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The activity ofa radioactive isotope falls to 12.5% in 90 days. Calculate the half life and decay constant. |
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Answer» Solution :`N_(0)=100 N=12.5 t=90 days` DECAYCONSTANT `LAMBDA=(2.303)/(t) log (N_(0))/(N)` `=0.02558 log 8` `=2.311xx10^(-2) days^(-1)` `t^(1//2)=(0.693)/(lambda)=(0.963)/(2.311xx10^(-2))=29.99 days` |
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