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InterviewSolution
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Obtain the relation `N=N_(0)e^(-lamdat)` for a sample of radio active material having decay constant `lamda` where N is the number of nuclei present at instant t. Hence, obtain the relation between decay constant `lamda` and half life `T_(1//2)` of the sample. |
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Answer» According to radion active law, the rate of decay at any instant is proportional to number of atoms present at that time. i.e., `-(dN)/(dt)propN" "implies" "(dN)/(dt)=-lamdaN" "....(i)` `(dN)/(N)=-lamdadt" "impliesint(dN)/(N)=-lamdaintdt" "....(ii)` `log_(e)N=-lamdat+C" at "t=N=N_(0)` Equation (ii) `implieslogN_(0)=C` Put in equ. (ii) `logN=-lamdat+logN_(0)logN-logN_(0)=-lamdat` `log""(N)/(N_(0))=-lamdat" "implies" "(N)/(N_(0))=e^(-lamdat)` `N=N_(0)e^(-lamda)t` Furthre `t=T_(1//2),N=(N_(0))/(2)` `N=N^(0)e^(-lamdat)or(N_(0))/(2)=N_(0)e^(-lamdaT_(1//2))` `implies(1)/(2)=e^(-lamdaT_(1//2))impliese^(-lamdaT_(1//2))=2` `lamdaT_(1//2)=elog2=.693impliesT_(1//2)=(0.693)/(lamda)`. |
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