Question

What is the half-life of an isotope that decays to 12.5% of its original activity in 19.8 hours?

Answer 1

$x= x_{0} e^{kt} ,where~ x ~is~ the ~ |amount\ of \material\ at\ any~~time~t$
$and~ x_{0} ~the~original~amount.$
$when~x= \frac{12.5}{100} x_{0} }= \frac{125}{1000} x_{0}$
$\frac{x}{ x_{0} } = \frac{125}{1000}= \frac{1}{8}$
$when t=19.8 hrs, x= x_{0} e^{19.8 k}, \frac{x}{ x_{0} } = e^{19.8 k}, \frac{1}{8} = e^{19.8k} , ln \frac{1}{8} =19.8 k,$
$ln1-ln8=19.8k, 0-ln 2^{3} =19.8 k, -3 ln2=19.8k k= \frac{-3 ln2}{19.8} = \frac{- ln 2}{6.6}$$x= x_{0} e^{ \frac{-ln2}{6.6}t } , \frac{x}{ x_{0} } = e^{ \frac{- ln 2}{6.6}t }$$\frac{1}{2} = e^{ \frac{- ln2}{6.6}t } = \frac{1}{ e^{ \frac{ln 2}{6.6} t} } , 2= e^{ \frac{ln2}{6.6}t } , ln2=ln e^{ \frac{ln 2}{6.6}t } = \frac{ln 2}{6.6} t~ln e= \frac{ln2}{6.6} t,$