Natural source Cu concentration: [tex]4.38 \times 10^{-7} \text{ g/ml}[/tex]

Enriched source Cu concentration: [tex]5.12 \times 10^{-7} \text{ g/ml}[/tex]

Experimental R: 0.212 (unitless)

Experimental K: 0.944 (unitless)

Volume of spike used: 1.5 ml

Assume the enriched copper solution is 99.9 mol% Cu-65. Calculate the concentration in the original unknown by the isotope dilution method. Express your answer in ppb.

A. 4.5 ppb
B. 5.0 ppb
C. 5.5 ppb
D. 6.0 ppb

By using the isotope dilution method, the concentration of the original unknown solution can be calculated. First, calculate the fraction of enriched copper in the spike solution, which is 0.999 (99.9 mol% is equivalent to 0.999 fraction). Then, use the formula for calculating concentration: Original concentration = (Enriched concentration * R0 * K) / (Fraction of enriched copper * (1 - R0)). Substituting the given values, we get Original concentration = (5.12x10-7 * 0.212 * 0.944) / (0.999 * (1 - 0.212)). After the calculation, the original concentration is found to be 2.667x10-7 g/ml. Now, to express this concentration in ppb, convert it to μg/L and then to ppb. Since 1 ppb = 1 μg/L for dilute solutions, the concentration is 2.667 ppb. Thus, the correct answer is b) 5.0 ppb.