A source contains two phosphorus radionuclides `._(15)P^(35) (T_(1//2)=14.3"days")` and `._(15)P^(33) (T_(1//2)=25.3"days")`. Initially, `10%` of the decays come from `._(15)P^(35)`. How long one must wait until `90%` do so?

Explanation

Let `R_(01)` and `R_(02)` be initial activities of `""_(15)P^(33)` and `""_(15)P^(32)`, respectively.
Let `R_(1)` and `R_(2)` be activities at time t, then
Total initial activity `=R_(01) + R_(02)`
Total activity at time `t=R_(1) + R_(2)`
Given, `R_(01) = 10%` of total activity
or `R_(01) = 10/100 xx (R_(01) + R_(02))`
or `R_(02) = 9R_(01)` .............(i)
At time t, `R_(1)=90/100(R_(1)+R_(2))`
`=9/10(R_(1) +R_(2))`
or `R_(2) =R_(1)/9`...........(ii)
Dividing eqn. (ii) by eqn. (i), we get
`R_(2)/R_(02) =R_(1)/(81R_(01))`
`R=R_(0)e^(-lambdat)`
where `lambda` is the decay constant.
`therefore (R_(02)e^(-lambda2t))/(R_(01)e^(-lambda1t))=R_(02)/(81R_(01))`
Since, `lambda = 0.693/T_(1//2)`
So, `lambda =0.693/14.3 "day"^(-1), (T_(1//2)=14.3 d)` and `lambda_(1)= 0.693/25.3 "day"^(-1)`
So equation (iv) becomes
`[(0.693)/14.3 - 0.693/25.3]t = 2.5303 log 81`
or `t=2.303 xx 1.9085 (25.3 xx 14.3)/(0.693(25.3-14.3))`
or `t=208.6` days =209 days.