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The mean lives of a radioactive substance are 1620 year and 405 yearfor alpha - emission andbeta -emission respectively . Find the time during which three-fourth of sample will decay if it is decayed both byalpha -emission and beta -emission simultaneously. |
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Answer» Solution : The decay constant `lamda` is 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 on both sides , we get `lamdat = log_(e) 4 " or " t = 1/lamdalog_(e) 2^(2) = 2/lamdalog_(e)2` `= 2 XX 324 xx 0.693 = 449` year . |
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