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Candlestick III

Dear Reader,

A lady from the USA who collects glass has been in communication with me regarding the uranium glass candlesticks, she has a pair of very similar candlesticks to my uranium glass one. It is possible that they all came from the same maker.

I did collect a gamma spectrum from my candlestick, as the activity level in a candlestick is low it took many hours to get a good spectrum. Here is the full range gamma spectrum from 10 keV photon energy to about 3 MeV photon energy. This covers almost all of the gamma spectrum. OK I can think of gamma events which are outside this range like the n + N-14 reaction to form C-14 which emits photons at about 10 MeV but such a high energy photon is very rare.

Full range gamma spectrum of the candlestick

Full range gamma spectrum of the candlestick

You should be able to see that almost all of the peaks are at low energies (below 500 keV), I have stretched the y axis by using a log scale so you can see some more of the minor peaks.

Candle stick full range log scale for counts

Candle stick full range log scale for counts

Now I have replotted the graph to show the lower energy end only, it should be clear that a series of peaks are present due to the gamma rays from the candlestick. Also on the left hand side is a big peak due to X-ray formation. What happens is that high speed electrons (both beta particles and photoelectrons) fly around inside the shielding and form X-rays by hitting the copper inner shielding on the lead castle in which the detector is placed. The copper inner shield reduces the fraction of photoelectrons which are ejected from the lead surfaces that are able to strike materials close to the detector crystal.

Also any X-rays formed in the surface layer of the lead nearest the radiation detector and the object being measured will tend to be stopped by the copper metal. By experience is that 1 mm of copper sheet is able to stop almost all copper K x-rays. And I imagine that it will stop many moderate energy photons like the moly K x-rays which are used for single crystal crystallography work and for mammograms. In the ideal world we would also have an aluminium layer inside the copper shield to further reduce the back ground at low energies due to secondary X-rays, and then a layer of a plastic like PMMA to finish the job. This is an example of graded Z shielding.

Even if the shielding is designed in such a way to prevent X-rays reaching the detector, it is possible to create X-rays inside the sample. The ideal sample for gamma counting would be a very small sample with a very high specific activity (radioactivity per mass or unit volume). If this was wrapped in plastic then any beta particles which come from the sample would lose their energy gently in the plastic without making horrible X-rays. But in our case the volume of the sample is large and the glass contains plenty of elements such as silicon which can form bremsstrahlung as the electrons pass close the atomic nuclei.

But we will not let a little bremsstrahlung spoil our fun. Here is the lower end of the gamma spectrum.

Candle stick spectrum lower end

Candle stick spectrum lower end

What we need to do next is to work out the relative efficiency for the detector for the different gamma lines, what we need to do is to measure the size of the peaks which we can identify and then use the fact that the yield of the different gamma lines are known. This will allow us to reconstruct the relative efficiency for the detector. In case anyone is thinking that we could do it with more easy by using a mixed photon energy radioactive source such as a radium-226 source. There is a problem, the odd shape of the sample makes it have a sum of many different counting geometries and a lot of self adsorption for the lower energy gamma photons.

I could make it more simple by crushing or melting my candle stick to make it into a more simple shape, but that would spoil my candle stick so I am not going to do that.

X-ray energy and getting the terms right

While reading the article entitled “the art detectives” in the RSC’s Chemistry World magazine I saw the statement that high energy X-rays are used for XRF of elements such as zinc. I strongly suspect that a misunderstanding has occurred, for example the zinc k lines will come at 8.6 keV which is hardly high when compared with the X-ray photons commonly for the industrial radiography of steel objects. To excite an atom in a X-ray fluorescence (XRF) experiment only moderate energy photons are needed (tube voltage of 40 kV is acceptable) while for industrial X-ray radiography it is common to use much higher accelerating voltages (100 kV and higher). For very thick metal objects photons in the MeV range are used.

What I think the article should have stated is that the object in XRF was illuminated with a high intensity of x-ray photons, to my mind intensity (photons cm-2 s-1) is very different to photon energy. But why would anyone use an expensive intense x-ray source rather than a weaker and cheaper one ?

