Can a large nucleus be made to spin fast enough to fall apart?

Question

Is it possible to make an atomic nucleus spin so rapidly that it ejects neutrons or protons because the force holding neutrons and protons together is no longer strong enough?

Answer 1

I don't think this question is answerable because when you speak of "a nucleus", it is a specific state.

First, we have $Z$ and $A$ (or $N=A-Z$). That is, the number of protons($Z$) and the number of neutrons ($N$).

The only reason we talk about isotopes is the fact that atomic properties are dominated by $Z$ and almost independent from $N$. That's a fact in the laboratory, but as far as nuclear physics is concerned: it really doesn't matter.

That's why we have mirror nuclei, such as $^3{\rm He}$ and $T= $$^3{\rm H}$, which are almost isospin symmetric.

Ok. Then you get into isomers. These are states that are not ground states. For example: $^{180 \rm m}{\rm Ta}$. It has the same neutron and proton content as tantalum-180, but is in a nuclear state that is higher energy. It decays with a half life ( which is longer than $^{180}{\rm Ta}$'s half life), to hafnium or tungsten.

OK, so what is the point? When you ask "if a nucleus can be made to spin fast enough [to disintegrate]", the question is not proper. A spinning nucleus is a different nucleus, so the question really should be:

"Is there a nuclear isomer with a spin so high that it doesn't exist?"

The answer is logically contradictory? If it exists does it not exist?

Also: on of the comments talk quantum, and it reminds me of a question of spinning an alpha particle.

If a particle is spin-0: you cannot spin it. A rotation operator is the same as multiplying by one. The question then becomes: is there a state with 2n2p that has angular momentum? idk.

Same for bigger nuclei: if there is state with a large orbital angular momentum, the only question is how unstable is it?

Answer 2

We want to be careful about what we mean by a phrase like "the nucleus spins fast."

Nuclear spin, like all quantum-mechanical spins, is a complicated phenomenon which can't always be interpreted in an intuitive, classical way. Remember that the intrinsic half-$\hbar$ angular momentum of a structureless particle like an electron can't be interpreted as the spinning of a little ball — not least because "a little ball" isn't structureless.

Nucleons show evidence of internal structure in high-energy interactions, which we describe as "the quark model." But at the energies where nuclei decay by nucleon or alpha emission, the quark degrees of freedom are completely frozen out, and the nucleon's intrinsic half-integer spin doesn't have any classical interpretation. When nucleons are bound into a nucleus, as when electrons are bound into an atom, their collective structure becomes nontrivial. There are some electronic states whose angular momentum can be parsimoniously described as electrons occupying a particular angular-momentum orbital around an inert electron core; we use this model when we talk about lithium or sodium as having "one valence electron." But there are other states where such an interpretation would force a person to say that the orbits of several electrons are highly entangled. In nuclear systems, where the spin-orbit coupling is larger, models where a single nucleon is excited relative to some core are less successful than such models are for electrons.

When we talk about "a rapidly-spinning nucleus," we have the image of collective motion. Collective motion would be a special case of highly-entangled motion, where every nucleon in the nucleus is rotating about some common axis with the same frequency. There is actually evidence for collective rotation in nuclei with mass number $A\gtrsim 10$, especially when those nuclei are produced in collision experiments. Think of a car crash where the collision is not quite head-on, and imagine the cars spinning out, suggested my first nuclear physics mentor; then remember that glancing collisions are much more probable than head-on collisions.

Consider the rotation of a classical object with moment of inertia $I$ and angular momentum $\vec L = I\vec \omega$, where the kinetic energy goes like $$ T = \frac12 I\vec\omega^2 = \frac1{2I}\vec L^2. $$ The analogous quantum operator $$ \hat T = \frac{1}{2I}\hat J^2 $$ would have eigenvalues like $\frac{\hbar^2}{2I}J(J+1)$. A quantum-mechanical "rigid rotor" with energy-independent moment of inertia $I$ would have an energy spectrum of several states with this $E_J\propto J(J+1)$ spacing, where the states tend to decay by emitting electric quadrupole $(E2)$ photons. There are a number of such mini-spectra in heavy nuclei, known as "rotational bands." There are even several nuclides where, when created at high angular momentum, the nucleus will emit a series of quadrupole photons corresponding to a spectrum $E \sim \frac1{2I_1}J(J+1) + E_1$ towards a $J\to0$ state, then undergo a transition to another high-$J$ state and another set of quadrupole photons corresponding to a $E\sim\frac{1}{2I_2} J(J+1) + E_2$ spectrum with a different moment of inertia. This can be interpreted as a rigid rotor spinning down, then undergoing some internal transition to a different rigid shape with the same total angular momentum, then rigidly spinning down some more. No discussion of changing moment of inertia $I$ is complete without inviting the reader to imagine a figure skater tucking in her arms to increase her rotational frequency without changing her angular momentum; a nuclear-specific search term is "yrast".

You asked,

Is it possible to make an atomic nucleus spin so rapidly that it ejects neutrons or protons because the force holding neutrons and protons together is no longer strong enough?

The Wikipedia article on yrast states suggests that this may in fact be the case:

In some instances, however, predictions based on [models where emission of strongly-interacting nucleons is preferred] underestimate the total amount of energy released in the form of gamma rays; that is, nuclei appear to have more than enough energy to eject neutrons, but decay by gamma emission instead.

That is, a nucleus with a lot of angular momentum might decay by the spindown $\to$ shape change $\to$ spindown $\to$ shape change $\to$ spindown process described above. But it might also decay by leaving the rigid rotor spectrum for a state which is better-described by a single nucleon orbiting a many-nucleon core, followed by a state described by that core plus a free nucleon. This is exactly the scenario you ask about, where a collective rotation results in the ejection of a single nucleon or an alpha particle; however in the collective-motion states the collective de-excitation is apparently preferred.

My undergraduate internship on highly-deformed high-spin rigid-rotor nuclear states was in a different century, so please have patience with me if I've mis-summarized anything.