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The activity of a radioactive sample is `1.6` curie, and its half-life is `2.5 days`. Its activity after `10 days` will beA. `0.8 curie`B. `0.4 curie`C. `0.1 curie`D. `0.16curie` |
Answer» Correct Answer - c Since four half-lives have elapesed `A=(A_(0))/(2^(4)) =(A_(0))/(16)=(1.6)/(16) curie=0.1 curie`. |
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