A sample of `.^(18)F` is used internally as a medical diagnostic tool to look for the effects of the positron decay `(T_(1//2) =110 min)` . How long does it take for `99%` of the `.^(18)F` to decay?

Answer

`12.2 h`

Explanation

Radioactive deacy equation is
`N =N_0 e^(- lambda t)=N_0 e^(-1n(2)t//T)" "( because lambda=(In2)/(T))`
After decay of `99%` of the initial sample, only `1%` will be left, i.e., `N//N_(0)=1%`
`:. (N)/(N_(0))=(1)/(100)=e^(-1n(2)t//T)`
If we take the natural logarithm, we have
`-1n 100= -1n 2 xx(t)/(T)`
Which on solving for t yields
`t=(1 n100)/(1n2)xx T =(log 100)/(log2) xxT`
`=(2)/(0.3010) xx 10 =731 min =12.2 h`.