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A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by : |
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Answer» t = T log (1.3) `N_a/N_A=3 RARR N_B = 3N_A` also let initially there are total `N_0` number of nuclei `N_A+N_B=N_0` `N_A=N_0/1.3` Also as we KNOW `N_A=N_0 e^(-lambdat) , N_0/1.3=N_0e^(-lambdat)` `1/1.3=e^(-lambdat) rArr ln(1.3) =lambdat` or `t=(ln(1.3))/LAMDA` `t=(ln(1.3))/((ln(2))/T) =(ln(1.3))/(ln(2))T` |
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