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Q.
The activity of a radioactive material is $2.56 \times 10^{-3}\, Ci$. If the half life of the material is $5$ days, after how many days the activity will become $2 \times 10^{-5} Ci$ ?
$\frac{ A }{ A _{0}}=\frac{ N }{ N _{0}}$
$\frac{2 \times 10^{-5}}{2.56 \times 10^{-3}}=\frac{ N }{ N _{0}} $
$\frac{ N }{ N _{0}}=\frac{1}{128} \Rightarrow N =\frac{ N _{0}}{128}$
After $7$ half life activity comes down to given value $T =7 \times 5$
$=35 $ days