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Depressing film

Dear Reader,

Me and my wife have been watching some disaster films recently, last night we choose to watch “On the Beach” which is a ‘oh so not so joyous film‘ about the end of the human race. The human race is snuffed in the film (and in the novel) out by very intense fallout from a nuclear war.

This made me think about it for a moment, several things were wrong with the physics and the chemistry of the film.

1. The fallout takes many weeks to arrive in Auz, looking at a paper on bomb fallout by some workers from Kyoto, Kanazawa and Hiroshima Universities it is clear that the fallout will be more than 200 times less radioactive gram for gram after ten days than it would have been 2 hours after it was deposited as ‘heavy’ local fallout very close to the bomb detonation.

So by the time that the fallout would have got down under it would have lost much of its strength. In the book on the beach it took much more than ten days for the fallout to arrive in Auz. This means that the reduction in the radioactivity due to simple decay would be even larger.

I had taken a very simple view that no chemical separation would have occurred during the transport of the radioactivity from the bomb detonation to Auz.

2. Around about 10 to 100 days after detonation the majority of the gamma dose will be due to Te-132/I-132, Ba-140 / La-140, Zr-95 / Nb-95, Ru-103 and I-131. While the iodine (I) and tellurium (Te) are very mobile the barium (Ba) and zirconium are less mobile. A good rule of thumb is that the chemical form of an isotope twenty seconds after the detonation will determine how the isotope behaves.

100 % of the zirconium-95 will be in the form of yttrium-95 (half life ten minutes) at twenty seconds after detonation. The yttrium is very non volatile and is likely to be in the form of an solid oxide at that point. I suspect that it will be very likely to become part of the larger particles from things like soil which will fall to earth more quickly. As a result the local fallout close to the bomb will be enriched in zirconium-95 while the distant fallout will have less of this isotope.

The barium-140 will be formed from cesium-140 (half life one minute) which will be the form of this isobar twenty seconds after the detonation. The cesium is only a semi volatile so about 30 % will be in a solid form.

The ruthenium-103 (Ru) is formed by the decay of molybedum-103 (Mo), as the Mo-103 has a half life of seventy seconds it will dictate much of the behaviour of the Ru-103.  The boiling point of MoO3 is about 1160 oC which is lower than that of ZrO2 and Y2O3 (both about 4300 oC) and higher than what I would expect for of Cs2O. So as a result I much of the ruthenium-103 to be lost into solid particles. So the enrichment process may tend to deposit it in the local fallout.

Now according to the graphs published by Japanese workers if all the involatile forms of the radioisotopes are lost from the fallout then the dose absorbed over the ten to one hundred days the will be only three quarters of the dose which would have been suffered if no chemical fractionation had occurred.

Both of these effects would tend to protect the population down under by lowering their dose, I hold the view that a nuclear war would be a horrible event. But I do not think that a total extermination of mankind by fallout would be a certainty.


One Response

  1. The great DX8 2XL has liked this post. By the way the Japanese paper I was reading was in Journal of Radiation Research, 2006, volume 47, A121-A127

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