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This should have been in Camera

Dear Reader,

The recklessness of some people drives me to a new level of distraction, I have chosen to not name the case or the newspaper in this blog article but I will tell you the story. Some irksome pest tries to build a homemade bomb, this homemade special fails. Then the news paper reports the court case explaining how and why it failed to detonate.

What I think is that any public discussion of a failed bomb is an educational moment which we are better off without. The problem I see is that building any complex gadget is hard when you have to make everything yourself. Having had an interest in electronics I can tell you that many gadgets have taken a lot of development work to get to the level which the public now considers “normal”.

Also having been in the business of developing chemical processes, I can tell you that the creation of a process (or even just the implementation of an existing process)  can take a lot of work. One of the things which always makes it more easy is a knowledge of how someone else did it successfully and an understanding of what went wrong the last time.

The knowledge that a particular method or material is unsuitable for a task is a great help, it saves a lot of time and effort. The reason is that a person will not go chasing after something which does not work. By publishing a truthful discussion of how a bomber failed the newspapers are helping the next generation of bombers by improving their knowledge.

Some years ago I was told something interesting at the ITU in Germany, they commented that a large fraction of illicit plutonium samples which have been intercepted contain red materials. They believe that the Soviet intelligence services leaked some misinformation about “red mercury”. The story is that “red mercury” greatly increases the ease of building a working nuclear weapon.

By releasing this crazy story about a spoof material, the Soviets were attempting to  waste the time and efforts of would-be nuclear terrorists. By encouraging them to chase a false lead it would have helped world peace. The reason is that every hour and dollar a terrorist spends looking for ways to obtain and use this material is a hour or dollar which they could not use one something which is more likely to provide them with a weapon. Very clever I think !

I have to ask why the facts of this type of case be discussed in public, and why the newspaper staff did not have the sense not to self censor what they published.

Electrochemistry calculations (redox potentials and cells)

Dear Reader,

I was recently teaching some electrochemistry to some students, now before we get going it is important to note that the conventions on writing electrode potentials changed years ago. So if thou gets one’s ye olde text book out or ye olde almanack of chemie be ready for some possible problems.

For example in the 1950s paper on plutonium redox chemistry by Sherman W. Rabideau and  Joe F. Lemon, Journal of the American Chemical Society, 1951, 73, 2895-2899 the redox couple for Pu(IV)/Pu(III) is listed as being -0.953 volts vs the standard hydrogen electrode. Now in modern text books it is listed as about + 0.95 volts. This modern use is an example of the “European” convention while the example I gave as an example of the “American” convention. I would say that both are equally valid but if you want to do a redox calculation be careful that you do not mix unwittingly mix data from both conventions up.

I have not seen the american convention being used much in modern text books, but please be aware that it does exist.

OK Health warning over lets get on with some chemistry. Now as a brain teaser (or brain expanded) I asked my students to consider the question of will a solution of iron(II) tend to reduce a solution of plutonium(IV) to plutonium(III) thus forming iron(III) in the process.

Now the redox couple of iron(II) / iron (III) is +0.77 volts (European convention). So we can calculate the cell voltage for our cell under “standard” conditions (1 mole per litre of everything, at 25 ºC and 1 bar).

We can combine the following two half equations

Fe2+ → Fe3+ + e

e + Pu4+ → Pu3+

To give us

Fe2++ Pu4+ → Fe3+ + Pu3+

The emf of the cell will be under standard conditions equal to 0.18 volts, but which thing will be reduced and what will be oxidized. The plutonium couple is higher (more positive) than the iron one in our European type text. So the plutonium will tend to oxidize the iron to form plutonium(III).

We get the emf of the cell from the difference between the two redox couples. All redox couples are expressed relative to hydrogen gas / hydrogen ions in 1 M acid. This choice of standard is simply a convention. Like the Greenwich meridian we need some arbitrary point to call zero. Hydrogen is a good choice as it features in so many reactions.

From the value of the emf of our cell we can get the ΔG of the reaction, to do this we use the following equation.

ΔG = -nFE

n is the number of electrons which are transferred in the cell reaction (1) and F is Faraday’s constant (charge on a mole of electrons) which is equal to 96485 C. Using this we can get a value for the Gibbs free energy of the reaction. Keep in mind that a cell with a positive emf is a cell which is able to do work, thus ΔG for the reaction in the cell must be negative.

This works out as -17367.3 joules per mole.

