I remember reading in another thread that the end result of fusion is ultimately iron (Hydrogen to helium to ??? to iron).
Why is that? What is so special about iron? I would expect the end result to at least be a noble gas.
Because the creation of elements up to and including iron releases energy. The creation of elements heavier than iron requires a net input of energy. When a star’s core has been transmuted to iron, no more energy creation is possible to resist the force of gravity. That’s why some stars go supernova. The star collapses at a velocity of up to a third of the speed of light,crushing the core to neutrons. The material impacting on this non-compressable core rebounds, creation a tremendous shock wave that blows much of the star into space and at the same time utilizing some of this energy for the creation the trans-iron elements.
Thanks…
3 follow-up questions
(A) Is a “trans-iron” element one with more protons in its nucleus than iron?
(B) Why is fusion of nuclei with quantity of protons more than iron a net energy sink but those with less than iron a net energy release?
© Is there more iron on the earth or silicon? I see lots of silicon everywhere but maybe that’s because the iron all sank to the center?
Ya know, I never bump threads, and I discourage the practice. But this is too good a question to languish on page three.
Anybody care to take a crack at it?
Chronos is probably the one to answer this, but he’s been away since Dec. 26.
One model of the nucleus that is pretty easy to envision is called “the liquid drop model”. In an actual drop of water, water molecules in the middle of the drop are attracted to all the neighboring molecules. Molecules near the surface have no neighbors on the outside, so the net force on surface molecules is toward the center. This is surface tension, which tends to keep the water drop spherical. Something similar keeps the nucleus together, but it gets a bit complicated.
In the nucleus, there are attracive forces (the strong nuclear force) between all the nucleons (protons and neutrons) but also repulsive forces (the electric force) between the protons. (Protons repel each other because they are all positively charged).
The stability of a nucleus depends on a lot of different things, which are encapsulated in a formula called Weizäcker’s semiempirical formula[sup]1[/sup], which has four terms.
[list=1][li]The first term depends only on the number of nucleons. It takes into account the attractive force between the all nucleons, the strong nuclear force. This term is positive because it tends to make the nucleus stable. This term tends to make heavier nuclei more stable than light ones.[/li][li]the second term depends on the number of nucleons [to the two-thirds power]. This term takes into account the surface tension. This term is also positive, because the nuclear force is attractive and tends to make the nucleus stable. This term also tends to make large nuclei more stable than small ones.[/li][li]The third term depends on the number of protons [squared] and the number of nucleons [to the one-third power.] it takes into account the electrical repulsion between the protons. (The reason it doesn’t depend solely on the number of protons is that with more neutrons, the protons are less closely packed and repel each other less). This term is negative because it tends to make the nucleus less stable. This term tends to make nuclei with many neutrons more stable than those with few neutrons.[/li][li]the fourth and final term is the one most diffcult to understand. It takes into account the “asymmetry energy”. Isotopes with equal numbers of neutrons and protons tend to be especially stable. For example, Carbon-12 is composed of exactly 6 protons and 6 neutrons, and is much more stable than any other nucleus composed of 12 nucleons.[/li]
The reason is that the protons can’t all fit into the same energy state, and the neutrons can’t all fit into the same neutron state. (Neutron energy states are separate from proton energy states). When you add an extra proton (or neutron), it goes into a higher energy state than the rest. Imagine a hypothetical molecule with six protons and no neutrons (Carbon-6). The sixth proton goes into a very high energy state, but there are a lot of empty neutron energy states that have gone unfilled, an unstable situation.
So the fourth term tends to make nuclei more stable that have an equal number of protons and neutrons. This term is zero for “balanced”, negative for all others. The bigger the difference between the number of protons and neutrons, the more negative the number will be.[/list=1]
Iron-56 (with 26 protons and 30 neutrons) is as stable as it is because it represents a compromise between these 4 considerations. Carbon-12, for example, loses fewer points for #3 and #4, but also gets fewer points for #1 and #2. Uranium-238 gets more points for #1 and #2, but loses more points for #3 and #4. It’s true that iron-56 is the most stable isotope, but there are a lot of isotopes with about 40 to 70 nucleons that are almost as stable.
[sup]1[/sup]The formula is
B = 15.753 A - 17.804 A[sup]2/3[/sup] - 0.7103 Z[sup]2[/sup]/(A[sup]1/3[/sup]) - 94.77(½A-Z)[sup]2[/sup]/A
where B is binding energy in MeV, Z is the number of protons, and A is the number of nucleons.
Question C - From CRC, 64th ED. (kind of old) - The Earths core is (as far as I know) still somewhat theoretical, but it is believed to be mostly iron (I’ve seen estimates on the order of 90% by weight). In the crust, iron is the fourth most abundant element by weight (approx. 25%) and silicon is the second most abundant (approx. 25.7%).
The formula is right, but my commentary on term #2 is wrong. Term two is negative, meaning that the greater the number of nucleons, the more negative this term will be (and the less stable the nucleus will be). I can’t say I understand that fully, so I’ll have to think about it.
Also, the total binding energy isn’t what makes the nucleus stable, it’s the binding energy per nucleon that matters (The higher the binding energy per nucleon is, the more stable the nucleus will be). If my calculations are right, the formula would predict that the binding energy per nucleon
for carbon-12: 7.0 MeV
for iron-56: 9.9 MeV
for uranium-238: 7.7MeV.
You might want to try this page. If I remember my astrophysics classes, then bibliophage is right. Basically, When stars go bang (or do their funky stuff after all the hydrogen is burnt), progressively heavier elements are produced, but those after Iron are increasingly unstable and less likely to hang around.
For relative abundances, try here. The bulk compostion of the Earth is tricky measure directly because the surface bits (which we can see) are nothing like the deep bits of the crust and mantle. The best guess we have is to say that it’s pretty close to a Chondritic meteorite.
Tran-Iron elements are those even and odd Z cosmic ray nuclei with Z between 26 and 83. The term is normally reserved for nucleii detected in cosmic ray experiments as far as I can tell
Correct on silicon. Here’s a quote from Cecil…
Thanks for the endorsement, bibliophage. I’m still away, and hence don’t have my books here (thanks for the formula; I probably wouldn’t even have bothered to dig that up), but there’s still a few more things to add. First off, noble gasses are irrelevant, because they deal with chemical stability (the arrangement of electrons around the nucleus), not nuclear stability (the arrangement of protons and nucleons within the nucleus). Not all elements heavier than iron are unstable, just because they have surplus potential energy: A brick sitting on a tabletop has more potential energy than one on the floor, but it’s not going to just spontaneously fall to the floor. Elements heavier than iron (and many atoms of lighter elements, as well) are produced exclusively in supernovae, where there’s so much extra energy around that a little extra in a few atoms of uranium and such will never be missed.
As for the sign of the second term in that formula, biblio, I seem to recall that the strong force is actually repulsive at distances greater than about the diameter of a nucleon, so that could account for it.
My CRC (71st ed, about 10 years old) lists Silicon as the 2nd most common element in Earth’s crust with 28.2% and Iron as fourth with 5.63%.
One other comment, howardsims wrote:
When evaluating the relative stability of nuclei, the chemical activity of individual elements is irrelevant. Certain elements are noble gases because of the stability of their electrons, while nuclear stability concerns itself only with the nucleus of an atom.