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A radio active isotope has a half-life of T years. How long will it take the activity to reduce to 1% of its original value ?

Answer» Here `(N)/(N_(0))=(1)/(100)`
From `(N)/(N_(0))=e^(-lambda t)=(1)/(100)`
`- lambda t=log 1- log_(e )^(100)`
`= 0-2.303 log_(10)^(100)= - 2.203xx2`
`= - 4.606`
`t = (4.606)/(lambda)=(4.606)/(0.696//T)=6.65 T`


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