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InterviewSolution
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The mean lives of a radioactive substance are 1620 year and 405 year for alpha- emission respectively . Find the time during which three - fourth of a sample will decay if it is decaying both by alpha - emission and beta - emission simultaneously . |
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Answer» Solution :The decay constant `lamda` is the reciprocal of the mean life `TAU ` THUS, `lamda_(alpha)=1/(1620)` per year and `lamda_(beta) = 1/(405) ` per year `:.` TOTAL decay constant , `lamda=lamda_(alpha)+lamda_(beta)` (or) `lamda=-(1)/(1620)+1/(405)=1/324` per year . We know that `N=N_(0)e^(-lamdat)` When `3/4` th PART of the sample has disintegrated , `N = N_(0)//4` `:.N_(0)/4=N_(0)e^(-lamdat)("or")e^(lamdat)=4` Taking logarithm of both sides , we get `lamdat = log_(e)4` (or) `t=(1)/lamdalog_(e)2^(2) =2/lamdalog_(e) 2 = 449` year |
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