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Two radioactive substance A and B have decay constants 5 lambda and lambda respectively. At t=0 they have the same number of nuclei. The ratio of number of nuclei of nuclei of A to those of B will be (1/e)^(2) after a time interval |
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Answer» `4lambda` At t=0, `(N_0)_A =(N_0)_B` At TIME t, `N_A/N_B=(1/e)^2` According to radioactive decay, `N/N_0=e^(-lambdat)` `THEREFORE N_A/((N_0)_A)=e^(-lambda_A t)`…(i) and `N_B/((N_0)_B )=e^(-lambda_B t)` ....(ii) Divide (i) by (ii) , we GET `N_A/N_B=e^(-(lambda_A-lambda_B)t)` or `N_A/N_B=e^(-(5lambda-lambda)t)` or `(1/e)^2 = e^(-4 lambdat) ` or `(1/e)^2 =(1/e)^(4 lambdat)` `rArr 4lambdat=2 ` or `t=2/(4lambda)=1/(2lambda)` |
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