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= (12.5)/(100) x_(0) }= (125)/(1000) x_(0)`

`(x)/( x_(0) ) = (125)/(1000)= (1)/(8)`

`when t=19.8 hrs, x= x_(0) e^(19.8 k), (x)/( x_(0) ) = e^(19.8 k), (1)/(8) = e^(19.8k) , ln (1)/(8) =19.8 k,`

`ln1-ln8=19.8k, 0-ln 2^(3) =19.8 k, -3 ln2=19.8k k= (-3 ln2)/(19.8) = (- ln 2)/(6.6)`
`x= x_(0) e^{ (-ln2)/(6.6)t } , (x)/( x_(0) ) = e^{ (- ln 2)/(6.6)t }`
`(1)/(2) = e^{ (- ln2)/(6.6)t } = \frac{1}{ e^{ (ln 2)/(6.6) t} } , 2= e^{ (ln2)/(6.6)t } , ln2=ln e^{ (ln 2)/(6.6)t } = (ln 2)/(6.6) t~ln e= (ln2)/(6.6) t,`