Talk:Avogadro constant

The following statement is incorrect:

Such an atom consists of 6 protons, 6 neutrons and 6 electrons, and N_A is therefore equal to 12 grams divided by the sum of the masses of a proton, neutron and electron.

While an atom of Carbon-12 does include the various subatomic particles mentioned above, it also has some mass due to the nuclear binding energy. Therefore, the sum of the masses of 6 free protons, neutrons and electrons will be different from that of an atom of Carbon-12. -- User:Matt Stoker

Ah yes, that makes sense. I originally added the above sentence, and I was a little worried that the numbers didn't come out quite right... :-)

But now you've got me thinking. The example on mole unit adds molecular masses like this: atomic mass of carbon is 12, atomic mass of hydrogen is 1, so molecular mass of C₂H₆ is 2*12+6 = 30. This is incorrect, isn't it? It neglects the mass in the binding energy. AxelBoldt

Strictly speaking, you are correct that the chemical binding energy contributes mass. However, this mass is very small compared to the mass due to the nuclear binding energy and so can usually be neglected. For example, the chemical binding energy in a hydrogen molecule contributes a mass of ~2.4E-9 g/mol. By contrast the nuclear binding energy in Deuterium contributes a mass of ~0.0029 g/mol (over six orders of magnitude greater than the chemical binding energy). For the purpose of defining Avogadros Number to arbitrary precision, both nuclear binding energy and chemical binding energy must be accounted for and I would assume the definition refers to 0.012 kg of free Carbon-12 atoms (ie. no chemical bonding), since graphite or diamond would have a slightly different mass. However, for most other applications the chemical binding energy contribution to mass can be neglected, so the example under mole unit should be fine. --User:Matt Stoker

I see, thanks. This is interesting stuff, too interesting to be hidden away in Talk:. It would be nice if we had this information either in mole unit or Avogadro's number.

Also, am I correct in assuming that free protons have a higher mass than nuclear ones? AxelBoldt

This last question is not as simple to answer as it sounds, since there is no way to independently determine the mass of a proton bound into a nucleus. However, it is true that the atomic mass of an atom is always less than the sum for a corresponding number of free electrons, protons and neutrons. This mass deficiency corresponds to the nuclear binding energy (E=mc^2) of the atom and represents the amount of energy that would have to be added to break the atom into its component subatomic particles. If you plot the mass per nucleon (neutrons and protons) vs. number of nucleons for all of the isotopes, you will notice that it initially decreases, reaches a minimum at 56 nucleons, then gradually increases. (This is why fusion is possible for light elements and fission for heavy elements). --User:Matt Stoker

It's a Good Thing that such discussions take place and result in encyclopedic material. However, I think that the section Connection to the mass of protons and neutrons has become a little unwieldy (like its title). Perhaps we could shorten this to maybe two or three well chosen sentences?

—Herbee 01:04, 2004 Apr 3 (UTC)

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just a detail about the name: after the change, it could now seem that it was discovered by Amedeo Avogadro, while it is instead named this way only in honour of him (and after his death). Wasn'it more correct before?

Ok, I'll put something in to that effect. AxelBoldt

True or false??

True or false: it has been proven that Avogadro's number is not an integer. 66.245.8.219 23:16, 13 Sep 2004 (UTC)

Currently it isn't an integer, however it would be possible to get an integer if we redefined the kilogram. This might sound stupid, but in fact it would be pretty reasonable to redefine the kilogram in a way that would give you an integer.

Avogadro's number is essentially just a conversion factor between the microscopic mass system (atomic mass units or Daltons) and the kilogram system. The microscopic mass system is based on the mass of Carbon-12, while the kilogram system is currently based on the mass of a particular "standard" lump of metal in France. So naturally there's no clean integer conversion relationship between the two. However, it wouldn't be totally outlandish to redefine the kilogram in terms of some particular number of atoms, rather than an arbitrary lump. If the atoms picked were Carbon-12, then you would end up with an integer relationship (because of the nuclear binding mass/energy, it would have to be Carbon-12, see above).

Essentially, this discussion shows that whether or not Avogadro's number is an integer isn't a very interesting question; Avogadro's number is simply a conversion factor, not some fundamental physical value, so whatever number it happens to be is just not that significant. --Chinasaur 01:12, 14 Sep 2004 (UTC)

Amp

I thought remembered this. Is it woorth adding?

A coulomb is Avagadros Number of electrons (6.023x10**23). AMPERE - The unit used for measuring the quantity of an electrical current flow.

One ampere represent a flow of one coulomb per second.--Jirate 01:48, 14 Sep 2004 (UTC)

No, if you check the Coulomb article you'll see that's incorrect. --Chinasaur 02:38, 14 Sep 2004 (UTC)

but not according to * http://fig.cox.miami.edu/~lfarmer/BIL265/BIL2001/neuron/tsld010.htm it is. It also corrisponds with my memory though the SI defintion of an amp invloves wires of negligable xsection and infinite length in vacuum.

No, the parenthetical statement on that slide is definitely wrong or confused. The rest of the slide is right though: F is equal to elementary charge (charge on one electron) times Avogadro's number. This is probably what you are thinking of. As you can see, F is not 1C, but rather ~96500C. --Chinasaur 01:07, 16 Sep 2004 (UTC)

I 'll go for that but their are several places on the WWW that are wrong. Not just my memory.--Jirate 12:39, 16 Sep 2004 (UTC)

A's # is just a convertion factor then?

In chemistry class, we always looked upon Avogadro's number as an almost magical physical universal cosmological constant. Yet I think this type of thinking could lead to a severe misunderstanding of subject (I didn't argue with the professor because I wanted good grades).

Isn't the number simply the number of amu's in a gram? That means that it is no more than a convertion factor, like 2.45 cm = 2 inch, or 1 pound = 4.45 newtons. There is nothing "special" about the gram; it's just an arbitrary unit, as is the amu (albeit less arbitrary).

So then, if you consider avogadro's number a "physical constant," that you might as well consider 2.45 and 4.45 "physical constants" too. The magical avogradro's number is in no way even close to being in the league of true constants, such as the gravitational constant, or the cosmological constant, or planck's constant, or the fine structure constant etc.

This artical does not reflect that. It seemes we are elevation this number to a place it does not belong. GWC Autumn 57 2004 13.10 EST

That's right, Avogadro's number is just as "thingys in an arbitary wossname" type of number, caused by the fact that the original definition of the kilogram had something to do with the density of water. There's no real reason why the mass unit should not be defined in some other way involving the other fundamental units and dimensionless constants: in fact, there are a number of efforts to do exactly that. -- Anon 18:36, 17 Nov 2004 (UTC)

You will find this discussed fairly exhaustively under the header immediately above. Sheesh; I'm starting to get the impression this misconception should be addressed in the article proper. --Chinasaur