What is the half-life period of a radioactive material if its activity drops to $1 / 16^{\text {th }}$ of its initial value in 30 years?
Solution
<p>$\because$ $A = {{{A_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$</p>
<p>$\Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{A_0}} \over A} = 16$</p>
<p>$\Rightarrow {t \over {{T_{1/2}}}} = 4$</p>
<p>$\Rightarrow {{30} \over {{T_{1/2}}}} = 4$</p>
<p>$\Rightarrow {T_{1/2}} = {{30} \over 4}$</p>
<p>$= 7.5$ years</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
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