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A source contains two phosphorus radio nuclides " "_(15)^(32)P(T_(1/2) = 14.3 d) and " "_(15)^(33)P(T_(1/2) = 25.3 d). 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 :Here at t=0 if activity of `" "_(15)^(32)P` be `R_(0)` and `R_(0)^(.)` that of `" "_(15)^(33)P` then it is given that `R_(0)`=10% of (`R_(0) + R_(0)^(.))` i.e., `R_(0)^(.) =9 R_(0) or R_(0)/R_(0)^(.)= 1/9` Let after a time of t days, new activities of `" "_(15)^(32)P` and `" "_(15)^(33)P` be R and R. such that R=9 R.. But from RADIOACTIVE decay law `R= R_(0)e^(-lambdat) and `R. = R_(0)^(.)e^(-LAMBDA^(.)t). `therefore R/(R.)=R_(0)/R_(0)^(.)e^(-lambdat)/e^(-lambda^(.)t)=R_(0)/R_(0)^(.)e^((lambda^(.)-lambda)t)=R/R_(0)^(.).e^(-(0.693/T_(1/2)^(.)-0.693/T_(1/2))t)` Substituting the values, we have `9=1/9.e((0.693/14.3-0.693/25.3)t) or 81=e^((0.693/14.3-0.693/25.3)t) or log_(e)81=(0.693(25.3-14.3))/(14.3xx25.3).t IMPLIES t=(log_(e)81xx14.3xx25.3)/(0.693xx11)=(2.303log_(10)81xx14.3xx25.3)/(0.693xx11)=204.2d`. |
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