Journal of Creation - 2018 Volume 32 Issue 1

where λU is the effective ‘decay’ rate of 238U to 14C. Substituting this in ( 4) with NB = NC14, and rearranging gives

NU 238 = ΣtλβNC14

2.07λfσaNN14 ( 15)

where NU238, NC14 and NN14 are, respectively, the number of 238U, 14C, and 14N atoms per cm³, λf is the ‘decay’ rate for the beta decay of 14C to 14N, λf is the ‘decay’ rate for the spontaneous fission of 238U, σa is the thermal neutron absorption cross-section for 14N, and Σt is the macroscopic scattering cross-section of the coal. Dividing the numerator and denominator on the righthand side by NC12, the number of 12C atoms per cm³, gives

NU 238 = ΣtλβNC14 NC12 ⎛⎝⎜ ⎞⎠⎟

2.07λfσaNN14 NC12 ⎛⎝⎜ ⎞⎠⎟

( 16)

NN14

NC12 is the ratio of 14C to 12C in the coal, which is the actual AMS measurement in pMC, and NN14

NC12 is, effectively,

the ratio of the amount of nitrogen in the coal to the amount of carbon in the coal. Thus,

NU 238 = Σtλβ pMC 100 MC ⎛⎝⎜ ⎞⎠⎟

2.07λfσaNN14 NC12 ⎛⎝⎜ ⎞⎠⎟ ( 17)

As noted earlier, the composition of coal varies as C: 75–90%, H: 4. 5–5.5%, N: 1–1.5%, so the maximum value for NN/NC12 (i.e. the ratio that would provide the largest amount of 14N, and therefore the highest rate of 14C production) would be 1.5/75 = 0.02.

Thus, for the situation under discussion, pMC = 0.235% = 0.00235, MC = 1. 2 × 10–12 atoms of 14C to atoms of 12C, 21, 22 λβ= ln( 2)/( 5,700 × 365 × 24 × 3,600) = 3.856 × 10–12 per second, λf = ln( 2)/( 8. 27 × 1015 × 365 × 24 × 3,600) = 2. 66 × 10–24 per second, σa = 1. 91 × 10–24 cm², 23 NN 14 NC 12 = 0.02, and Σt = 0.7012 (table 1). Solving for NU238 gives NU238 = 3. 6 × 1022 atoms per cm³.

Since a mole of 238U consists of 6.022 × 1023 atoms, this corresponds to approximately 0.06 moles of 238U per cm³. A mole of 238U weighs very nearly 238 g. Therefore, 0.06 moles of 238U would weigh 14. 3 g, which would mean that the density of the ‘coal’ would be at least 14. 3 g/cm³. However, the bulk density of coal is about 1. 35 g/cm³. This means that the coal does not contain the amount of 238U required to produce the observed number of 14C atoms.

Moreover, uranium generally occurs in the form of
uraninite, a.k.a. pitchblende, which is mainly UO2. Thus
0.06 moles of uranium implies 0.06 moles of uraninite.
Since the molecular weight of oxygen is essentially 16 g, the
molecular weight of uraninite is 270 g. Thus a 1 cm³ volume
containing 0.06 moles of uraninite would have a density 16. 2
g/cm³. However, the density of pure uraninite is only about
10. 8 g/cm³. So, material containing sufficient uranium atoms
to sustain the amount of 14C observed in the coal would need
to be more than 100% pure uraninite.
Alternately, given the measured density of coal, 1 cm³ of
coal, if pure carbon, would contain 0.11 moles of carbon.
Putting this amount of carbon with the required amount
of 238U would mean that the ‘coal’ would be about 35%
238U (0.06/(0.06 + 0.11)). The richest uranium deposits in
the world (at Cigar Lake and McArthur River, both in the
Athabaskan Basin in Canada) have ore grades (uranium
concentrations) of 18% and 17% respectively. Thus, if the
levels of 14C measured in the coal were the result of the
presence of uranium, the ‘coal’ would more correctly be
called ‘top-rate uranium ore’ and mined for its uranium
content rather than its coal content.

Finally, one can take the observed concentration of uranium in coal and use this to calculate the number of 14C atoms that would be present once secular equilibrium had been achieved and compare this to the observed concentration of 14C in the coal. Coal contains about 1–2 ppm uranium. 24 Thus, per cm³ of coal, there will be about