We can now get the equilibrium constant for this reaction, when a cell has a emf of zero then it has reached equilibrium. Now we need to use a different equation.

ΔG = ΔGº + RT Ln Q

In our case

Q = aPu3+ aFe3+ / aPu4+ aFe2+

Now in the ideal world (nice place I wish I lived there) the activity coefficient is equal to one, the closest we get to an ideal world is a dilute solution. So for nice and dilute solutions we can write

Q = [Pu3+] [Fe3+] / [Pu4+] [Fe2+]

If ΔG = 0 (zero) then

ΔGº = -RT ln ([Pu3+] [Fe3+] / [Pu4+] [Fe2+])

Rearrange to

-ΔGº / RT = ln ([Pu3+] [Fe3+] / [Pu4+] [Fe2+])

exp (-ΔGº / RT) = ([Pu3+] [Fe3+] / [Pu4+] [Fe2+])

Now do the maths

exp (7) = ([Pu3+] [Fe3+] / [Pu4+] [Fe2+]) = k

Now k = 1107

we can rewrite the equations to give us

ΔGº = -RT ln k

Now it should be clear to my readers that a solution of iron(II) will reduce tetravalent plutonium into trivalent plutonium. You might be interested to read that the classic method of adjusting the oxidation state of plutonium from +4 to +3 in the PUREX process is to use ferrous sulfamate.

Trinitite II

Dear Reader,

I have reexamined the gamma spectrum from the trinitite, and I have some news for my loyal readers. What I did was to look at someone else’s gamma spectrum of trinitite and then try to match peaks.

Here is the spectrum


Gamma spectrum of trinitite

What we can now see are two peaks (51.7 and 129.3 keV) which are due to the gamma emissions from plutonium-239. Also we can see a set of three lines due to uranium L lines X-rays.

We might ask why are we seeing uranium x-rays coming from a sample which contains so little uranium. One explanation which I think is very reasonable is that the alpha decay of the plutonium-239 forms uranium-235 which is formed in an electronically excited state. The uranium-235 then undergoes a rearrangement of the electrons to form the X-rays. This has been observed by others during XRF studies on plutonium metal.

This is further evidence that the sample contains the radionuclides which should be expected from the trinitiy test. So now I have managed to prove that the sample contains plutonium.

As the sample also contains americium-241 I think it would be reasonable to next make an attempt to find the lines for neptunium X-rays. These could be a further sign that the sample contains americium. I can not think of any other alpha emitters which will be present in large / moderate or even less than tiny amounts in the trinitite.

I will have to think further about the sample.

It is back in action

Dear Reader,

It has come to my attention that the space probe which landed on the comet has started working again. I have to ask the question why did the probe lack a RTG (RadioThermal Generator) based on something such as Pu-238 or Am-241. If it had been equipped with such a nuclear battery then it would have been able to operate without a need for sunlight.

While these battery packs are not very PC, I am aware that some of these radioactivity powered generators have survived launch accidents. If they are well designed then even in the event of a rocket blowing up on the lauch pad then no threat is posed to the general public.


While the general public get both excited and concerned about plutonium, some of the other actinides are equally important. Americium because of its higher stability of the +3 oxidation state has chemistry which is very different to plutonium.

I have seen predictions which suggest that the americium-241 either released from Chernobyl or formed in the environment as a result of the beta decay of the plutonium-241 released by the accident will become the radionuclide of greatest importance near Chernobyl after the cesium-137 has decayed away.

Fukushima will be a different matter as far less of the americium or plutonium in the fuel was released during the accident.

Palomares and the H-bombs

Dear Reader,

Now some doomsayers may have tried to tell you that once radioactivity appears in soil that you should give up all hope, also on the otherhand some false prophets of insincere reassurance will just tell you to stop worrying and that “everything will be OK”. My advice is not to trust either of these two false friends.

The story of the air crash which involved four H-bombs has popped up again, the BBC report that the local people in Spain are fifty years after the air crash unhappy about what has been done.

The BBC report suggests that the local farmers have a problem getting a good price for their produce at market. I would like to point something out.

The plutonium in the H-bombs would have been in the form of the metal, during the accident this would have been burnt into plutonium dioxide. Now the thing to note about plutonium dioxide is that it is very hard to dissolve in acid, also it is not mobile in soil. Any plutonium which was in a water soluble form is likely to have bonded to the soil minerals thus making it impossible for plants to absorb it via their roots.

