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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 ? |
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Answer» Solution :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` |
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