# How can the nucleus of an atom be in an excited state?

An example of the nucleus of an atom being in an excited state is the Hoyle State, which was a theory devised by the Astronomer Fred Hoyle to help describe the vast quantities of carbon-12 present in the universe, which wouldn't be possible without said Hoyle State.

It's easy to visualise and comprehend the excited states of electrons, because they exist on discrete energy levels that orbit the nucleus, so one can easily see how an electron can excite from one energy level into a higher one, hence making it excited.

However, I can't see how the nucleus can get into an excited state, because clearly, they don't exist on energy levels that they can transfer between, but instead it's just a 'ball' of protons and neutrons.

So how can the nucleus of an atom be excited? What makes it excited?

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First you say

It's easy to visualise and comprehend the excited states of electrons, because they exist on discrete energy levels that orbit the nucleus

By way of preparation, I'll note that in introductory course work you never attempt to handle the multi-electron atom in detail. The reason is the complexity of the problem: the inter-electron effects (screening and so on) mean that it is not simple to describe the levels of a non-hydrogen-like atom. The complex spectra of higher Z atoms attest to this.

Later you say

[nuclei] don't exist on energy levels that they can transfer between

but the best models of the nucleus that we have (shell models) do have nucleons occupying discrete orbital states in the combined field of the all the other nucleons (and the mesons that act as the carriers of the "long-range" effective strong force).

This problem is still harder than that of the non-hydrogen-like atoms because there is no heavy, highly-charged nucleus to set the basic landscape on which the players dance, but it is computationally tractable in some cases.

See my answer to "What is an intuitive picture of the motion of nucleons?" for some experimental data exhibiting (in energy space) the shell structure of the protons in the carbon nucleus. In that image you will, however, notice the very large degree of overlap between the s- and p-shell distributions. That is different than what you see in atomic orbitals because the size of the nucleons is comparable to the range of the nuclear strong force.

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It is easy to think of the nucleus as a simple ball because it is so small (about 100,000 times smaller than an atom), but even at this level there is structure. While a different force mediates the interaction among nucleons (the parts of a nucleus), it is analogous to the interactions between nuclei and electrons that gives rise to the structure of atoms. These structures are a consequence of the quantum mechanical rules governing the interactions between particles and fields.

The nucleus of an atom is bound together through the strong nuclear force. This is one of the four fundamental forces, of which electromagnetism is a member. The many states of an given atom are governed by electromagnetic interactions between the electrons and the nucleus of an atom, with a ground state that represents the lowest possible energy configuration of the system, and excited states that are also allowable, but with higher energy values.

Similarly, there are many nuclear states for a given configuration of nuclei. Although mediated through a different fundamental force, there is still a ground state representing the lowest energy configuration of a particular collection of neutrons and protons, and there are many possible excited states as well. These excited nuclear states follow essentially the same rules that excited atomic states, except that the form of the potential term when writing down the Hamiltonian is different.

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Usually, it is after a nucleus has decayed via a $\alpha$- or $\beta$- decay that it is left in a excited state. To see this let us consider the decay of Co-60 into Ni-60.

Co-60 is an unstable nucleus because it has to many neutrons, and therefore it decays in the most probable channel by turning a neutron into proton according to the decay mode

$^{60}$Co$_{27}\rightarrow ^{60}$Ni$_{28} + e^-+\bar{\nu_e}$

or at the nucleon level

$p\rightarrow n+e^-+\bar{\nu_e}.$

The nucleus of Ni$-60$ finds itself in the new configuration of protons and neutrons which does not correspond to the lowest energy level, because the newly created neutron is at a higher energy level due to the decay. So at this stage Ni$-60$ is in an excited state Ni$^*-60$

From this point on the nucleus of Ni$^*_{60}$ will decay to attain its lower energy state. Imagine the similar situation in an atom, where an electron from an inner shell, $E_i$ say, is shot out by a laser beam. The remaining atom is in an excited state because there can be an electron in an outer shell, $E_o$ say, and it will therefore make a transition to the energy level that has just been evacuated.

So Ni$^*-60$ will decay in a cascade with 2$\gamma$ photon emission as shown in the following decays

$^{60}$Ni$^*_{28}\rightarrow$ $^{60}$Ni$^{\prime*}_{28} + \gamma_1$ still an excited state,

$^{60}$Ni$^{\prime*}_{28}\rightarrow$ $^{60}$Ni$_{28} + \gamma_2$ and this is the ground state of Ni-60.

where $E(\gamma_1)$=1.333 MeV and $E(\gamma_2)$=1.173 MeV.

The nucleus of an element can be left in an excited state after an $\alpha$-particle emission as well, and the reasons for the $\gamma$-particle decay are very much the same.

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It is also relatively easy to kick nuclei into excited states with a gentle nudge as in $n\text{(at least 4.5 MeV kinetic energy)} + ^{12}\mathrm{C} \to n + ^{12}\mathrm{C}^*$ followed by the decay of the excited carbon. –  dmckee Mar 3 at 19:42
@dmckee Absolutely. Excited nuclei (nuclear isomers) can be manufactured using the mechanism you are describing, as well as using synchrotron irradiation of the nucleus. A very interesting part of applied nuclear physics. –  JKL Mar 3 at 20:52