A radioactive isotope has a life of T years. How long will it take the activity to reduce to (a) `3.125% (b) 1%` of its original activity?

Explanation

(a) Half-life of radioactive isotope = T years.
Let `A_(0)` be the initial activity. If t is the disintegration time, then activity after time t is
`A=3.125/100 A_(0)`
Also, decay constant, `=0.6931/T"year"^(-1)`
Since, `A=A_(0)e^(-lambdat)`
`therefore rArr (3.125)/100 A_(0) = A_(0)e^(-lambdat)`
or `e^(-lambdat) = 3.125/100`
or `e^(lambdat) =100/3.125 = 32`
Taking natural log on both sides, we get
`lambdat = log_(e)32 =2.302 xx log_(10)32`
or `t=(2.302 xx log_(10)32)/0.6931`
or t=5T years.
(b) Here `A = 1/100 A_(0)`
So `e^(lambdat) =100/1`
`lambdat = log_(e)100 = 2.302 xx log_(10)100`
`t=(2.3026 xx 2T)/0.6931`
`=(4.6052 xx T)/0.6931 = 6.65` T years.