Why can we not produce a "giant" nucleus?

[fxdung]

In Condensed Matter Physics there are "giant" molecules that are macro bodies(e.g crystals).But why in Nuclear Physics we can not produce a "giant" nucleus?

 

[A. Neumaier]

In Condensed Matter Physics there are "giant" molecules that are macro bodies(e.g crystals).But why in Nuclear Physics we can not produce a "giant" nucleus?

They exist and are called neutron stars.

 

[fxdung]

But why in the Earth we only see a finite number of neutrons in atoms?What is the condition to exist a "giant" atom?

 

[hilbert2]

There needs to be a binding force that cancels the Coulomb repulsion between protons. In small nuclei it is the strong nuclear force and in neutron stars it's gravitation.

 

[fxdung]

Why there are not too many neutrons in small nuclei?

 

[PeterDonis]

There needs to be a binding force that cancels the Coulomb repulsion between protons. In small nuclei it is the strong nuclear force and in neutron stars it's gravitation.

Neutron stars have no protons so there is no Coulomb repulsion to cancel. However, Coulomb repulsion is not the only effect involved. There is also degeneracy pressure, which is what balances gravity in a neutron star.

 

[PeterDonis]

Why there are not too many neutrons in small nuclei?

The strong nuclear force is short range, so there is a limit to how large a nucleus can be if it is held together by the strong force.

 

[fxdung]

The vinicity neutrons have strong force,why is there a limit to how large nucleus can be(despite of short range force)?

 

[hilbert2]

Neutron stars have no protons so there is no Coulomb repulsion to cancel. However, Coulomb repulsion is not the only effect involved. There is also degeneracy pressure, which is what balances gravity in a neutron star.

But isn't the coulomb force the first thing that has to be overcome when squeezing ordinary matter to neutron matter?

 

[A. Neumaier]

But isn't the coulomb force the first thing that has to be overcome when squeezing ordinary matter to neutron matter?

The Coulomb force only acts between two charged particles.

It is the van der Waals force between neutral matter. This is still more long range (namely $O(r^{-6})$) than the exponentially decaying strong force. [Correction: It is the repulsive exchange force, see post #31.]

 

[Dale]

The vinicity neutrons have strong force,why is there a limit to how large nucleus can be(despite of short range force)?

That was already answered: The range of the strong nuclear force is short and Coulomb repulsion (Which has infinite range) is strong.

 

[fxdung]

What is the degeneracy pressure?

 

[A. Neumaier]

That was already answered: The range of the strong nuclear force is short and Coulomb repulsion (Which has infinite range) is strong.

This is not sufficient to answer the query. It only explains why nuclei with many protons cannot exist. But why adding many neutrons to a stable nucleus makes it unstable needs an additional explanation - beta decay. Wikipedia has a good synopsis.

 

[fxdung]

What is the condition for beta decay easily happens?

 

[Dale]

But why adding many neutrons to a stable nucleus makes it unstable needs an additional explanation - beta decay.

That is true, both the strong force and the weak force are important.

 

[A. Neumaier]

What is the condition for beta decay easily happens?

A nucleus with p protons and n neutrons will beta decay if and only if the binding energy of a nucleus with p+1 protons and n-1 neutrons is smaller by more than the mass of an electron.

 

[fxdung]

And with neutron star this condition does not happen,does it?

 

[A. Neumaier]

And with neutron star this condition does not happen,does it?

...because gravitation changes the binding energy balance. (For terrestrial nuclei gravitation is negligible.)

 

[fxdung]

With too many neutron terrestrial nuclei, why is there not the process that neutrons detach the nuclei, but it must there be beta decay?

 

[Dale]

why is there not the process that neutrons detach the nuclei, but it must there be beta decay?

There is alpha decay where two neutrons and two protons detach. There can also be neutrons emitted. All that matters is that the sum of the masses of the products be less than the original nucleus

 

[ZapperZ]

In Condensed Matter Physics there are "giant" molecules that are macro bodies(e.g crystals).But why in Nuclear Physics we can not produce a "giant" nucleus?

This is comparing apples to oranges. A solid state crystal is nowhere near similar to a nucleus. You might as well claim that a cow looks like a Frank Gehry building.

