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Derive an expression for the half-life of a radio active nuclide. |
Answer» SOLUTION :Radioactive law : The rate of disintegration of radio active ATOMS present in the SAMPLE of an element is directlyproportional to the number of radioactive atoms present at that instant.  i.e.,`(dN)/(dt) prop N` or `(dv)/(dt)=-lambdaN` where `lambda` is known as the disintegration constant .The -ve sign indicates that thenumber of radioactive nuclei/atomsdecreases with the passage of TIME. Hence,`(dN)/N=-lambdadt` Integrating both sides we get `log_e=-lambda t +C` Applying the initial conditionfor `t=0 , N=N_0` We get , `C=log_e N_0` i.e.,`log_e (N/N_0)=-lambdat` or `N/N_0=e^(-lambdat)` Hence,`N=N_0e^(-lambdat)` A graph.N. v/s time .t. gives the exponential curve as shown in the figure. |
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