Elemental Puzzle

[lavalamp]

Originally Posted by 10metreh

Looking at the original puzzle, I think this is one hell of a

Is it possible to hijack ones own thread?

[cheesehead]

Originally Posted by lavalamp

Is it possible to hijack ones own thread?

From http://www.merriam-webster.com/dictionary/hijack we have "1 a : to steal by stopping a vehicle on the highway ... c : to stop and steal from (a vehicle in transit) ..."

So (assuming we can treat an active thread as a vehicle in transit on the mersenneforum.org highway) the question is:

Did davar55 stop his own thread and steal (the original topic, or something like that) from it?

I vote "yes". However, unless he were to file a false report that someone else did the hijacking or a fraudulent claim to some thread-insurance company, I see no prosecutable crime.

- - -

Besides, isn't the thread title "Elemental Puzzle" an acceptable category for an alternative cosmology?

From http://www.merriam-webster.com/dictionary/elemental we have "1... b (1) : of, relating to, or being the basic or essential constituent of something : fundamental ..."

There aren't many things more fundamental than cosmology ... and this is a puzzling one.

Shucks, I'd nominate davar55 for "Best Pun of the Week".

[davar55]

Originally Posted by cheesehead

From http://www.merriam-webster.com/dictionary/hijack we have "1 a : to steal by stopping a vehicle on the highway ... c : to stop and steal from (a vehicle in transit) ..."

So (assuming we can treat an active thread as a vehicle in transit on the mersenneforum.org highway) the question is:

Did davar55 stop his own thread and steal (the original topic, or something like that) from it?

I vote "yes". However, unless he were to file a false report that someone else did the hijacking or a fraudulent claim to some thread-insurance company, I see no prosecutable crime.

- - -

Besides, isn't the thread title "Elemental Puzzle" an acceptable category for an alternative cosmology?

From http://www.merriam-webster.com/dictionary/elemental we have "1... b (1) : of, relating to, or being the basic or essential constituent of something : fundamental ..."

There aren't many things more fundamental than cosmology ... and this is a puzzling one.

Shucks, I'd nominate davar55 for "Best Pun of the Week".

Well, I started with what I considered a Puzzle about five elements,
and when everyone starting contributing info about other elements,
I couldn't resist mentioning my theory about the finiteness of the
number of elements and that I thought I could show this limit to be 200.
The rest of this cosmological viewpoint - which I knew hard questions
would force me to defend and which I've barely started (and I admit
I may have to modify, though not yet) is not a thread hijacking, IMO,
because of Sherlock Holmes - it's elementary.

[CRGreathouse]

Originally Posted by cheesehead

Besides, isn't the thread title "Elemental Puzzle" an acceptable category for an alternative cosmology?

I was thinking "Miscellaneous Science", but whatever floats your boat.

[xilman]

Originally Posted by cheesehead

Then how does your theory explain the factual observation that when slow neutrons are falling in vacuo under the influence of gravity, they fall at only certain discrete rates

So far, so good.

Originally Posted by cheesehead

-- that is, that gravitational force has been physically demonstrated to occur only in discrete quanta, not continuously as Newtonian mechanics has it?

However, this conclusion is not the one I draw from the results of the experiment. Mt conclusion is that the energies of the neutrons in a gravitational potential well are quantised in accordance with theory.

The experiment says nothing about the quantum structure of the potential well itself. Exactly the same result would have been predicted if gravity is a quantum field or a classical continuous field.


Paul

[davieddy]

Originally Posted by xilman

So far, so good.
However, this conclusion is not the one I draw from the results of the experiment. Mt conclusion is that the energies of the neutrons in a gravitational potential well are quantised in accordance with theory.

The experiment says nothing about the quantum structure of the potential well itself. Exactly the same result would have been predicted if gravity is a quantum field or a classical continuous field.


Paul

Do you remember Joe Hatton from Exeter?

D.

[cheesehead]

Originally Posted by xilman

However, this conclusion is not the one I draw from the results of the experiment. Mt conclusion is that the energies of the neutrons in a gravitational potential well are quantised in accordance with theory.

Thanks! Upon careful rereading, I see that's indeed what the physicsworld.com article says.

The experiment says nothing about the quantum structure of the potential well itself. Exactly the same result would have been predicted if gravity is a quantum field or a classical continuous field.

Ah, I see now.

I actually hadn't seen the physicsworld.com article before, and was careless in reading it. The past articles I recalled had distinctly claimed that the "quantum of gravity", or some similar phrase referring to a quantum amount of force exerted by the gravitational field on the neutron, had been measured. Also, I'm distinctly rusty in this area and probably should stop doing interpretations from memory.

Or maybe it was just that James Bond film title. :-)

[davar55]

Currently, the first 120 elements (not all yet generated) are thought to
occupy parts or all of the first eight electron shells.

