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After 82.5 seconds, 1/8 of a radioactive sample remains. Identify the radioisotope.

A) Carbon-14 (¹⁴C).
B) Uranium-238 (²³⁸U).
C) Strontium-90 (⁹⁰Sr).
D) Technetium-99m (⁹⁹mTc).

1 Answer

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Final answer:

After 82.5 seconds, 1/8 of a radioactive sample remains, which is consistent with the decay pattern of Technetium-99m (Tc-99m), which has a half-life of 6 hours and would show significant decay within that time frame. The other options, including Carbon-14 and Uranium-238, have much longer half-lives and thus would not fit this decay pattern.

Step-by-step explanation:

The student asked which radioisotope would have only 1/8 of its original sample remaining after 82.5 seconds. To identify the radioisotope, we must consider the half-lives of the given options. It's given that Technetium-99m (Tc-99m) has a half-life of 6 hours, which means that it decays relatively quickly compared to the others listed, such as Carbon-14 with a half-life of 5730 years or Uranium-238 with a half-life of 4.5 billion years.

Since the decay of Uranium-238 occurs at a rate that is significantly slower, and it would take billions of years for only a fraction of it to decay, option B (Uranium-238) is clearly incorrect. Similarly, Carbon-14 decays too slowly for 1/8 of the sample to remain after only 82.5 seconds, eliminating option A (Carbon-14). Strontium-90's decay is not specifically given, but considering the context of vast timescales associated with Uranium-238 and Carbon-14 decays, it is unlikely.

Given that Technetium-99m decays much faster and has a half-life suitable to show significant decay within 82.5 seconds, option D (Technetium-99m) is the correct radioisotope in this case.

User Tyger Guzman
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