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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 form `._(15)P^(35)`. How long one must wait until `90%` do so? |
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Answer» Suppose, initially the source has `90% ._(15)P_(1)^(32)` and `10% ._(15)P_(2)^(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_(1)^(32)` i.e, y gram of `P_2` and 9y gram of `P_1`. we have to calculate t. form `N/(N_0)=(1/2)^n=(1/2)^(t//T)=2^(-t//T)` `N=N_(0) 2^(-t//T) :. y=(9x) 2^(-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^(-11t//25.3xx14.3)` `log 1-log 81=(-11t)/(25.3xx14.3) log 2` `0-1.9085=(11t)/(25.3xx14.3)xx0.3010` `t=(25.3xx14.3xx1.9085)/(11xx0.3010)=208.5 days` |
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