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State radioactive decay law. Derive N=N_0e^(-lambdat)for a radioactive element |
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Answer» Solution : Statement : In any RADIOACTIVE sample which undergoes `alpha m beta , GAMMA - `DECAY it is found thatthe number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. If N is the number of nuclei in the sample and `Delta N`UNDERGO decay in the time `Delta t`then `(Delta N)/(Delta t) prop N rArr (Delta N)/(Delta t) = lamda N` Where `lamda`is CALLED radioactive decay constant (or) disintegration constant The change in the number of nuclei in the sample is `dN = -Delta N` in time `Delta t ` Thus the rate of change of N is (in the limit `Delta tto 0` ) `(dN)/(dt) = - lamda N rArr (dN)/(N) = - lamda dt` Now integration on both sides `int_(N_0)^(N) (dN)/(N) = - lamda int_(t_0)^(t) dt` `ln N - ln N_0 = - lamda (t - t_0)` Here `N_0`is the number of radioactive nuclei in the sample at some arbitarary time `t_0`and N is the number of radioactive nuclei at any subsequent time t, but `t_0 = 0`, then `ln((N)/(N_0)) = - lamda t` `rArr (N)/(N_0)= e^(-lamda t)` `N = N_0 e^(-lamda t)` |
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