If 'f' denotes the ratio of the number of nuclei decayed (Nd) to the number of nuclei at t = 0 (N0) then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as :
[$\lambda$ is the radioactive decay constant]
Solution
N = N<sub>0</sub>e<sup>$-$$\lambda$t</sup><br><br>N<sub>d</sub> = N<sub>0</sub> $-$ N<br><br>N<sub>d</sub> = N<sub>0</sub> (1 $-$ e<sup>$-$$\lambda$t</sup>)<br><br>${{{N_d}} \over {{N_0}}} = f = 1 - {e^{ - \lambda t}}$<br><br>$\Rightarrow$ ${{df} \over {dt}} = \lambda {e^{ - \lambda t}}$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
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