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(a) Derive the law of radioactive decay N=N_(0)e^(-lamdat) (b) The half- life of ""_(92)^(238)U undergoing alpha - decay is 4.5 xx10^(9) years. Find its mean life. (c) What fraction of the initial mass of a radioactive substance will decay in five half - life periods ? |
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Answer» Solution :(a) According to the law of radioactive decay , we know that rate of disintegration of a radioactive sample is directly proportional to the actual quantity of that material at that instant. Mathematically , `(-dN)/(dt)=lamdaN` where `lamda` is the disintegration (or decay) constant of given radioactive substance `:. (dN)/N=lamda dt ` On INTEGRATION , we have `int_(N_0)^(N_t)(dN)/N=-lamdaint_(0)^(t)dtimplies[log_eN]_(N_0)^(N_t)=-lamda[t]_0^(t)` `implies log_(e) N_(t)-log_(e)N_(0)=-lamdat" or " log_(e)(N_t/N_0)=-lamdat` or `N_t/N_0=e^(-lamdat)impliesN_t=N_(0)e^(-lamdat)` (b) As half life period of `""_(92)^(238)U,T_(1/2)=4.5xx10^9` years `:.` Mean life `tau=(T_(1/2))/(0.693)=(4.5xx10^9)/(0.693)=6.5xx10^(9)` years (c) FRACTION of radioactive subsance LEFT intact after n = 5 half LIVES is = `[1/2]^n=[1/2]^(5)=1/32` `:.` Fraction of the initial mass of a radioactive substance decayed in 5 half lives `=1-(1)/32 = 31 / (32)` or 96.875% |
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