Is it true that the nucleus of all atoms (including radioactive isotopes) contain at least one proton? Is there an atomic nucleus consisting entirely of neutrons? (Let's exclude neutron stars for the moment.) If so, how does one name them? (since Periodic Table starts from atomic number 1, not 0.)
Nuclear Physics – Can an Atomic Nucleus Consist Only of Neutrons?
atomsnuclear-physics
Related Solutions
This is really a comment, since I don't think there is an answer to your question, but it got a bit long to put in as a comment.
If you Google for "Why is technetium unstable" you'll find the question has been asked many times in different forums, but I've never seen a satisfactory answer. The problem is that nuclear structure is much more complex than electronic structure and there are few simple rules.
Actually the question isn't really "why is technetium unstable", but rather "why is technetium less stable than molybdenum and ruthenium", those being the major decay products. Presumably given enough computer time you could calculate the energies of these three nuclei, though whether that would really answer the "why" question is debatable.
Response to comment:
The two common (relatively) simple models of the nucleus are the liquid drop and the shell models. There is a reasonably basic description of the shell model here, and of the liquid drop model here (there's no special significance to this site other than after much Googling it seemed to give the best descriptions).
However if you look at the sction of this web site on beta decay, at the end of paragraph 14.19.2 you'll find the statement:
Because the theoretical stable line slopes towards the right in figure 14.49, only one of the two odd-even isotopes next to technetium-98 should be unstable, and the same for the ones next to promethium-146. However, the energy liberated in the decay of these odd-even nuclei is only a few hundred keV in each case, far below the level for which the von Weizsäcker formula is anywhere meaningful. For technetium and promethium, neither neighboring isotope is stable. This is a qualitative failure of the von Weizsäcker model. But it is rare; it happens only for these two out of the lowest 82 elements.
So these models fail to explain why no isotopes of Tc are stable, even though they generally work pretty well. This just shows how hard the problem is.
- This is could be considered chemistry question and might receive a more interesting response on Chem.SE. But it's probably fine here.
- According to ChemWiki at UC Davis, "Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are added to the same shell. However, at the same time, protons are being added to the nucleus, making it more positively charged. The effect of increasing proton number is greater than that of the increasing electron number; therefore, there is a greater nuclear attraction. This means that the nucleus attracts the electrons more strongly, pulling the atom's shell closer to the nucleus. The valence electrons are held closer towards the nucleus of the atom. As a result, the atomic radius decreases.
- Nuclear charge increases due to protons NOT electrons. Greater number of electrons doesn't increase the strength of the nucleus. There are no electrons in the nucleus! As you move across the periodic table, the number of protons increases, increasing the charge of the nucleus by $+1$ for each proton added. For atoms with neutral charge, this implies same number of protons and electrons. And since the periodic table lists atoms in their neutral state, this has the effect that the number of electrons happens to correlate with the number of protons and thus the charge of the nucleus. But if I remove two electrons from a calcium atom, I still have a calcium atom because the nucleus is the same. Also, just to be thorough, two atoms can have the same number of protons and electrons and still be different in the number of neutrons. They would still be the same element, but we call them different "isotopes."
- There's a difference between nuclear charge and effective nuclear charge. In a hydrogen atom, the electron experiences the full charge of the positive nucleus, which is just a proton. A hydrogen atom is thus like two point charges, so the effective nuclear charge can be calculated from Coulomb's law. However, in an atom with many electrons the outer electrons are simultaneously attracted to the positive nucleus and repelled by the negatively charged electrons. Each electron (in the n-shell) experiences both the electromagnetic attraction from the positive nucleus and repulsion forces from other electrons in shells from 1 to n. (Remember, electricity has both attraction and repulsion!) This causes the net force on electrons in outer shells to be significantly smaller in magnitude. Thus, these electrons aren't as strongly bonded to the nucleus as electrons closer to the nucleus. This is known as the 'shielding effect.'
- There are many interesting periodic trends. You can learn more here.
EDIT:
I was able to narrow down an answer I think. We define effective nuclear charge as $Z_{\mathrm{eff}} = Z - S$, where $Z$ is number of protons and $S$ is the average number of electrons between the nucleus and the electron in question. Only the 1s orbital electrons have $Z_{\mathrm{eff}} = Z + 0 = Z$, ie $S = 0$ only for neutral hydrogen and helium. For all other standard elements, we have additional orbitals like 2s and 2p and 3s, etc and these all experience $S \neq 0$ so $Z_{\mathrm{eff}} < Z$.
Apparently, there is something called the Slater's rules. The shielding constant for each group is formed as the sum of the following contributions:
So for iron, here is the effective nuclear charge for different electrons.
Now let me try to address the radius issue according to Slater's rules. I make no guarantee Slater's rules are foolproof. You should investigate that yourself. I'm just going to take these rules and apply them. Let's consider fluorine. I like fluorine as an example because I always think of it as the hungriest of the elements. It wants an electron to fill it's outer shell badly. Why? So for the 2nd row, the effective nuclear charge increases as you go across the periodic table as follows:
- $Z_{\mathrm{eff,Li}} = 3 - (0.85 \times 2 + 0.35\times 1) = 0.95$
- $Z_{\mathrm{eff,Be}} = 4 - (0.85 \times 2 + 0.35\times 2) = 1.60$
- $Z_{\mathrm{eff,B}} = 5 - (0.85 \times 2 + 0.35\times 3) = 2.25$
- $Z_{\mathrm{eff,C}} = 6 - (0.85 \times 2 + 0.35\times 4) = 2.90$
- $Z_{\mathrm{eff,N}} = 7 - (0.85 \times 2 + 0.35\times 5) = 3.55$
- $Z_{\mathrm{eff,O}} = 8 - (0.85 \times 2 + 0.35\times 6) = 4.20$
- $Z_{\mathrm{eff,F}} = 9 - (0.85 \times 2 + 0.35\times 7) = 4.85$
So the effective nuclear charge does increase across the table! And thus, the electrons in the outer orbital experience a greater nuclear charge for elements on the left than on the right.
The basic idea is,
- Electrons in the same orbital don't shield as well as those of lower energies. So lower energy orbitals contribute more to shielding.
- Different orbitals contribute different shielding amounts, which presumably breaks down to the intricacies of the chemistry and physics.
and the result is the radii decrease as you go from right to left as a result because effective nuclear charge is increasing.
EDIT 2:
I learned on Chem SE that Slater's rules are just an approximation.
Best Answer
Well, actually there exists a nucleus which contains no protons. It has the atomic number 0, the mass number 1, and consists of one neutron, zero protons and zero electrons. It is called neutron. It is an unstable nucleus which decays via beta decay.
If you think that calling the neutron a nucleus is not proper, then think of the following: The hydrogen nucleus is just a proton. And chemists have no problem to talk about $\mathrm{H}^+$ ions, which are also nothing but protons, without any electrons around them.
According to the Wikipedia page DarioP linked to, also a di-neutron ($Z=0$, $A=2$) has been observed, which is extremely unstable. While the decay channel is not stated there, I guess the two neutrons just separate from each other; whether you call that neutron emission of spontaneous fission is a question of semantics. I guess in principle there would be also the possibility of beta decay to deuterium, but I'd not expect that to happen in observable rates.
Higher isotopes have not been observed, and are not to expect from theory.