13.17 Radioactive isotopes decay according to first-order kinetics. For one particular isotope, 1.00 nmole registers \( 1.2 \times 10^{6} \) decays in 1.00 min. (a) How many decays will occur in 1.00 min if 5.00 nmoles of this isotope are present? (b) What fraction of the isotope decays per minute in each case? (c) Explain the relationship between your answers to (a) and (b).
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If 1.00 nmole of the radioactive isotope produces \( 1.2 \times 10^{6} \) decays in 1.00 minute, then for 5.00 nmoles, you simply multiply that number by 5. Thus, the expected decays in 1.00 minute for 5.00 nmoles will be \( 5.00 \times 1.2 \times 10^{6} = 6.0 \times 10^{6} \) decays. To find the fraction that decays per minute in each case, we take the number of decays and divide it by the total number of isotopes present. For 1.00 nmole, the fraction decaying is \( \frac{1.2 \times 10^{6}}{1.00 \text{ nmole}} \). For 5.00 nmoles, it will remain the same since decay rate is independent of the amount, yielding the same fraction. The relationship between the total decays in (a) and the decay fraction in (b) highlights a core concept of first-order kinetics: while the absolute number of decays increases with the amount of isotope, the decay fraction remains constant, reflecting that each individual isotope has an equal likelihood of decaying regardless of the quantity present.