Let's solve both questions step by step, with explanations after each step. --- ### **First Question: Decay of 99mTc and 99Tc to 99Ru** #### **Step 1: Calculate time in hours (3 weeks)** - **1 week = 7 days** - **3 weeks = 21 days** - **1 day = 24 hours** - **Total time = 21 × 24 = 504 hours** **Explanation:** We need the total time elapsed in hours to use it in half-life calculations. --- #### **Step 2: Fraction of original 99mTc left after 504 hours** The fraction remaining after time \( t \) is: \[ \text{Fraction remaining} = \left(\frac{1}{2}\right)^{t/t_{1/2}} \] For \( ^{99m}\text{Tc} \), \( t_{1/2} = 6.03 \) hours: \[ \frac{504}{6.03} \approx 83.75 \] \[ \text{Fraction of } ^{99m}\text{Tc} = \left(\frac{1}{2}\right)^{83.75} \approx 7.2 \times 10^{-26} \] **Explanation:** Almost all \( ^{99m}\text{Tc} \) will have decayed after 3 weeks. --- #### **Step 3: Fraction of 99Tc and 99Ru** Now, \( ^{99m}\text{Tc} \) decays to \( ^{99}\text{Tc} \), which then decays to \( ^{99}\text{Ru} \) (with a much longer half-life, so it builds up slowly). The fraction of \( ^{99}\text{Tc} \) left after 504 hours: - \( t_{1/2} = 213,000 \) years = \( 213,000 \times 365.25 \times 24 = 1.87 \times 10^9 \) hours - \( \frac{504}{1.87 \times 10^9} \approx 2.7 \times 10^{-7} \) - \( \left(\frac{1}{2}\right)^{2.7 \times 10^{-7}} \approx 1 - 1.87 \times 10^{-7} \) (essentially unchanged) Nearly all the original atoms are now \( ^{99}\text{Tc} \), except for the tiny amount that has decayed to \( ^{99}\text{Ru} \) (which is negligible). So: - \( ^{99m}\text{Tc} \) fraction ≈ \( 7.2 \times 10^{-26} \) - \( ^{99}\text{Tc} \) fraction ≈ 1 (almost all nuclei) - \( ^{99}\text{Ru} \) fraction ≈ 0 (almost none, since very little \( ^{99}\text{Tc} \) has decayed) **Final Answer:** - \( ^{99m}\text{Tc} \) fraction = \( 7.2 \times 10^{-26} \) - \( ^{99}\text{Tc} \) fraction ≈ 1 - \( ^{99}\text{Ru} \) fraction ≈ 0 --- ### **Second Question: 40K Decay for 98% to Decay** #### **Step 1: Understand what fraction remains** If 98% has decayed, 2% remains. \[ \text{Fraction remaining} = 0.02 \] **Explanation:** We're looking for the time when only 2% of the original \( ^{40}\text{K} \) is left. --- #### **Step 2: Use decay formula** \[ \text{Fraction remaining} = \left(\frac{1}{2}\right)^{t/t_{1/2}} \] \[ 0.02 = \left(\frac{1}{2}\right)^{t/1.26 \text{ billion years}} \] Take logarithms: \[ \ln(0.02) = \frac{t}{1.26 \times 10^9} \ln(0.5) \] \[ -3.912 = \frac{t}{1.26 \times 10^9} \times (-0.693) \] \[ 3.912 = \frac{t}{1.26 \times 10^9} \times 0.693 \] \[ t = \frac{3.912}{0.693} \times 1.26 \times 10^9 \] \[ t \approx 5.645 \times 1.26 \times 10^9 \] \[ t \approx 7.12 \times 10^9 \text{ years} \] **Explanation:** We used the logarithm form of the decay equation to solve for time. --- #### **Step 3: Final Answer** \[ \boxed{7.1 \text{ billion years}} \] **Explanation:** It would take about 7.1 billion years for 98% of \( ^{40}\text{K} \) to decay. --- ## **Summary of Final Answers** **1. Tc/Ru decay question:** - \( ^{99m}\text{Tc} \) fraction = \( 7.2 \times 10^{-26} \) - \( ^{99}\text{Tc} \) fraction ≈ 1 - \( ^{99}\text{Ru} \) fraction ≈ 0 **2. 40K decay question:** - **Time to wait = 7.1 billion years** (for 98% decay)

The concepts used involve radioactive decay principles, primarily the half-life formula, which relates the remaining fraction of a radioactive isotope to the elapsed time. For the first question, we applied the exponential decay law to determine how much of the short-lived technetium-99m remains after several weeks, and how it transitions to longer-lived technetium-99 and ultimately to ruthenium-99. In the second question, we used logarithmic calculations of decay to find the time required for a specific percentage of potassium-40 to decay, based on its known half-life. These concepts rely on understanding exponential decay and the relationship between half-life and elapsed time.