If we assume that the increase measured above background is directly proportional to the concentration of an element and the intensity of the incoming exciting x-ray beam, then if the background is 10 cps, then with a weak x-ray source then we could get a reading of 20 cps on a spot on a painting. As for a random events the standard deviation on the count number is the square root of the count number after 1 second then the sum of the two SDs is 7.634 which is close to the difference between the two count numbers. If we were to use a source ten times brighter then the sum of the standard deviations (10.49 + 3.16 = 13.65) is small compared to the difference in counts after 1 second.

The great problem is that people writing about science sometimes tend to throw words about, almost randomly, without thinking about the fact that the word already has a meaning. To write clearly about science we must first avoid confusion.


Dear Reader,

Those of you who live in Sweden may have heard of a body called SSM (StralSakerhetsMyndigheten), this is a state body in Sweden which has the task of protecting people and the environment from the adverse effects of radiation both today and in the future. In common with the now defunct NRPB their task includes X-rays, “Nuclear” radiation (α, β, γ and neutrons), radiowaves (cell phones etc) and UV light.

One of the great problems we as a society is that for the good radiological protection of future generations living far in the future we need to make predictions based on experiments which only last a short time.

One important issue is the formation of organic complexing agents in low and intermediate level radioactive waste. If a substance forms which is able to bind to metals and form water soluble complexes which do not bind to mineral surfaces then the rate at which radioactivity leaks out of a waste store could be increased.

One such compound which has been considered by many people is isosaccharinic acid (ISA) which is formed from cellulose when it is exposed to calcium hydroxide. The cellulose can come in the form of wood, paper or cloth while many cements contain calcium hydroxide.

The classic way to make ISA is to treat lactose with calcium hydroxide, I have done this several times and the mixture soon turns brown and after boiling it down you are rewarded with a dark brown mixture which smells strongly of cooking. By careful filtration of the dark brown mixture a brown solution can be obtained which is then evaporated to a dark solid. This is then extracted with water and recrystalized to give a white solid, due to the insolubility of the calcium salt of the alpha isomer of ISA this is possible. The calcium salt of the beta isomer is water soluble and stays in the mother liquor with a lot of other compounds.

As a result it is relatively easy to obtain alpha ISA, the beta ISA is harder to obtain, so as a result almost all work done on ISA has been done with the alpha isomer. As the properties of the two isomers are not exactly the same it may not be safe to assume that alpha ISA can be used to model a mixture of alpha and beta ISA. Within this project we will explore beta ISA and determine if it poses a special threat in nuclear waste stores.

Now you might wounder why I mention SSM, the reason I mention them is that they are funding this research. One of the ways that SSM protect society is to fund research which allows them to make better predictions about the future. Now my SSM work is about to start, the plan is that I will try to publish papers as well as writing a report for SSM. I also want to bring you some updates about the work here on my blog.

Candlestick II

Dear Reader,

I took my candlestick to work and I quickly found it was radioactive, it was emitting beta particles according to a quick check with a contamination meter. As it was emitting that nice yellow/green light when exposed to UV light and it was emitting beta particles I quickly decided it was genuine uranium glass.

The next step in the characterization of the candle stick was to use gamma spectroscopy on it, now before we get going I would like to point out that gamma spectroscopy is not a press the button and get the result type of machine. For those of you who are proper traditional chemists / scientists you will be aware that for a new type of sample it is very hard with most machines to create a method with a spectrometer where you just put in the sample and press go before getting the final answer.

One of the problems is the issue of self adsorption, for the lower energy gamma lines many of the photons will never escape from a large sample. The ideal sample for gamma spectroscopy would be a tiny spec (a point source) which would be at a well defined distance from the detector.

The candlestick is anything but well defined in distance from the detector and it is far from being a point source. I did not want to melt it down to make a lump with a more simple shape so I decided that we should measure it in its native form.

One of my questions about the candle stick was “is the uranium a depleted uranium, or is it a natural uranium which is likely to predate the nuclear age ?”

