InterviewSolution
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InterviewSolution
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For a certain radioactive substance, it is observed that after `4h`, only `6.25%` of the original sample is left undeacyed. It follows that.A. the half-life of the sample is `1h`B. the mean life of the sample is `(1)/(1n2)h`C. the decay constant of the sample is `1n(2) h^(-1)`D. after a further `4h`, the amount of the substance left over would by only 0.39% of the original amount |
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Answer» Correct Answer - a,b,c,d We have, `6.25%=(6.25)/(100)=(1)/(16)` The given time of `4h` thus equals `4 half-life is `1 h. Since half-life `=(1n 2)/(decay constant) and mean life =(1)/(decay constant)`, after further `4 h`, the amount left over would be `(1)/(2^(4)) xx 1/2^(4)`, i.e., `(1)/(256)` or `(100)/(256)` or `39%` of original amount., |
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