Sedit
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Elemental Radioactivity Question.
What makes some elements radioactive even though there mass does not suggest they should be unstable.
There are two main ones I can think of first being Technetium. Why is this unstable even with its low molecular weight. Yet Molybdenum and Ruthenium
are stable.
The other element is Promethium where as even with its much higher weight then Technetium it still does not seem to fit the bill of heavy unstable
elements like everyother radioactive element does.
Also is there a pattern to the radioactive strengths in the heavier elements or is there flukes like these two that are just inherantly unstable even
if it where not for there large mass?
Knowledge is useless to useless people...
"I see a lot of patterns in our behavior as a nation that parallel a lot of other historical processes. The fall of Rome, the fall of Germany — the
fall of the ruling country, the people who think they can do whatever they want without anybody else's consent. I've seen this story
before."~Maynard James Keenan
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The Fountain of Discordia
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It is, simply put due to QCD: All possible combinations of protons and neutrons with 43(or 61) can find a much more favorable equilibrium state in a
diferent configuration. It is quite similar to why some unstable energetic materials (e.g. Manganese Heptoxide) will decompose. As to the stability of
radioactive elements, highly stable nucleii are related to magic numbers, these number being proton-neutron balances resulting in perfectly or
near-perfectly spherical shell configurations of one or both. In heavier atoms this is less true and it depends more on the shell level being filled.
[Edited on 9-7-2009 by The Fountain of Discordia]
Chemistry is art by other means.
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12AX7
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Has to do with nuclear energy levels. In a nutshell, it turns out there are "magic numbers", akin to electron shell numbers (8, 12, 20, 28, ...),
which have particularly high stability (akin to atoms' tendency towards stable filled shells). Conversely, there are "valleys of instability"
inbetween, where the nuclei just don't stick together well. Technetium and promethium are practically victims of circumstance; a shame, because
technetium's electron shell structure makes a good element in corrosion resistant coatings and steel alloys (like its neighbors chromium and
molybdenum), while promethium might be useful for magnets, phosphors and other rare-earth applications like its healthy brothers.
As for the heavy elements, keep in mind first of all that binding energy per nucleon falls past iron, so (after infinite time) none of these heavy
elements should exist. If the energy level is low enough, the nucleus won't be stable and will decay to something that is more energetically
favorable. That said, all elements between bismuth and thorium are very short lived (radium is the longest, just a few thousand years). Thorium is
the first 'magic number' nucleus in this area, landing a whopping 14Ga half life, quite stable by most measures. Next, protactinium has an odd number
of protons, so loses the gamble on magic numbers; it decays in 24ka. Uranium is almost as stable as thorium, U-238 having a fairly stable number.
All subsequent elements known so far are less and less stable, though another island of stability is theorized for the next set of elements -- 118 and
up. These isotopes must be very neutron-rich, and it's notoriously difficult to synthesize atoms of this scale and proportion from smaller atoms of
much smaller neutron count.
Tim
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Polverone
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Does the nucleon binding energy actually say anything about decay? If I set a marble in a coffee mug next to a deep pit there's a lower-energy state
nearby but barring some help from rough weather or living creatures the marble will never cross that barrier.
I know that only a handful of elements can sustain a chain reaction, but do (for example) mercury or lead release net energy on bombardment with the
right energy neutrons? Do all heavier-than-iron elements show net energy release when bombarded appropriately?
PGP Key and corresponding e-mail address
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JohnWW
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The instability of ALL isotopes of Pm, element 61 and a rare-earth metal (enough of it has been isolated to be able to examine the properties of the
pale-pink salts of Pm(III)), is also related to the group of long-lived naturally-occurring alpha-emitting isotopes of several other rare-earth
elements, and of the following elements from Hf to Re. These result in mixed crude rare earths extracted from pegmatite (formerly deeply-buried
granite, the last part of the granite to have crystallized) being slightly radioactive.
These long-lived alpha-emitting isotopes of rare earths arise because their incomplete shells of protons and neutrons are furthest from being either
empty or complete, while at the same time the radii of their nuclei are such that their internal bonding due to nuclear binding force comes close to
being exceeded by the electrostatic repulsion of the protons. But in the following elements, from Os to Pb, both a proton shell and a neutron shell
approaches completion, increasing internal binding with little or no increase in nuclear radius, up to Pb-208, which has both completed proton (82)
and neutron (126) shells, and enabling Pb to have 4 stable isotopes and Tl to have 2 stable isotopes. The limit of total stability is reached when
just one further proton is added to the completed shells, in Bi-209, in spite of its having an odd number of protons, although even this is thought to
have a finite half-life, of the order of 10^17 years.