I reason that as DU is less valuable than natural uranium it would be the logical uranium to use if you were making a uranium glass candlestick in the 1950s or later. But if it was a more early candlestick then it would be more likely to have a natural isotope signature for its uranium.

We need to consider three uranium isotopes

238U which is the bulk of natural uranium, this does not have any useful gamma lines but its daughter (234Th) which emits gamma rays, as the half life of 234Th is short when compared with the age of the candle stick it can be treated as an extension of the radioactive decay of the parent 238U. 70% of the 234Th will decay to the meta stable state of 234Pa (234mPa). It is important to note that the 234Pa (both forms) give a forest of gamma lines (hedgehog spectrum).

Nuclide Half life Decay mode Main gamma lines
238U 4.468 x 109 years alpha No gamma
234Th 24.1 days beta 63.3 (4.8 %), 92.4 (2.8 %) and 92.8 (2.8 %)
234mPa 1.17 minutes beta 258.3 (0.73 %), Hedgehog spectrum
234Pa 6.7 hours beta Hedgehog spectrum


If the uranium had been a depleted uranium then I would expect that almost all the 234U and 235U would have been removed. As the 234U has a long half life it serves to block the decay chain of 238U if the sample is not old on a geological time scale.

I reasoned that by looking for the decay products of 234U that I could test the hypothesis that the uranium was a prenuclear age natural mixture of isotopes.

This uranium will decay to form a long lived radium (226Ra) which will then slowly on the timescale of the candlestick’s age decay further.

234U –> 230Th –> 226Ra –> 222Rn –> 218Po –> 214Pb

Nuclide Half life Decay mode Main gamma lines
234U 245500 years alpha No gamma
230Th 75380 years alpha 67.7 (37 %)
226Ra 1600 years alpha 186 (3.6 %)
222Rn 3.8 days alpha No gamma
218Po 3.1 minutes alpha No gamma
214Pb 26.8 minutes beta 242 (7.4 %), 295 (19.3 %), 352 (37.6 %),
214Bi 19.9 minutes beta Forest of lines
214Po 0.1643 ms alpha No gamma


The 214Pb will decay by beta emission to form 214Bi and then 214Po which then decays to form 210Pb. As after 226Ra no nuclide has a half life longer than a few days until you reach 210Pb we can treat these decays as extensions of the radium decay if we make a kinetic model of the candlestick.

The fissile 235U does have a useful gamma emission of its own, this can be used to confirm if the uranium was natural or depleted.

It will decay by alpha emission according to the following mechanism.

Nuclide Half life Decay mode Main gamma lines
235U 703800000 years alpha 109 (1.5 %), 144 (11 %), 163 (5.1 %), 186 (57 %), 205 (5%),
231Th 25.52 hours beta No gamma
231Pa 32760 years alpha Forest of lines
227Ac 21.773 years beta No gamma
227Th 18.72 days alpha Forest of lines


I hope to now be able to go through the spectrum and then hunt for lines, I recall that the 186 keV line for 235U was present. So far I think the uranium is from before the nuclear age.

Uranium glass

Dear Reader,

For some time I have been looking for some uranium glass, this is a type of glass which contains uranium. because of the uranium it is green in colour. The glass also fluorescent, when exposed to UV light it emits green light. I was in a second hand shop today and I spotted some green glass, I happened to have a long wavelength UV light in my pocket so I did a quick test of the glass with UV light. As the 20 SEK candlestick emitted green light I bought it.

I have made a short film of the candlestick being exposed to UV light. I hope to upload to youtube soon a film of the candlestick in a dark room being exposed to a UV light which I turn on and off.

20 SEK (circa £ 2) candlestick made of uranium doped glass

20 SEK (circa £ 2) candlestick made of uranium doped glass

I also hope to take the glass candlestick to work and measure its radioactivity. When I do I will post the details in another blog post.

All Things Bright and Beautiful

Dear Reader,

It has come to my attention that the BNP have released a bigoted song using the tune of “All Things Bright and Beautiful”, this is a horrid thing. To my mind the song “All Things Bright and Beautiful” is a celebration of all things which are bright and beautiful. To my mind their use of the tune is like defacing an attractive oil painting with roofing tar.

