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InterviewSolution
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The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) ` |
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Answer» We know that `lambda =0.693//T_(1//2)` Here, `T_(1//2) =3.8 day` `:. Lambda =(0.693)/(3.8)=0.182 per day` If initially `(at t=0)` the number of atoms present be `N_(0)`, then the number of atoms N left after a time given by `N =N_(0) e^(-lambda t)` `(N)/(N_(0))= e^(-lambda t)` or `(1)/(20) =e^(- lambda t)` Taking log, `lambda t=log_(e) 20 =2.3026 log_(10) 20` `:. t=(2.3026 log_(10)20)/(lambda) = (2.3026 log_(10)20)/(0.182) =16.45 days`. |
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