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watson.fawkes
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Yes, in an analogous way that the
chemistry terms "energetically favored" and "reaction rate" are related, which is to say, somewhat but not deterministically. When you're talking
about bombardment, an analogue of reaction rate is the notion of cross section, which is a measure of how reactive the target nucleus is with respect
to the bombarding particle. Some reaction cross sections are low, some high. Another aspect is selectivity. Sometimes a bombardment in total has
reasonable cross section, but you get out one of several reactions, fission being a rare outcome.
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12AX7
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Quote: Originally posted by Polverone  | | Does the nucleon binding energy actually say anything about decay? If I set a marble in a coffee mug next to a deep pit there's a lower-energy state
nearby but barring some help from rough weather or living creatures the marble will never cross that barrier. |
There are indeed nuclear examples of this very phenomenon. Some nucleons are stable simply because there isn't anything else more stable nearby.
Such nuclei usually undergo a double beta decay at unreasonably long intervals.
http://en.wikipedia.org/wiki/Double_beta_decay
Also like how singlet to triplet oxygen is "one of the most forbidden transitions".
Tim
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JohnWW
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I have read that article. As I expected, it occurs only in the most relatively neutron-rich naturally-occurring isotopes, with half-lives greater than
10^19 years, such as especially Ca-48, which is the only one of the isotopes in which it has been observed which is lighter than the maximum in
nuclear binding energy occurring at around Fe or Ni. It, having a completed proton shell, would very slowly decay to Ti-48, not to a Sc isotope (of
which element the only stable isotope is Sc-45, substantially heavier than by far the most common Ca isotope, "doubly magic" Ca-40 with completed
proton and neutron shells)
[Edited on 9-7-09 by JohnWW]
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entropy51
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Sedit, there are some very good answers to your question in the thread, but here's a somewhat more intuitive (to me) explanation.
There is a balance of forces between neutrons and protons in the nucleus. If this balance is not maintained the nucleus is unstable and will decay in
such a way as to restore the balance of forces.
The forces are (1) the repulsive force between the positively charged protons, which repel each other. This force is proportional to Z squared, where
Z is the atomic number. (2) the attractive force between neutrons and neutrons or neutrons and protons. This is the so-called Strong Force, and is
roughly proportional to A, the sum of the neutrons N and protons Z (atomic mass no. A = N + Z).
Since the electrostatic repulsion (1) increases as Z squared and attractive force (2)increases as A, as Z increases from element to element, there
must be more neutrons than protons to balance the repulsive and attractive forces. That is N is approximately equal to Z for low Z, but N > Z at
high Z. So there is a ratio of N/Z required for stability that increases at higher atomic number.
The stable number of neutrons N for a given Z, the region of stability, is shown in a graph here:
http://books.google.com/books?id=ws8QZ2M5OR8C&pg=PT70&am...
If N/Z is lower or higher than necessary to balance the forces, then N and Z are changed to bring N/Z back to the region of stability.
If there are too many neutrons N for stability, then N/Z is too high for that particular element (isotope) and a neutron is converted to a proton to
achieve the stable N/Z ratio. This is beta decay, N -> proton + electron and the emitted electron is the beta. Because a proton is created in
beta decay, the atomic number Z increases.
If there are too many protons for stability, then N/Z is too low for that isotope and a proton is converted to a neutron. This is positron decay:
proton -> neutron + positron and the positively charged electron (positron) is emitted. Because a proton is lost, the atomic number Z decreases.
Generally the daughter nucleus that is left after decay is excited (has excess energy) and it releases the excess energy as electromagnetic radiation,
a gamma ray.
When an element is irradiated by neutrons in a reactor the nucleus can absorb a neutron, increasing N beyond the region of stability. Eventually a
beta decay will get rid of this extra neutron.
If this is less clear than the explanations above, never mind.
[Edited on 10-7-2009 by entropy51]
[Edited on 10-7-2009 by entropy51]
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watson.fawkes
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| Quote: Originally posted by entropy51 | | If N/Z is lower or higher than necessary to balance the forces, then N and Z are changed to bring N/Z back to the region of stability.
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To extend this answer, I wish to note that α emission, taking N/Z to (N-2)/(Z-2) has the effect of
increasing the neutron ratio. This mode is particularly common amongst the heavy nuclides.
In addition to positron emission, there is the fairly common electron capture. These two have the same N/Z effect.
For completeness, there are also proton and neutron ejection modes immediately following beta decays or electron capture decays, but these are fairly
rare.
As a further illustration, for elements with multiple stable isotopes, the unstable nuclides with mass numbers in between stable ones generally have
both electron and positron emission modes.
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JohnWW
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That is mostly right, Entropy51, although with the following qualifications. In a nucleus with too many protons, K-electron capture, followed by
emission of X-ray and short UV photons of various wavelengths as the higher electrons cascade downwards to fill the vacancy in the K (lowest) electron
shell and subsequent ones, is more usual, although positron emission has been observed.
