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A source contains two phosphorous radio nuclides ._(15)^(32)P (T_(1//2)=14.3d) and ._(15)^(33)P (T_(1//2)=25.3d). Initially, 10% of the decays come from ._(15)^(33)P. How long one must wait until 90% do so ? |
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Answer» Solution :Suppose INITIALLY the source has 90% `._(15)Pt^(32)` and 10% `._(15)P_(1)^(32)`, say 9x gram of `P_(2)` and x gram of `P_(1)`. After t days, suppose the source has 90% `._(15)P_(2)^(33)` and `10 % ._(15)P_(t)^(32)` i.e., y gram of `P_(2)` and 9y gram of `P_(1)` we have to calculate : from `(N)/(N_(0))=((1)/(2))^(n)=((1)/(2))^(t//T)=2^(-i//T)` `N=N_(0)2^(-t//T)` `y=9x2^(-t//14.3` for `P_(2)` and `9y=x 2^(-t//25.3)` for `P_(1)` Dividing we GET `(1)/(9)=9xx2(t//25.3-t//14.3)` or `(1)/(81)=2^(-11 t//25.3)xx14.3` log 1 - log `81 = (-11 t)/(25.3xx14.3)` log 2 `0-1-9085=(-11 t)/(25.3xx14.3)xx0.3010`. `t=(25.3xx14.3xx1.9085)/(11xx0.3010)=208.5` days. |
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