A crystal depends on the bonding state that can be formed between atoms, or more specifically, the nature of the valence shell that the atoms have and how they are able to form bonds with one another. No such thing is relevant in the nucleus because you have a very short range and asymptotic nature of the nuclear forces. There are no valence shell to "tie" one nucleon to the other.

Zz.

 

[Dale]

You might as well claim that a cow looks like a Frank Gehry building.

That is a defensible claim! Although I usually think they look more like pasta or butter, but something food-like.

 

[Vanadium 50]

I think that was an SAT question. "Crystal is to nucleus as cow is to ______."

 

[hilbert2]

Even a hypothetical piece of matter composed of only protons would probably keep at constant volume if compressed below its Schwarzschild radius. But it would produce a huge electric field around it and polarize matter at a very large distance.

 

[PeterDonis]

Even a hypothetical piece of matter composed of only protons would probably keep at constant volume if compressed below its Schwarzschild radius.

No, it would collapse. A static object made of ordinary matter can't remain static at a radius below 9/8 of the Schwarzschild radius for its mass.

 

[PeterDonis]

What is the degeneracy pressure?

https://en.wikipedia.org/wiki/Degenerate_matter#Neutron_degeneracy

 

[hilbert2]

No, it would collapse. A static object made of ordinary matter can't remain static at a radius below 9/8 of the Schwarzschild radius for its mass.

That was what I meant, it would stay confined in the form of a black hole, with the electric field produced by the protons probably extending outside that BH.

 

[mfb]

Neutron stars have no protons so there is no Coulomb repulsion to cancel. However, Coulomb repulsion is not the only effect involved. There is also degeneracy pressure, which is what balances gravity in a neutron star.

Neutron stars have protons - just not as many as neutrons. The interior has many neutrons and a few protons and electrons. The outer parts have nuclei and even regular atoms.

 

[PeterDonis]

Neutron stars have protons

But few enough that their Coulomb repulsion is negligible compared to the neutron degeneracy pressure. Also, the protons are balanced by electrons, which cancels at least part of the Coulomb repulsion anyway.

The outer parts have nuclei and even regular atoms.

But again, the effects of these are negligible compared to neutron degeneracy pressure.

 

[TeethWhitener]

It is the van der Waals force between neutral matter. This is still more long range (namely $O(r^{-6})$) than the exponentially decaying strong force.

The $O(r^{-6})$ part of the van der Waals potential is the attractive portion of it (induced dipole-induced dipole). The repulsive portion is far more complicated, but an exponential works pretty well as an approximation.

 

[A. Neumaier]

The $O(r^{-6})$ part of the van der Waals potential is the attractive portion of it (induced dipole-induced dipole). The repulsive portion is far more complicated, but an exponential works pretty well as an approximation.

The repulsive force between neutral atoms is the exchange force due to the Pauli principle. It grows for $r\to 0$ only like $r^{-1}$. An exponential is a traditional but incorrect approximation.

But I just realized that all this onöy applies to atoms and not to nuclei. It is still relevant for the formation of neutron stars since gravitation must overcome the repulsive force between atoms...

 

[Xforce]

But why in the Earth we only see a finite number of neutrons in atoms?What is the condition to exist a "giant" atom?

Actually, when gravity is holding them together instead of strong nuclear force (I’m talking about neutron stars) that’s the condition of having a giant atom... because gravity decays much slower than nuclear forces over distances. However, try not to make it too big because it will collapse and becomes a black hole.

 

[fxdung]

What is an exchange force as Prf Neumaier said?

 

[Vanadium 50]

https://en.wikipedia.org/wiki/Exchange_interaction

 

[Nik_2213]

Reminds me of an article on the Russian 'heavy isotope' factory, and their extraordinary work to push further and further beyond known nuclei, into 'super-heavy country'. Part of the problem is the lower periodic table's 'magic numbers' no longer hold. To put it politely, the nucleus becomes too big for neat 'shells' to hold sway. The hoped-for 'Island of Stability' may prove nothing more than a 'shoal'...

IMHO, chemical analogy may be gold, mercury and lead, whose nuclei are so massive they significantly alter the behaviour of their inner electron 'shells', the relativistic effects causing remarkable changes in properties. Metals, yes, but anomalous colour, melting point etc. IIRC, copper almost qualifies...

And, I learned about the 'drip line', beyond which lesser nuclei shed surplus neutrons they cannot stabilise...