InterviewSolution
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InterviewSolution
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The half-life of the radioactive nucleus `._(86)^(226)Ra` is `1.6 xx10^(3)yr`. If a sample contains`3.0 xx10^(16)` such nuclei, determine the activity at this time. |
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Answer» First, let us convert the half -life to seconds. `T_(1//2)=(1.6xx10^(3) years) (3.16xx16xx10^(7)s//years)` `=5.0 xx 10^(10) s` `because lambda=(0.693)/(T^(1//2))=(0.693)/(5.0xx10^(10)s)=1.4 xx10^(-11)s^(-1)` we case calculate the activity of the sample at `t=0` using `R_(0) =lambda N_(0)` , where `R_(0)` is the decay rate at` t=0` and `N_(0)` is the number of radioactive nuclei present at `t=0`. `R_(0) = lambda N_(0)=(1.4 xx 10^(-11) s^(-1))(3.0 xx 10^(16))` `=4.1 xx10^(5) decays s^(-1)` Beacuse `Ci=3.7 xx10^(10) decays s^(-1)` , the activity or decay rate at `t=0` is `R_(0) =11.1 mu Ci` . |
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