nuclear chemistry

Calculating the Ratio (semimajor-axis/semiminor-axis) from the Deformation

As chemists we view the nucleus as some arbitrary positively charged sphere that keeps the electrons bound to the atom. As a nuclear chemist we know the only nuclei that are perfectly spherical are nuclei that contain a magic number(2,8,20,28,50,82,126) of protons, or neutrons, or both. The other nuclei, not associated with magic numbers, will be deformed by some degree. The measurement of the deformity is called the deformation parameter. The deformation parameter is typically interchangeable with what we discuss as the quadrupole deformation, although higher order deformations (hexadecapole, hexacontatetrapole) can and will be calculated by others. A quadrupole deformation can have two different effects on a nucleus. 1) It can cause a prolate deformation, making the spherical nucleus actually appear more like a football, positive quadrupole deformations. 2) It can cause an oblate deformation, making the spherical nucleus actually appear more like a doorknob, negative quadrupole deformations. Both are shown below

Prolate

Oblate

All that is nice, but lets get to the point. Often in the literature, authors will list the deformation of a nucleus and will leave it at that. I often find myself wanting to know physically what does it all mean. Simply put, the deformation is a fancy equation nuclear scientists use that relates the length of the semimajor-axis to the length of the semiminor-axis. Think of a 2d-ellipse, turn it into 3-dimensions (ellipsoid) and you have the generic picture I’m trying to convey. In order to translate a deformation into reality (i.e. the ratio of the longest side to the shortest side) you can use the following equation I painfully solved yesterday during night-shift. I purposefully didn’t simplify it and left it in the form of the quadratic formula in case any curious readers wanted to play with it, or attempt to back-track to the original logic behind it. Why they do not give these equations in any text-book, or journal article that I have come across since joining this field is beyond me. Below are the equations.

For a prolate nucleus use the following:

For an oblate nucleus use the following:

Where beta is the quadrupole deformation.

Have fun with it, if anyone ever needs it.

Mitch

By October 23, 2006 0 comments nuclear chemistry

Discovery of Element 118 by Oganessian, Don’t Call it Ununoctium

Element 118 has had a sordid past. It literally is a wonderful case study in scientific fraud and how the culture of science always eventually corrects and catches the dishonest individual. Disregarding the juicy aspects of that salacious back-story and moving forward to the recent discovery claim, by Oganessian for element 118. Surprisingly the article wasn’t published in Nature or Physical Review Letters, but in Phys Rev C (PRC). PRC is a journal for nuclear scientists and it just seems odd to me that they didn’t pick a journal with a higher impact factor. Then again Oganessian is already the man in heavyelement’s research so I guess he doesn’t need to be in the lime-light, although he does seem to have a disproportionate number of first author papers. The 118 decay chain is shown below:

Usually, nuclear scientists in the field will make a new element and check to see if it was really made by seeing if the daughters have the same decay characteristics as those already published, we call this a genetic correlation. In this paper we would check to see if the decay characteristics of element 116, 114, and 112 matches with what we have seen previously. The problem is, no one else but the Russians have ever made those isotopes of 116, 114, and 112. So as of yet, they are unconfirmed and still need to be further investigated themselves! That’s the problem of taking a blind leap of faith and running a 118 experiment, the daughters haven’t been investigated, so you end up trying to prove you made what you made by using various theoretical models that point to similar decay characteristics as you expected. To Organessian’s great credit they did not publish their results on 118, which they technically discovered in 2002, but waited until they did further investigations of 116 and 114 and even several chemistry runs for 112. So, in this paper it all matches up, the decay of 118 matches what they saw for 116 and so on.

Unfortunately, the paper has no smoking gun for 118. I define a smoking gun as being an EVR-a-aasf, the bold indicating that the decay occurred when the beam was turned off. Even without that, the background is low enough, that the probability of random correlations seems low and the data still good. One worrisome thing about the 118 claim are the two 118 alphas with 11.65MeV in energy. See their table below highlighted in red.

212Pom has the exact same alpha energy of 11.65MeV. One would also expect to see 212Pom as a common transfer product contaminant in these reactions. Also, from what I understood from their data analysis code, the first alpha in strip 3 should of turned off the beam, and I didn’t see where they explained why it didn’t. Also, having a mean life-time, for 116, ranging from 98ns to 42ms seems too large, even for just 3 events.

In summary, the problem with the work is that 118 doesn’t decay into the known and investigated regions of the periodic table, like element 110(Ds) and 108(Hs). So we can’t say with 100% certainty they made what they said they did. Until the rest of the nuclear community “catches-up” to the Russians with thorough systematic studies all the way from Roentgenium to element 118, we will not be able to evaluate the claim thoughtfully. And since the Americans and the Germans haven’t been able to prove Russia’s 112 element claim thus far, I doubt anyone will be running a 118 confirmation experiment in the near future.

Note 1: You can download the paper here: Oganessian’s element 118 claim
Note 2: No one in the field will ever call element 118 “ununoctium”, so please don’t embarrass yourself.
Note 3: Edited some of my nonsense because of Grejak’s comment below.
Note 4: I think the Swiss did do a 112 chemistry experiment from the short-lived 114 alpha decay, but that is still unpublished to my knowledge.

Mitch

By October 16, 2006 9 comments nuclear chemistry

How to calculate a coulomb barrier with a prolate deformed nucleus

Trying to calculate a coulomb barrier for a prolate deformed nucleus with a spherical projectile has consumed my attention since Friday; I could not find anyone in the literature who did this without using fancy quantum mechanics. Anyone can look in an E.M. textbook to learn how to calculate a coulomb potential for two spherical charged balls, but once you perturb the spherical geometry of one of the balls it’s no longer a simple problem.

Disregarding the classical physics aspects of the problem and only caring about the nuclear chemistry part, the formula you would use to calculate the coulomb barrier will vary slightly depending on how heavy the two nuclei are (ie. where you lie in the periodic table). Since I deal with the heaviest elements humans can make, I finally decided to use Swiatecki’s coulomb barrier equation but modified his coulomb barrier parameter, the original formula is shown below.

z is Swiatecki’s coulomb barrier parameter, Zt atomic number of the target nucleus, Zp atomic number of the projectile, At and Ap are the mass numbers of the target nucleus and projectile nucleus respectively.

To calculate the coulomb barrier parameter with a prolate deformed target nucleus, to a first order approximation, I used the following expansion around the At term.

Where b2t is the quadrupole deformation parameter(degree of non-sphericalness), which one can look up for any target atom of interest. The x term is the angle of contact and ranges from 0 to pi/2. An angle of 0 corresponds to an equatorial touching, while pi/2 corresponds to polar touching, shown respectively in the next figure.

From staring at the equation all day, this is most likely only physically valid at an angle of zero and at an angle of pi/2, which were the only two conformations I was interested in anyways. The above seems to hold up well against authors who used a quantum mechanical approach, so it does hold some physical merit.

Note1: All units are in MeV
Note2: After solving for the coulomb barrier parameter you have to plug this back into Swiatecki’s coulomb barrier equation, to determine the actual coulomb barrier.
Note3: You can download Swiatecki’s paper from here: Fusion by Diffusion II.
Note4: I do realize I posted this right after making fun of p-chemist in the previous entry.

Mitch

By October 10, 2006 0 comments nuclear chemistry