Question

The activity of a radioactive isotope falls to $12.5\%$ in $90days$. Compute (lie half life and decay constant of isotope.

Answer 1

$Ao=100$, $A=12.5$, $t=90$ days

$t1/2=?$

$K=?$

$90=\frac{2.303\times t1/2}{0.693}\times log\left(\frac{100}{12.5}\right)$

$t1/2=30$ days

$K=\frac{0.693}{30}=0.0231{days}^{-1}$