Isotope Abundances in Godiva Criticality Calculations

 

The PFC Godiva criticality calculation shown in Example 8.4 of Volume 1, pages 259-267, used the metal density and isotopic fractions given in Los Alamos report LA-4208 (reference 8 of chapter 8) to compute the system multiplication.  The isotopic abundances presented in the Los Alamos reference are mass fractions.  Although not explicitly stated in the text, because of the way the isotope mass fractions were used in the example calculation it is clear that the assumption was made that the mass fractions and atom fractions of the uranium isotopes in Godiva are the same.  This is not quite correct.

 

For an element containing a mixture of isotopes the number density of isotope i is given by

 

            Ni = ρiN0/Ai                                                                                                     (1)

 

where ri is the partial density of isotope i in the mixture, No is Avagadro’s number, and Ai is the atomic mass of isotope i.  The atom fraction of isotope i is thus

 

            fia = Ni/∑Ni = ρi/Ai/∑(ρi/Ai)

 

The mass fraction, on the other hand, is given by

 

fim = ρi/∑ρi

 

where Sri = r is the total density of the mixture and the sum is taken over all of the isotopes of the element in the mixture.  We then have

 

            Ni = fimρN0/Ai                                                                                                  (2)

 

Comparing eqns 1 and 2 we have

 

fim = fiaAi/∑ fiaAi

 

For polyisotopic elements it is clear that the mass fractions are equal to the atom fractions only if Ai/SfiaAi = 1.  In practice the latter is approximately true for high atomic mass elements.  However, even for uranium, with Ai values very close to 235 and 238 for the dominant isotopes, the mass fractions differ slightly from the atom fractions.  The mass and atom fractions for Godiva uranium, obtained using the above equations and the data given in Example 8.4, are shown in Table 1.

 

 

 

Table 1. Atom and Mass Fractions for Godiva Uranium

 

i

fim

fia

235

0.9386

0.9393

238

0.0614

0.0607

 

 

The number density of uranium atoms in Godiva metal is given on p. 261 of Volume 1 as 0.04815 atoms/barn-cm.  This result was obtained assuming the mass fractions were equal to the atom fractions, atomic masses of exactly 235 and 238, and using 0.6023 x 1024 for Avagadro’s number:

 

N = N0ρ/A ≈ 0.6023 x 18.80/(0.9386 x 235 + 0.0614 x 238) ≈ 0.04815

 

For this number density and the mass fractions of Table 1 the calculation of the number densities shown in Table 8.33 gives

 

For U-235:  0.9386*0.04815E24 = 0.04519359E24 atoms/cm3

For U-238:  0.0614*0.04815E24 = 0.00295641E24 atoms/cm3

 

The mass percent enrichment used is thus:

 

0.04519359*235.0439/(0.04519359*235.0439 + 0.00295641*238.0508) = 93.79%

 

The calculation of a critical mass of 52.58 ± 0.44 kg in Example 8.4 is therefore the result for a mass percent enrichment of 93.79% rather than the actual 93.86% enrichment for Godiva. 

 

The “correction” for the use of 93.79% mass enrichment rather than 93.86% can be estimated by using an empirically derived relationship[1] which is that the critical mass of a bare sphere of enriched uranium is inversely proportional to the enrichment in atom percent to the 1.7th power.  Thus, since the critical mass calculated for 93.79% was 52.58 ± 0.44 kg, the correct value for 93.86% would be:

 

CritMass = 52.58*0.93861.7 / 0.93931.7 = 52.51 kg,

 

since 93.86 mass percent enrichment corresponds to 93.93 atom percent enrichment.  The statistical uncertainty remains ± 0.44 kg but this does not reflect any uncertainty resulting from the approximation in reference 1.  Thus the “correction” for increasing the enrichment from 93.79% to 93.86% is to reduce the critical mass by 0.07 kg.

 

Another correction can be calculated based on a more accurate calculation of the atomic number density for the uranium.  The figure used in the text is 0.04815E24 atoms/cm3.  At the time the Hanson-Roach cross sections were calculated, the atomic masses and Avogadro’s number were based on the “Physical Scale” whereby the mass of the oxygen-16 isotope was assumed to be exactly 16 AMU.  The data of the time (for example from Semat[2]) gave the atomic mass of U-235 as 235.12517 and the atomic mass of U-238 as 238.13232.  Avogadro’s number on the physical scale was calculated as 6.02472E23.

 

Using the above, the atomic mass of the 93.86% enriched uranium (again assuming incorrectly that the number is atom percent) would then be:

 

0.9386*235.12517 + 0.0614*238.13232 = 235.30981

 

The number density would then be:

 

18.80*6.02472E23/235.30981 = 0.048134E24

 

 

Thus the number density for the calculation in Example 8.4 should have been 0.048134E24 rather than 0.04815E24.  The effect of this can also be estimated.  As is intuitively obvious, especially from Monte Carlo calculations of criticality, the size of a critical sphere of uniform material is dependent on the (number density) times (radius), or as is often written, on ρR.  The distances can be expressed in terms of mean free paths.  The actual dimensions do not affect the criticality calculation.  For example, if the density of the material in a critical sphere were doubled, then the radius would need to be divided by two to maintain a critical configuration.  The amount of material in the critical sphere is a variable, however, since it depends on the density times the volume, which is proportional to the radius to the third power.

 

The critical radius obtained in the calculation was 8.74 cm.  With the above correction it would have been 8.7429 cm and this would have added about 0.05 kg to the critical mass.

 

Note:  The use of more current data has little effect on the calculations.  Using current data from NIST:

            The atomic mass of U-235 is 235.0439.

            The atomic mass of U-238 is 238.0508.

            Avogadro’s number is 6.0221415E23. 

The above are all based on the atomic mass of C-12 as 12.

 

Assuming 93.86% enrichment based on atom percent would give the following:

The effective atomic mass would be

 

0.9386*235.0439 + 0.0614*238.0508 = 235.2285

 

The number density would then be:

 

18.80*6.0221415E23/235.2285 = 0.04813E24

 

 

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[1] LA-1958, “Critical Masses of Fissionable Metals as Basic Nuclear Safety Data,” H.C. Paxton, Jan 1955

[2] Henry Semat, “Introduction to Atomic and Nuclear Physics,” Fourth Edition, March 1962