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I'm currently applying information theory to physics, and have begun considering the nature of isotopes, and in particular, how much information is contributed to the properties of an element by its subatomic components.

One initial conclusion I've reached is that there's a window of ionization for most materials within which the properties of the material don't change in any macroscopically meaningful way.

For example, if I rub a balloon on my sweater, I will change the local electrostatic charge of my sweater, but it won't change in appearance in any meaningful way. This is obviously also true of conductive wires, and everyday compounds and elements generally.

We can say, therefore, that electrons don't contribute much information to the properties of a typical substance, within some window of ionization.

My question is whether this is true of neutron isotopes, and in particular, Gold 197. I understand that this is a naturally occurring, stable isotope, and I'm wondering if it's any different than ordinary gold in terms of appearance, conductivity, etc.

As a general matter, it seems possible that neutron isotopes are actually not physically similar to their elements, since, for example, a given isotope might not be stable. This suggests that neutrons generally contribute more information to the behavior and properties of a substance than electrons.

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    $\begingroup$ "electrons don't contribute much information to the properties of a typical substance" makes no sense... Charges on everyday stuff are so small in comparison with with their respective amounts of atoms that from chemical point of view they're as good as non-existent. $\endgroup$
    – Mithoron
    Commented Nov 9, 2019 at 23:55
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    $\begingroup$ Given there is only one naturally occurring isotope of gold, you have picked an odd element to focus on. Au-197 IS naturally occurring gold. $\endgroup$
    – Jon Custer
    Commented Nov 10, 2019 at 0:10
  • $\begingroup$ @Mithoron - add electrons to a surface. Nothing about the surface really changes, other than its net charge. Change its proton number. You have a element. That is the different between contributing a lot of information and not. $\endgroup$
    – Feynmanfan85
    Commented Nov 10, 2019 at 1:56
  • $\begingroup$ @JonCuster Thanks, but that didn't answer my question. Is the neutron isotope any different than the element. $\endgroup$
    – Feynmanfan85
    Commented Nov 10, 2019 at 1:56
  • $\begingroup$ Well, you said "I'm wondering if it's any different than ordinary gold" - and it is ordinary gold. Your best bet for a difference is protium vs deuterium vs tritium. $\endgroup$
    – Jon Custer
    Commented Nov 10, 2019 at 3:18

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As a general matter, it seems possible that neutron isotopes are actually not physically similar to their elements, since, for example, a given isotope might not be stable.

It was very well known (pre-1940s) that the number of protons (Z) distinguish one element from the other. If we talk about the simplest element with Z=1, whether it contains 0, 1 or 2 neutrons it is still a colorless gas which burns in air and forms a colorless liquid- water. Yet some of the physical properties of that "water" with 0,1 or 2 neutrons change such as refractive index, boiling or melting point etc, their chemical properties remain almost the same. The chemical properties are determined by the number of valence electrons. Hence if bromine is red, no matter how many neutrons its isotope has- they will all appear red.

Your observations about the number of neutrons are a classical example of casual relationship "A causal relation between two events exists if the occurrence of the first causes the other. The first event is called the cause and the second event is called the effect. A correlation between two variables does not imply causation." https://www.varsitytutors.com/hotmath/hotmath_help/topics/correlation-and-causal-relation

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  • $\begingroup$ This is very helpful thank you. So proton number is obviously the most important, and determines the element. Then, neutron number can determine some characteristics such as refractive index, boiling point, etc., but do not change the element. Finally, electrons seem to not change much at all about a given element. $\endgroup$
    – Feynmanfan85
    Commented Nov 10, 2019 at 2:02
  • $\begingroup$ Well it is difficult to say neutron number determines the refractive index, boiling & melting point, we should rather say that it having an additional neutron for an element with Z protons "affects" some of the physical properties even reaction rates. The chemical properties of the elements with Z protons is determined by the distribution of Z electrons around the nucleus. $\endgroup$
    – ACR
    Commented Nov 10, 2019 at 2:14
  • $\begingroup$ That's an interesting point, so you're saying that the electrons affect the reactivity of certain elements. But for everyday materials, charge has absolutely not affect at all on a substance. Where do these electrons live in the material? For example, if I rub a balloon on my sweater, where are the electrons associated with that additional charge? $\endgroup$
    – Feynmanfan85
    Commented Nov 10, 2019 at 3:19
  • $\begingroup$ How many molecules are there in a drop of water? About 500000000000000000000 molecules. Now count the number of electrons (1 from hydrogen and 8 from O each). Remove a few electrons from this number. This would a highly charged droplet. What fraction of electrons have we removed? The key message is that we have to differentiate between bulk and molecular scale. The electrons on a balloon/sweater indeed come the atoms making up the polymer chains of your sweater and the polymer of your balloon. A physicist might be able to explain better as how dielectrics (=insulators) develop a charge. $\endgroup$
    – ACR
    Commented Nov 10, 2019 at 4:48
  • $\begingroup$ On the contrary if we had a single water molecule in gas phase and you removed an electron, it would be positively charged. It would behave just like a positively charged particle and easily manipulated in electric and magnetic fields. This type of ionization is routinely done in mass spectrometers. What is your background in science and what is meant by applying information theory to physics? $\endgroup$
    – ACR
    Commented Nov 10, 2019 at 5:11

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