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The half-life of Cu-61 is approximately 3 hours, meaning that after each half-life, half of the remaining sample decays. After 9 hours, which is three half-lives (3 hours x 3 = 9 hours), the original 2 mg sample would have gone through three decay cycles. Thus, the amount remaining would be (2 , \text{mg} \times \left(\frac{1}{2}\right)^3 = 2 , \text{mg} \times \frac{1}{8} = 0.25 , \text{mg}).

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1w ago

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