In an experiment on two radioactive isotopes of an element (which do not decay into each other), their number of atoms ratio at a given instant was found to be `3`. The rapidly decaying isotope has large mass and an activity of `1.0muCi` initially. The half-lives of the two isotopes are known to be `12` hours and `16` hours. what would be the activity of each isotope and their number of atoms ration after two days?

Explanation

we have ,`t=nt_(1//2),`
here , t=2 days
and after n half -life ,number of nuclei left undecyed is ,
` N=N_(0)((1)/(2))^(n)`.
for first isotope ,
`4xx2=n_(1)xx12implies n_(1)=(48)/(12)`
half -lives of first isotopes .
`N_(A)=(N_(A))_(0)((1)/(2))^(n_(1))=(N_(A))_(0)((1)/(2))^((48)/(12))=(N_(A))_(0)((1)/(2))^(4)`
forsecond isotope ,
` 24xx2=n_(2)xx16implies n_(2)=(48)/(16)`
half -lives of second isotope ,
`N_(B)=(N_(B))_(0)((1)/(2))^(n_(2))=(N_(B))_(0)((1)/(2))^((48)/(16))=(N_(B))_(0)((1)/(2))^(3)`
`therefore `ratio of half -lives of two isotope ,
`(N_(A))/(N_(B))=((N_(A))/(N_(B)))_(0)((1)/(2))=(3)/(2)`