Table 1: A numerical example showing the mass-dependent relations between the different reaction rates in the oxygen isotopic system. In successive columns: (1) the isotopic reaction iO+jOM $\to {}^{i}$OjOM; (2) the reduced mass of the reactants for M=5; (3) the reduced mass ratio $\Delta \mu _{\rm Y-MX}^{a }$ normalized to the reaction 16O+16OM for a=1; (4) the product of the relative abundance with $^{16}\rm O=0.9976$, $^{17}\rm O=4\times 10^{-4}$ and $^{18}\rm O=2 \times 10^{-3}$; (5) the product of $(3)\times (4)$. The 17O/16O ratio ( x'17/x'16) of the OOM molecule (noted [17O/16O] $_{\rm OOM })$ is calculated from the values of the column (5); that is: $x'_{16} = \{1/2 [4.128 \times 10^{-4} + 2.129 \times 10^{-3} $ $+ 4.070 \times 10^{-4 }+ 2.073 \times 10^{-3}] + 9.952 \times 10^{-1}\}$. Similarly for x'17 and x'18. This gives: [17O/16O] $_{\rm OOM} = (x'_{17}/x'_{16}) = 4.119 \times 10^{-4}$and [18O/16O] $_{\rm OOM} = (x'_{18}/x'_{16}) = 2.111 \times 10^{-3}$corresponding to $\delta ^{17}$O = ([ $4.119 \times10^{-4 }/ (4 \times10^{-4}/0.9976)]-1)\times1000 = +27.3\mbox{\textperthousand}$ and similarly $\delta ^{18}\rm O = +53.2\mbox{\textperthousand}$. The slope in the diagram $\delta ^{17}$O versus $\delta ^{18}$O is equal to 0.513.
[1] [2] [3] [4] [5]
16O + 16OM 9.081 $\equiv $1 0.9952 0.9952
17O + 16OM 9.395 1.0345 $3.990\times10^{-4 }$ $4.128\times10^{-4}$
18O + 16OM 9.692 1.0673 $1.995\times10^{-3 }$ $2.129\times10^{-3}$
16O + 17OM 9.263 1.0201 $3.990\times10^{-4 }$ $4.070\times10^{-4}$
17O + 17OM 9.590 1.0560 $1.600\times10^{-7 }$ $1.690\times10^{-7}$
18O + 17OM 9.900 1.0902 $8.000\times10^{-7 }$ $8.721\times10^{-7}$
16O + 18OM 9.436 1.0391 $1.995\times10^{-3 }$ $2.073\times10^{-3}$
17O + 18OM 9.775 1.0764 $8.000\times10^{-7 }$ $8.611\times10^{-7}$
18O + 18OM 10.10 1.1119 $4.000\times10^{-6 }$ $4.448\times10^{-6}$


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