M.I. Sheppard and D.H. Thibault, Health Physics, 1990, 59, 471 to 482 gives the binding constants for most metals to the four common soil types. It lists for plutonium the following Kd values.

Sand, 150 L/kg

Loam, 1200 L/kg

Clay, 5100 L/kg

Organic, 1900 L/kg

This means in a bucket containing a mixture of clay type soil and water that the plutonium content of the soil (Bq per kilo) will be 5100 times higher than the plutonium content of the water (Bq per litre).

Hence when 1000 Bq of plutonium is added to a litre of water mixed with a kilo of clay type soil, then the soil will absorb 999.8 Bq of plutonium while 0.2 Bq of plutonium will stay in the water. This calculation is for a static batchwise experiment but it will help experts in the field make predictions about the mobility of plutonium solutions in soil.

Another good bit of news is the fact any plutonium dioxide in the dust will not be well absorbed if it is swallowed (dust on the surface of the food), so orally the plutonium dioxide is not a great threat to life and limb. If you were to swallow a well sintered particle of plutonium dioxide it will pass unchanged through your digestive system.

However plutonium dioxide in the lungs is very dangerous to a persons health, I think that a key thing to do in Spain is to keep the plutonium in the most contaminated soils from entering the air as a dust. I think that the ban on building, farming or walking in the contaminated area is a good idea. But I think that it might be a good idea to pour concrete or asphalt onto the worst hot spots to try to fix the soil to keep it from becoming mobile again.

One of the problems with plutonium is that the colloidal particles of clay can make the plutonium mobile, while the plutonium does not move freely through the soil in aqueous solution the colloidal particles can move through the cracks in the soil. Thus sealing the soil would help to stop the plutonium from reaching the surface again in the form of dust.

Alpha decay part II

Dear Reader,

As I sit typing in a railway carriage on the way home sitting near a young lady who is sporting a ”Nuclear power no thanks” badge, I sit here thinking about nuclear processes hopeful that the young lady does not notice what I am typing.

It is interesting to note that one of the physical effects which regulate the reactions which go on inside the red sun of the “Karnkraft nej tak” badge are the electrostatic forces which oppose fusion. The same forces have an effect on the reverse reactions (alpha emission, fission and all the cluster emissions which come between those two extremes).

Now since I had a rather short hair cut recently I can not demonstrate electrostatic attraction using a comb dragged through my hair. I will let you try that at home, also I do not have a cat to rub on a bit of plastic so I can not use that either.

But back to nuclear processes and electrostatics, to a first approximation the atomic nucleus can be treated as a charged sphere. The size is given by the following equation.

R = Ro (A)0.3333

Where Ro is equal to 1.2 x 10-15 m, while A is the total number of nucleons (the sum of the number of protons and neutrons) in the nucleus.

So radius of a plutonium-238 nucleus is 7.44 fm, while its daughter (uranium-234) has a nuclear radius of 7.39 fm while an alpha particle has a radius of 1.90 fm. Using these radii we can calculate the energy required to push the alpha particles from plutonium-238 back inside the nucleus.

To do this we need a few more equations from A-level physics.

As the capacitance of a sphere is given by

C = 4 π ε r

Where ε is equal to the permittivity of free space which is 8.854187817620 × 10−12 F m−1 or just 8.85 × 10−12 F m−1

We can use this to estimate the electrostatic energy required to hold an alpha particle on the surface of the uranium-234 nucleus. As soon as the alpha particle is taken out of the nucleus it is no longer being strongly bonded by the very short ranged but very strong attraction between the protons and neutrons in the nucleus (the strong force). So suddenly only the electrostatic forces apply to the system (the weak force and gravity are far far weaker)

This energy (28.5 MeV) is far greater than the decay energy of the plutonium-238 (5.593 MeV), as a result the alpha particle, what has to happen is that the alpha particle must overcome this energy barrier before it can leave the nucleus of the atom. What happens is that by quantum tunneling the alpha particle leaves the nucleus of the atom and then goes on its merry way. Here is a graph of the electrostatic energy in MeV vs the distance from the centre of the nucleus for both the alpha particle and the carbon-12 nucleus.

Electrostatic energy as a function of distance from the centre of the daughter nucleus

When the calculation is repeated for the loss of a carbon-12 nucleus from plutonium-238 to form radium-226 then I have estimated that the energy barrier is 74.5 MeV, while the decay energy is now higher at 22.5 MeV you now have a bigger barrier and some other things also help slow down the release of C-12 nuclei.

I hope to get onto these things later.

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