If you have 200.0 g of radioisotope with a half-life of 5 days, how much isotope would remain after 15 days?

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BlackTom AI · Accepted Answer

We start by identifying the key values: initial amount = 200.0 g, half-life = 5 days, and elapsed time = 15 days.

First, determine how many half-lives pass in the given time: 15 days / 5 days per half-life = 3 half-lives.

Each half-life reduces the remaining quantity by half. After 3 half-lives, the remaining amount is 200.0 g × (1/2)^3 = 200.0 g × 1/8 = 25.0 g.

Now evaluate the answer options:
- 13.3 g: This would correspond to a reduction factor of about 200.0 g × 0.0665, which does not match any standard half-life sequence for 3 half-lives; it’s not consistent with the given half-life model.
- 12.5 g: This equals 200.0 g × (1/2)^4.0, i.e., four half-lives, not three, so it would require 20 days elapsed, not 15.
- 40.0 g: This is 200.0 g × (1/2)^? that would correspond to less than two half-lives (for instance (1/2)^2 = 0.25 gives 50 g; (1/2)^? giving 40 g would not align with a clean integer number of half-lives under the provided half-life period).
- 25.0 g: This exactly results from applying three half-lives to the original amount, which matches the calculation 200.0 g × (1/2)^3 = 25.0 g.

Thus, the option that aligns with the correct exponential decay after three half-lives is the one yielding 25.0 g. In the context of half-life decay, the remaining quantity must be a factor of the initial amount by a power of two corresponding to the number of half-lives elapsed; here, three half-lives give the factor 1/8, producing 25.0 g.

CHE 1V_SS1 2025 Final Exam_SS1_2025_Proctored

If you have 200.0 g of radioisotope with a half-life of 5 days, how much isotope would remain after 15 days?

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