Also, if the numbers of protons and neutrons are near optimal, but if the number of protons exceeds 83 as in Bi (or if the nuclear radius is of
comparable size to such nuclei like the long-lived rare-earth alpha-emitters), then decay modes involving the emissions of protons and/or neutrons
occur until the protons number either 83 or 82 (most commonly) or 81, most commonly emissions of alpha particles (helium nuclei, which accounts for
the occurrence of helium in natural gas deposits in sedimentary rocks derived from granite). This also partly accounts for why Pb is relatively common
for its atomic weight. Much rarer decay modes in such heavy elements are proton emission and neutron emission, which can happen in some cases of
excess protons or neutrons, particularly where there are odd numbers of them, and where the energy gain is comparable to those of the more common
decay modes.
Some radioactive nuclei, mostly with odd numbers of protons and with a combined total of neutrons and protons which are the same as combined numbers
which are stable for the two neighboring elements, have competing modes of decay, to two different products, with different half-lives for each mode.
One such nucleus is K-40, which is a long-lived naturally-occurring isotope, which is unusual for a nucleus with odd numbers of both protons and
neutrons and heavier than N-14 (the only other such naturally-occurring isotope is V-50). It owes its unusual stability to its proximity to the
tightly-packed doubly-magic isotope Ca-40 (the heaviest stable isotope with equal numbers of protons and neutrons). Most K-40 decays to Ar-40 by
K-electron capture and/or positron-emission, the basis of K-Ar dating of rocks and artifacts, but some of it instead decays to Ca-40 by beta-emission.
In the least stable heavy nuclei, spontaneous fission can compete with the above other modes of decay, shortening overall half-lives still further.
This occurs in some super-heavy artificially-produced elements, substantially heavier than uranium, which are all neutron-deficient because they are
formed by collisions of nuclei having lower neutron/proton ratios (the ratio of which for greatest stability steadily increases with atomic number),
with the neutron-rich naturally-occurring nucleus Ca-48 being a projectile of choice for such syntheses. In addition, chain-reaction spontaneous
fission induced by liberated neutrons and alphas occurs in the lesser-stable isotopes of Th and heavier elements, if collected in amounts larger than
their "critical masses", most commonly in the U-235 obtained by enrichment of natural uranium and in its byproduct Pu-239 (from absorption of neutrons
by natural U-238), utilized for bombs and power generation. Spontaneous fission usually results in asymmetric division of nuclei, the products being
neutron-rich isotopes of elements mostly around Ba and around Sr.
[Edited on 10-7-09 by JohnWW]
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entropy51
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Watson and John,
Thanks for filling out some of the details that I omitted. I was aware of the additional complications you discuss, but I wanted to present a simple
intuitive explanation that would be satisfying to the non-nuclear chemists among us.
Nuclear physics often becomes unclear physics to the non-specialist.
And neither of you covered Auger electrons!
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12AX7
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Actually, several isotopes of uranium (U233 from Th232 + n, the small amount of U234 which occurs naturally, and the small but useful amount of U235
that occurs naturally) are important for their spontaneous fission (SF) decay mode, which releases neutrons, resulting in a sustainable fission
reaction, not just the heavier (artificial) elements. Although, heavier isotopes are much more prone to SF.
Tim
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entropy51
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Yes, spontaneous fission is very important in some, ahem, applications.
For example, Californium-252 has such a high SF rate that it makes an excellent neutron source. Unfortunately you can't turn it off when you're not
using it.
The SF rate of the U isotopes is too low to be a dependable source of neutrons in a reactor or other chain reacting assembly, so an external neutron
source is always provided.
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Vogelzang
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I didn't see the word tunneling in this thread. From what I remember from radiochemistry, the alpha or beta particles undergo tunneling when emitted
from the nucleus. The activity depends on the probability of tunneling, IIRC.
See http://en.wikipedia.org/wiki/Quantum_tunnelling
[Edited on 11-7-2009 by Vogelzang]
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entropy51
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Vogelzang, I noticed the absence of tunneling too. I didn't add it because I didn't think Sedit's question was begging us to go Quantum on him. But
right you are!
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12AX7
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Indeed, alphas are a direct result of quantum tunneling; high energies lead to proportionally fast decay rates and vice versa. Alpha energies are, as
far as I know, monochromatic (i.e. single energy) and range from about 3-6 MeV.
Beta decay requires the weak nuclear force, and usually invokes neutrinos (since an electron is emitted where there was no electron to begin with, or
absorbed and none remains to balance). Due to the different mechanism involved, decay rates are different (the weak boson has an RME of about 80 GeV,
so it's fairly unlikely to be produced), and due to the extra particle involved (the neutrino), beta decay energies cover a continuum up to some
maxima, similar to bremsstrahlung radiation spectra.
Tim
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