I hold the view that the BNP are bad for british society.

What is solvent extraction ?

Dear Reader

You might ask the questions of what is the liquid-liquid extraction of metals and why is it important ?

I would say that the liquid-liquid extraction of metals is the more correct term for what is normally understood as “solvent extraction”. Solvent extraction is not the extraction of a solvent from a thing, but instead it is the extraction of a substance from one solvent into another. Commonly one of the liquids is water but there is no reason why the ideas of solvent extraction should not be applied to non aqueous systems such as silver being extracted from molten lead into molten zinc or the extraction of an organic species from methanol into hexane.

A typical metal ion in an aqueous phase is bonded to water molecules and/or chloride anions, this forms a water soluble metal complex. The metal complex normally has a very low solubility in an organic layer, but like many rules some exceptions do exist.

For example osmium(VIII) forms an organic soluble oxide (OsO4) which can be thought of as having formed from a Os8+ cation and four water molecules. The waters would have bonded onto the osmium before undergoing hydrolysis to hydroxyl ligands (OH-) which then react further to form oxide ligands. I suspect however that [Os(H2O)n]8+ is an impossible complex for several reasons so I do not think it will be possible to make it.

But I know that compounds such as [OsO3(OH)2] are well known, and this can be thought of as a hydrated form of OsO4, or alternatively OsO4 can be viewed as the acid anhydride of [OsO3(OH)2]. Yet another alternative is to stop thinking for a moment about osmium and then move onto some other things.

Well back to the solvent extraction of metals, if we take a hooks and eyes view of chemical bonding the electron poor metal ions can bond to the electron rich parts of solvent molecules such as the oxygens in water. These bonds to water can be broken and replaced with bonds to molecules with big long fatty groups, for example charged cobalt atoms in water are normally found with six waters around them. For example using 3D x-ray crystalovision (known to chemists as X-ray crystallography) we can glimpse a charged cobalt bearing six waters in the way it is thought to be in aqueous solution.

Cobalt with six waters

Cobalt with six waters

This cobalt ion was seen in a paper by some gifted chemists who made coordination complexes with tetrahydrofuran-1,2,3,4-tetracarboxylic acid.[i] I have to admire their compound, in some ways I wish I had done that chemistry myself.

Now it is well known that cobalt(II) salts in dilute water solutions are pale pink in colour, this is a sign that the cobalt is in a octahedral complex with six waters attached. However under some conditions deep blue complexes of cobalt are formed, this is normally a sign that a tetrahedral complex of cobalt as formed. It is well known that the solution of cobalt in bis-2-ethylhexyl hydrogen phosphate in a hydrocarbon solvent is deep blue. This made me want to look to see if X-ray crystallography has ever been used to characterize the idea mononuclear (complex with a single metal atom) of cobalt with two dialkyl phosphate ligands.

When I did the search I was in for a shock, I found a coordination polymer of cobalt and bis tert-butyl phosphate in which tetrahedral cobalts were linked by the dialkyl phosphates into a 1D chain. I also found bis-(((4,7,7-Trimethyl-3-oxobicyclo[2.2.1]hept-2-yl)phosphinato)-((4,7,7-trimethyl-3-oxobicyclo[2.2.1]hept-2-yl) hydrogen phosphinato))-cobalt(II) which is made from one cobalt(II) ion, two molecules of bis-(4,7,7-trimethyl-3-oxobicyclo[2.2.1]heptan-2-yl) phosphinic acid and two molecules of the conjugate base of this rather long named phosphinic acid.

I looked in the Cambridge database for metal complexes in which two dialkyl phosphate ligands chelate to a single metal atom to form a complex. None were found, I soon concluded that the oxygen oxygen distance is too large as the phosphorus atom is tetrahedral, this is different to a carboxylate where the central carbon is trigonal planar.

