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A sample of copper-64 was used in a procedure to study copper metabolism 65 hours ago, and now 0.50 mCi remain. What was the size of the original sample? (Copper-64 has a half-life of 13 hours.)

Answer :

Final answer:

Given that the half-life of copper-64 is 13 hours, we determine that five half-lives have passed in 65 hours. By doubling the current amount for each past half-life, we find the original sample size is 16.00 mCi.

Explanation:

To find the original amount of Copper-64 in the sample, we'll use the concept of half-life. The half-life of a substance is the time it takes for half of that substance to decay.

Given that the half-life of copper-64 is 13 hours, and the elapsed time is 65 hours, we can find how many half-lives have passed. 65 hours divided by 13 hours/half-life equals 5 half-lives.

By the end of each half-life, the substance has decayed to half of its previous mass. Since 0.50 mCi remain after 5 half-lives, we backtrack the half-lives to find the initial sample size.

Double the amount for each half-life:

  1. 0.50 mCi to 1.00 mCi (one half-life)
  2. 1.00 mCi to 2.00 mCi (two half-lives)
  3. 2.00 mCi to 4.00 mCi (three half-lives)
  4. 4.00 mCi to 8.00 mCi (four half-lives)
  5. 8.00 mCi to 16.00 mCi (five half-lives)

Therefore, the size of the original sample was 16.00 mCi.

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