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

Half-life of the radioactive isotope = T years Original amount of the radioactive isotope `= N_(0)`
(a) After decay, the amount of the radioactive isotope = N It is given that only `3.125%` of `N_(0)` remains after decay. Hence, we can write :
`(N)/(N_(0))=3.125%=3.125/100=1/32`
`But (N)/(N_(0))=e^(-lambdat)` Where,
`lambda=Decay constant`
t = Time
t = Time
` -lambdat=In 1-In 32`
` -lambdat=0-3.4567`
` t=3.4567/lambda`
`Since lambda=0.693/T`
`therefore t=3.466/underset(T)0.693underset(t)=5T years`
Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.
(b) After decay, the amount of the radioactive isotope = N
It is given that only 1% of N0 remains after decay. Hence, we can write:
`N/N_(0)=1%=1/100`
`But N/N_(0)=e^(-lambdat)`
`therefore e^(-lambdat)=1/100`
`-lambdat=In 1-In 100`
`-lambdat=0-4.6052`
`t=4.6052/lambda`
`Since, lambda=0.693//T`
`therefore t=4.6052/underset(T)0.693=6.645T years`
Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.