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A sample of copper-64 was used in a procedure to study copper metabolism 52 hours ago, and now 3.50 mCi remain. What was the size of the original sample? (Copper-64 has a half-life of 13 hours.)

Answer :

Final answer:

The original sample size of Copper-64, given it has a half-life of 13 hours and 52 hours have passed from original time leaving 3.50 mCi, was 56 mCi. This is calculated by doubling the remaining amount for each half-life that has passed.

Explanation:

In order to solve this question, we need to understand that the half-life of a radioactive isotope, in this case copper-64, is the time it takes for half of the isotope in a sample to decay. Given that the half-life of copper-64 is 13 hours, and 52 hours have passed, we have had 4 half-lives (52 hours / 13 hours per half-life).

Beginning with the remaining amount of copper-64, we can calculate the original amount by doubling for each half-life that has passed (since each half-life means there was twice as much present before it). So, starting with 3.50 mCi, doubling for 4 half-lives gives us 3.50 * 2^4 = 56 mCi. Therefore, the original sample size was 56 mCi.

This is a practical application in understanding copper metabolism, as copper-64 is a radioactive isotope of copper that can be used in medical testing to follow the path of copper in the body.

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