contrast, involve energies of the order of MeV (a million times greater): the emitted photon is so energetic that it is a $\gamma$ ray, not visible light.

$\ev{\text{nuclear de-excitation, }\Delta E\sim 0{,}1\text{–}10\text{ MeV}\Rightarrow\gamma}$

Collegamenti

Argomenti: Fisica nucleare Concetti: Decadimento radioattivo