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A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125%, (b) 1% of its original value ? |
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Answer» Solution :(a) Here `T_(1/2)` =T YEARS and `R/R(0)`=3.125% = `3.125/100=1/32` Let after n half-lives the activity is reduced to R. Then, we have `(1/2)^(n)=R/R_(0)=1/32 implies n=5` HENCE, after a time 5T years, the activity is reduced to 3.125% of its original activity. (b) Let after n half-lives the activity becomes 1% of its original value, `THEREFORE (1/2)^(n)=R/R_(0)=1/100 implies 100=(2)^(n) or LOG2 implies n=(log100)/(log2)=2.000/0.3010=6.65` `therefore time " "t=6.65T_(1/2)=6.65 T` years. |
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