I am given a table that describes an ancient spillage of spent nuclear fuel:

Pu-232  0.1150%
Pu-239  0.3730%
Pu-241  2.2900%
*Sr-90 44.3000%
U-234   0.1290%
U-235   0.0112%
U-235   0.0079%
U-238   0.0496%
Am-241  0.1920%
Cs-137 52.5000%

Time since the spill happened until measurement: 36 years

The slightly high proportion Sr/Cs could be because an extra spillage of 90Sr that happened time after in the same installation. But note the proportion of 241Pu, is it very high, isn't it?

So I think that this table shows a measurement of relative activity. How should one proceed to transform it to a table of isotope proportions?

According to notes, the spent fuel come from the SAPHIR reactor, but given that this reactor was sharing installations with the DIORIT, I wonder if it could be a mix from both. Also, the spillage did not happened in the origin lab, so some extra time, perhaps six months or one year, should be added to compare it with original "cask strength" fuel.

My first approach has been simply to use half-life to factor out the activity, then normalizing again... Then obviously I get a result where U-238 is the main component. Does it fit with the spillage of burned fuel? Or perhaps the proportions are not Becquerels but Sieverts or Grays? I wonder because, as Jacobson points out, even after correcting for activity the ratio Am/Pu seems to be out by a factor two (or perhaps it is just that it has been washed away during periodic cleaning of the room after the years)


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  • $\begingroup$ Relative activity will depend on the half-life of the given nuclei. But if they all have very long lifetimes you are probably correct. If it's a thick sample you'd need to account for the self-absorption of the gamma rays. There are computer codes to do this. $\endgroup$ – jmh Jul 6 at 19:50
  • $\begingroup$ @jmh well it is actually a sample spread in a concrete floor. And surely washed in multiple occassions during 36 years, so probably self-absortion is irrelevant, the most complicated thing could be the Americium. $\endgroup$ – arivero Jul 6 at 22:53
  • $\begingroup$ (to be clear, the spill is not there anymore, the data is from a report when the floor was finally removed) $\endgroup$ – arivero Jul 6 at 22:56
  • 1
    $\begingroup$ Pu241 decays to Am241 with a 14 year half life, so this balance of Pu241 and Am241 after 38 years seems odd. $\endgroup$ – Bob Jacobsen Jul 7 at 0:42
  • $\begingroup$ @BobJacobsen Indeed, this is one of the reasons I have to think that the table shows activity instead of weight or isotope proportions. $\endgroup$ – arivero Jul 7 at 12:43

After some inspection, I think the answer is trivial: multiply by half-life (or divide by disintegration coef) and normalize again the relative proportions.

But let me address the question of guessing first what kind of measurement the table has.

A first hint, as @BobJacobsen pointed out, is that a ratio Pu/Am tell us how old the spillage is, at most (and assuming no chemical manipulation later)

$$T={1 \over \lambda_p - \lambda_a} \ln (1 + {Am \over Pu} K_o {\lambda_p - \lambda_a \over \lambda_p})$$

with $K_o=0.9999754$ accounts for the branching ratio. So, putting numbers in: $$ T (years)= 21.414 \ln (1 + 0.96676 {Am \over Pu})$$

and the spillage would be only one or two years old at most, which doesn't fit with the hypothesis at all. Joining this calculation to the low proportion of Uranium itself, one can conclude that the table does not show a distribution of isotopes.

if, on the other hand, I rescale using the half-lifes: $${Am \over Pu} = {0.1920 \times 432\over 2.29 \times 14.35} = 2.524$$ I go to a more reasonable 26.45 years, still not inside the expected limits, but at least inside the working years of the installation, which was closed about 23 years before of the sampling.

The general rule of re-scaling with the half lives gives us a more sensical burned fuel, as for the proportion of U 238

Pu 238  0,000004%
Pu 239  0,003899%
Pu 241  0,000014%
Sr 90   0,000555%
U 234   0,013654%
U 235   3,725690%
U 236   0,080189%
U 238   96,175269%
Am 241  0,000036%
Cs 137  0,000688%

and from here we could try to recover the original proportions, if we were sure about the date.


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