The half-life of radioactive Technetium-99 is 6.0 hours.Find the number of ⁹⁹Tc nuclei necessary to produce a sample of activity 1.9 µCi.(1 Ci = 3.7 × 10¹⁰ decays/second) A)5.9E-2 B)2.2E+9 C)1.5E+9 D)6.1E+5 E)4.2E+5

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Quizplus · Accepted Answer

The activity (A) of a radioactive sample is given by the formula $A = \lambda N$, where $A$ is the activity in decays per second, $\lambda$ is the decay constant, and $N$ is the number of radioactive nuclei. The decay constant $\lambda$ can be calculated from the half-life ($T_{1/2}$) using the formula $\lambda = \frac{\ln(2)}{T_{1/2}}$. Given that the half-life of Technetium-99 is 6.0 hours, we first convert this to seconds: $6.0 \, \text{hours} \times 3600 \, \text{seconds/hour} = 21600 \, \text{seconds}$. Then, we calculate the decay constant: $\lambda = \frac{\ln(2)}{21600} \, \text{seconds}^{-1}$.Given an activity of 1.9 µCi, we convert this to decays per second: $1.9 \, \mu\text{Ci} \times 3.7 \times 10^{10} \, \text{decays/second/Ci} = 7.03 \times 10^4 \, \text{decays/second}$.Now, we can solve for $N$: $N = \frac{A}{\lambda} = \frac{7.03 \times 10^4}{\ln(2)/21600}$.Plugging in the values, we find $N \approx 2.2 \times 10^9$, which corresponds to choice B.

The Half-Life of Radioactive Technetium-99 Is 6