Do nuclei emit photons?

This answer addresses the title rather than the comment text, but all nuclei, regardless of state, emit photons, as they contain charged particles, which create electric fields, and (sometimes harder to measure) magnetic fields.

Electromagnetic interactions are mediated by photons, so anything that contains charged particles will emit photons.

The simple answer is yes. Here is a real life example where this is important.

To produce 'signal', the MRI scanner interacts with protons in the body. Randomly orientated protons become aligned with the powerful magnetic field in the bore of the scanner. A rapidly repeating sequence of radiofrequency pulses – produced by the scanner – then causes 'excitation' and 'resonance' of protons. As each radiofrequency pulse is removed, the protons ‘relax’ to realign with the magnetic field, and as they do so they give off radiofrequency 'signal' which is detected by the scanner and transformed into an image.

Source: https://www.radiologymasterclass.co.uk/tutorials/mri/mri_signal

It may be interesting as well to mention as an additional layer of information that quarks inside the proton also excite similar to an electron. They don't carry the same SPDF class, but instead form [Nucleon] Resonances which are excited states of protons/neutrons. This was the basis behind my PhD work. In this case they will primarily emit particles instead of photons (so pions, Kaons, new nucleons, etc)

It is good to say that nuclei emit not just photons, but also in some cases, their pretty close relatives - W bosons. That's how beta-decay happen (W boson is quick to decay into electron and neutrino).

Yes, many experiments of the form X(n,n'g)X are performed as a way to measure the energy levels of a nucleus. An incident neutron scatters off the target nucleus and if the incident neutron energy is high enough energy it will leave the target nucleus in an excited state. These states will often decay by emmision of a gamma-ray. The energy level of a nucleus can be determined by measuring the gamma-ray energies.

When nuclei are in an excited state and decay, they often release photons (almost always in the gamma/X-ray energies). A good example of this is the decay of metastable nuclear isomers.

For instance, the decay of Hafnium-178m2 (the 'm' represents a nuclear isomer of Hafnium-178) yields a photon of 2.5 MeV. There are attempts to use this isomer as a weapon. One of the lowest energy nuclear isomers is Thorium-229m, which decays to Thorium-229, releasing a photon with only around 7.6 eV(far ultraviolet!)

Another interesting nuclear isomer is the exceedingly long-lived Tantalum-180m. It is a primordial nuclei (i.e. if you get a sample of Tantalum, it will contain some Tantalum-180m), and has a half-life of over $10^{15}$ years. One decay mode is to drop to Tantalum-180, releasing a 77 keV photon. What is funny about this isomer is that once it drops to the ground state (Tantalum-180), the ground state nucleus is less stable than the excited state isomer, with a 8.1 hour half-life!

Yes, excited nuclei emit photons in the form of the highly energetic $gamma$-rays.

That these emissions are much more energetic than the VIS photons emanating from excited electron clouds can be somewhat heuristically understood by looking at the energy spectrum of a particle in a closed $text{1D}$ box, of length $L$:

$$E_n=frac{n^2h^2}{8mL^2}$$

So the emission energies are inversely proportional to system size:

$$Delta E propto frac{1}{L^2}$$

It follows that smaller systems, like nuclei, will emit much more powerful photons than atoms (electron clouds).

The energy content of a VIS photon is about $2text{ }mathrm{eV}$, that of a mid-range $gamma$-photon about $1.2times 10^6text{ }mathrm{eV}$.

Of course neither system above are $text{1D}$ boxes and in reality it is just a useful analogy.

Nuclei emit gamma rays, which are high energy photons. The photons emitted when electron in an atom changes its energy state are usually in optical spectrum, which are more frequently encountered in technology and real life, which is why they receive more attention in textbooks.

Remarks
To underscore why photons are less "important" for nuclei than for atoms:

• Nature of interactions It is worth noting that atoms, solids and molecules are held together by the Coulomb interaction (i.e. by electromagnetic forces), which is why their structural dynamics is strongly coupled to photons - the particles carrying this interaction. Nuclear forces are of different nature - although photons play a role, they are but one of many particles involved.
• Size Being charged particles protons should be coupled to EM field. The strength of this interaction is proportional to the dipole moment $d=r_ne$, where $r_napprox10^{-15}m$ (one Fermi) is the nuclear radius that is much smaller than the radius of an atom $r_aapprox10^{-10}m$ (one Angstrom). In other words, the coupling of protons to photon field is $10^{5}$ times weaker.
• Mass protons and neutrons both carry spin and could couple to electromagnetic field via Zeeman coupling. However, their mass is about thousand times bigger than that of electrons, resulting in a thousand times smaller gyromagnetic ratio (i.e. nuclear magneton is a thousand time smaller than Bohr magneton), i.e. the coupling is weak.
• Finally, here is an authoritative reference on the subject: Interaction of nuclei with electromagnetic radiation

Quote
The following quote is from book "Fundamentals in nuclear physics" by Besdevant, Rich and Spiro:

While the numbers (A, Z) or (N, Z) define a nuclear species, they do not 
determine uniquely the nuclear quantum state. With few exceptions, a nucleus 
(A, Z) possesses a rich spectrum of excited states which can decay to the ground 
state of (A, Z) by emitting photons. The emitted photons are often called 
gamma-rays. The excitation energies are generally in the MeV range and their 
lifetimes are generally in the range of 10^{−9}–10^{−15} s. Because of their 
high energies and short lifetimes, the excited states are very rarely seen on Earth 
and, when there is no ambiguity, we denote by (A, Z) the ground state of the 
corresponding nucleus.