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

Half life of `""_(15)^(32)P,T1//2=14.3 " days"`
Half life of `""_(15)^(33)P',T1//2=25.3 " days"`
`""_(15)^(33)` nucleus decay is 10% of the total amount of decay.
The source has initially 10% of `""_(15)^(33)P` nucleus and 90% of `""_(15)^(32)P` nucleus.
Suppose after t days, the source has 10% of `""_(15)^(32)P` nucleus and 90% of `""_(15)^(33)P` nucleus.
`underline("Initially "):`
Number of `""_(15)^(33)P` nucleus =N
Number of `""_(15)^(32)P` nucleus =9N
`underline (" Finally"):`
Number of `""_(15)^(33)P` nucleus =9N'
Number of `""_(15)^(32)P` nucleus=N'
For `""_(15)^(32)P` nucleus, we can write the number ration as :
`(N')/(9N)=((1)/(2))^((t)/(T_(1//2)))`
`N'=9N(2)^((-t)/(14.3))` .....(1)
For `""_(15)^(33)P` , we can write the number ration as :
`(9N')/(N)=((1)/(2))^((t)/(T'_(a//2)))`
`9N'=N(2)^((-t)/(25.3))`....(2)
On dividing equation (1) by equation (2), we get:
`(1)/(9)=9xx2^(((t)/(25.3)-(t)/(14.3)))`
`(1)/(81)=2^-((11t)/(25.3xx14.3))`
log1-log81`=(-11t)/(25.3xx14.3)log2`
`(-11t)/(25.3xx14.3)=(0-1.908)/(0.301)`
`t=(25.3xx14.3xx1.908)/(11xx0.301)~~208.5 days`
Hence, it wall take about 20.8.5 days for 90% decay of `""_(15)P^(33)`.