The activity of a radioactive material is $6.4 \times 10^{-4}$ curie. Its half life is 5 days. The activity will become $5 \times 10^{-6}$ curie after :
Solution
<p>$\because$ $A = {{{A_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}$</p>
<p>$$ \Rightarrow {2^{t/5}} = {{6.4 \times {{10}^{ - 4}}} \over {5 \times {{10}^{ - 6}}}} = 128 = {2^7}$$</p>
<p>$\Rightarrow {t \over 5} = 7$</p>
<p>$\Rightarrow t = 35$ days</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Radioactivity
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