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InterviewSolution
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A radioactive sample has activity of 10,000 disintegrations per second after 20 hours. After next 10 hours its activity reduces to 5,000 disintegrations per second. Find out its half-life and initial activity. |
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Answer» Solution :At time `t_(1) = 20h` the ACTIVITY of radioactive SAMPLE `R_(1) = 10,000 Bq` and after next 10h i.e., at time `t_(2) = 20 + 10 = 30 H` the activity `R_(2) = 5,000 Bq=R_(1)/2`. `implies` Half-life period `T_(1/2) = t_(2) - t_(1)=10h` If initial activity be `R_(0)`, then `R_(1) = R_(0)(1/2)t_(1)/T_(1/2) implies 10,000-R_(0)(1/2)^(20/10)=R_(0)(1/2)^(2)=R_(0)/4 implies R_(0)=4xx10,000=40,000Bq` |
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