InterviewSolution
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InterviewSolution
| 1. |
At time `t=0`, activity of a radioactive substance is `1600 Bq`, at `t=8 s` activity remains `100 Bq`. Find the activity at `t=2 s`. |
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Answer» Activity of the sample after n half n half life is `R=R_(0)((1)/(2))^(n)` Given , `R=(R_(0))/(16)` `(R_(0))/(16)=R_(0)((1)/(2))^(n)` or number of half - lives , n=4 four half are equivlent to 8 s. Hence ,2 s is equal to one half - life , so in one half - life activity will ramain half of 16000 Bq. i.e., 800 Bq. |
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