A souree contains two phosphorus radio nuclides `._(15)^(33)P (T_(1//2)=14.3d)` and `._(15)^(32)P (T_(1//2)=25.3d)`. Initially, `10%` of the decays come from `._(15)^(33)P`. How long must one wait unit `90%` do so ?

Explanation

Initially the source has `90%` of `._(15)^(33)P` nuclides. After (say) t days, the source has `10%` of `.(15)^(33)P` nuclides and `90% ._(15)^(32)P` nuclides.
`therefore` Initial number of `._(15)^(32)P` nuclides `=9x and ^(33)P=x`
Final number of `.^(32)P` nuclides `=y and of ^(33)P=9y`
Now, `(N)/(N_(0))=((1)/(2))^(n)=((1)/(2))^(1//T)=2^(-1//T)`
For the `.^(32)P` isotope, `N_(0)=9x, N=y, T=14.3` days
`y=9x(2)^(-1//14.3)` ..........(1)
For the `.^(33)P` isotope, `N_(0)=x,N=9y,T=25.3` days
`9y=x(2)^(-1//25.3)` ..........(2)
`{:((1)+(2)" ",(1)/(9)=(2^(-t//14.3))/(2^(-t//25.3))):}`
`2^(-t//25.3)=81xx2^(-t//14.3`
taking log on both sides,
`(-t)/(25.3)log2=log81-(t)/(14.3) log 2`
`-t((0.3010)/(25.3))=1.9085-t((0.3010)/(14.3))`
`-0.01190t=1.9085-0.02105t`
`0.00915t=1.9085`
`t=(1.9085)/(0.00915)=208.5d`