Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as :
(where $\lambda$ is the decay constant)
Solution
${{3{N_0}} \over 4} = {N_0}{e^{ - \lambda {t_1}}}$<br><br>${{{N_0}} \over 2} = {N_0}{e^{ - \lambda {t_2}}}$<br><br>$\ln (3/4) = - \lambda {t_1}$ ..... (i)<br><br>$\ln (1/2) = - \lambda {t_2}$ ..... (i)<br><br>$\ln (3/4) - \ln (1/2) = \lambda ({t_2} - {t_1})$ ....(i)<br><br>$\Delta t = {{\ln (3 /2)} \over \lambda }$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
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