Element 118 is [6]7s2f14d10p6, completing the seventh shell.
Element 119 is [7]8s1, beginning the eighth shell with an s electron.
Element 120 is [7]8s2, completing the 8s sub-shell.

Some propose that the next few elements' electrons will occupy
the 8g sub-shell (of 18 electrons), e.g. element 121 = [7]8s2g1.

My theory that there are exactly 200 (possible) elements implies and
is based on the next electron instead occupying the 8f sub-shell,
there not being any g electrons until the ninth shell.

My question is: how would we know, when we generate unbiunium
whether the last electron is g or f? What do we already know about
s,p,d,f, and g electrons that would let us determine this?

Obviously, the fate of my theory rests on this answer.

[sichase]

Assuming that only a few atoms of it are produced, and that the lifetime of the nucleus is very short compared to the electromagnetic decay time of any electronic excited states, it will be just about impossible to tell, since the nucleus will come and go before a full set of electrons is captured. In other words, we will produce ions rather than atoms of any new element. So the only basis for knowledge about its electronic structure would be a theoretical calculation based upon a multi-electron relativistic Schrodinger equation for the atom.

The existence of a new element does not depend its electronic structure*, so any counting argument about electron shells cannot have a bearing on how many elements exist. It depends only on the orbital dynamics of the protons and neutrons in the nucleus, which determines the overall binding energy of a particular combination of them and whether they form a bound or unbound assembly. To make a model of how many elements exist, you should look at the nuclear shell structure, not the electronic one.

*It is theoretically possible for the electronic contribution to the total energy of an atom to make the difference as to whether the overall assembly is bound or unbound. But the odds that it will matter are astoundingly low since nuclear binding energies are on the order of MeV, while electronic binding energies are a million times smaller, on the order of eV. So you'd need to find a combination of protons and neutrons that are almost exactly unbound, with only a few eV of negative binding energy so that the addition of orbital electrons would tip the balance in favor of binding. If such a thing existed, then you could induce nuclear decay merely by ionizing the atom. No such case is known to exist.

[xilman]

Originally Posted by sichase

Assuming that only a few atoms of it are produced, and that the lifetime of the nucleus is very short compared to the electromagnetic decay time of any electronic excited states, it will be just about impossible to tell, since the nucleus will come and go before a full set of electrons is captured. In other words, we will produce ions rather than atoms of any new element. So the only basis for knowledge about its electronic structure would be a theoretical calculation based upon a multi-electron relativistic Schrodinger equation for the atom.

The existence of a new element does not depend its electronic structure*, so any counting argument about electron shells cannot have a bearing on how many elements exist. It depends only on the orbital dynamics of the protons and neutrons in the nucleus, which determines the overall binding energy of a particular combination of them and whether they form a bound or unbound assembly. To make a model of how many elements exist, you should look at the nuclear shell structure, not the electronic one.

*It is theoretically possible for the electronic contribution to the total energy of an atom to make the difference as to whether the overall assembly is bound or unbound. But the odds that it will matter are astoundingly low since nuclear binding energies are on the order of MeV, while electronic binding energies are a million times smaller, on the order of eV. So you'd need to find a combination of protons and neutrons that are almost exactly unbound, with only a few eV of negative binding energy so that the addition of orbital electrons would tip the balance in favor of binding. If such a thing existed, then you could induce nuclear decay merely by ionizing the atom. No such case is known to exist.

Good post!

AIUI, Davar has previously used the word "element" in a rather non-standard way. You and I use the conventional term, where it refers to the number of protons bound in the nucleus. He seems to think an element must also be stable when electrically neutral with a full complement of electrons.

Please be a little careful when saying that orbital electrons have "only a few eV of negative binding energy". That is true of electrons which experience only a small effective nuclear charge. So it's true for valence electrons but emphatically wrong for high-Z inner core electrons where binding energies can be many keV. It doesn't alter your conclusion, AFAIK, but I think it pays to state such things precisely.

(Added in edit: actually, K-capture is a well-known mechanism by which a nucleus is stabilized by extra-nuclear electrons. However, K-capture converts one element into another and the electron doesn't remain extra-nuclear. It's inverse, beta-decay, has analogous properties. I suspect that your hypothetical element would also be unstable with respect to K-capture.)

Paul

[davar55]

Thank you, very helpful and informative.

According to my theory, elements 116,118,120,121, and 122
each have a stable isotope when surrounded by a
complete complement of electrons. IF you peel away enough
of the outer electrons, their nucleii become unstable.
The stability of the nucleus IN ISOLATION is not the only
criterion that determines atomicity.

There is a complementarity between protons and electrons,
and it is the mass difference that induces the fact that the
protons are central AND their electric field property generates
the electron shell structure.

One shouldn't look only at the nucleus in trying to determine
what elements will be stable. The increasing number of f and
eventually g electrons will add stability to higher atomic numbers.

Or so I think.