A radioacitve isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125% , (b) 1% of its original value?

Explanation

a `(A)/(A_0) = 3.125% = (3.125)/(100)`
`(A_0)/A = 100/(3.125) = 32`
`((A_0)/A) = 2^n i.e., 32 = 2^n`
`n log 2 = log 32, n = (log 32)/(log 2) = 5`
`t/T = n= 5, t = 5T`
b. `A/(A_0) = 1/100, (A_0)/A = 100 = 2^n`
`n log 2 = log 100`
`t/T = n = ((log 100)/(log 2)) , :. t = 6.65 T`.