Here are the XYZ coordinates of a typical model phosphorus compound

O(1) -1.368 1.517 -0.000
P(2) -0.171 0.647 -0.000
O(3) 1.160 1.562 -0.000
C(4) -0.185 -0.423 1.516
C(5) -0.185 -0.424 -1.516
H(6) 1.876 0.950 -0.000
H(7) 0.715 -1.078 1.517
H(8) -1.102 -1.054 1.516
H(9) -0.177 0.220 2.424
H(10) -1.091 -0.199 -2.122
H(11) 0.726 -0.221 -2.122
H(12) -0.199 -1.495 -1.213


The two oxygen atoms are 2.5 Å apart,  now we can compare the distance with that in acetic acid. Again here is the xyz table, and in acetic acid the oxygens are 2.2 Å apart which makes it much more easy to form a chelate ring.

C(1) -7.031 -0.130 -0.000
C(2) -5.522 -0.130 -0.000
O(3) -4.918 -1.176 -0.000
O(4) -4.853 1.029 -0.000
H(5) -7.403 -1.179 -0.000
H(6) -7.402 0.394 0.909
H(7) -7.402 0.395 -0.909
H(8) -3.910 0.796 -0.000


Now some of my readers may ask “what is a chelate ring”, the idea of chelation is where a molecule bonds to a metal with more than one atom. I would like you to do a thought experiment, we have gone to the beach and a fish has bitten my foot, now how do I remove it ?

I have to open its mouth and then I can pull it off and throw it back in the sea.

Now imagine that the crab has grabbed by foot, now to remove the crab I must release both of its pincers at the same time. If I remove only one at a time while I am dealing with the second one the first one will grab me again. The crab is able to chelate to me.

Things get even worse with a scorpion which has two front grabbers and a tail to grab with, and frankly if the octopus gets me with all eight tentacles then it is going to be very hard to escape from its grasp.

If a ligand is able to chelate to a metal then it normally makes it very hard to remove the metal from the ligand’s grasp.

Thinking about the extraction of cobalt we might suspect that it should be given by the following equation

DCo = k[(RO)2PO2H]2[(RO)2PO2-]2

I suspect that the (RO)2PO2H will form hydrogen bonded dimers in the organic phase which could alter the equation to

DCo = k[{(RO)2PO2H}2]1[(RO)2PO2-]2

This might be true when the concentration of (RO)2PO2H is very low, the problem will be that when the concentration of (RO)2PO2H is high then it increases the dielectric constant of the organic phase, this will disfavor the extraction of the lipophilic (fat loving) cobalt complex [Co((RO)2PO2)2 ((RO)2PO2H)2]

This will make the equation which relates the distribution ratio closer to

DCo = k[{(RO)2PO2H}2]0[(RO)2PO2-]2 = k[(RO)2PO2-]2

As the concentration of (RO)2PO2- depends on the proton concentration we can make the equation slightly more complex and more useful.


Ka = [(RO)2PO2-][H+]/[(RO)2PO2H]


Ka / [H+] = [(RO)2PO2-]/[(RO)2PO2H]


Ka [(RO)2PO2H] / [H+] = [(RO)2PO2-]

So next

DCo = k{[(RO)2PO2H]}2 ka2 / [H+]2


DCo = k{[(RO)2PO2H]}2 ka2 [H+]-2

This equation does explain how to change the distribution ratio by altering the concentration of acid, by making a change of pH we can adjust our D value. Now as my physics teacher told me we should use dimensional analysis to decide if the equation is good or bad.

Now DCo = [Co]organic / [Co]aqueous

So the D value has no units

So the k value has the units of mol-2 dm6 as [(RO)2PO2-] has the unit of mol dm-3

OK what next.

Ka has the units of mol dm-3

So we have

mol-2 dm6 . mol2 dm-6 . mol2 dm-6 . mol-2 dm6 = mol-4 dm12 . mol4 dm-12 = mol0 dm0

Which finally confirms that the equation

DCo = k{[(RO)2PO2H]}2 ka2 [H+]-2

Is a viable equation which is at least mathematically correct.

At this point I think we can call it a day and have another solvent extraction lesson another day.


[i] Liang-Fang Huang, Chang-Chun Ji, Zhen-Zhong Lu, Xiao-Qiang Yao, Jin-Song Hu, He-Gen Zheng, Dalton Trans, 2011, 40, 3183.


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