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Half life of ""^(131)I is 8 days. Sample has an activity 6.4 mCi at a certain time. What is the activity 40 days later? |
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Answer» Solution :DATA SUPPLIED, `A_0 = 6.4 xx 10^(-3) Ci = 6.4 xx 10^(-3) xx 3.7 xx 10^(10) BQ = 2.368 xx 10^8 Bq` `LAMBDA = (0.6931)/(T_(1//2)) = (0.6931)/8 = 0.0866` per day `A_t = A_0 e^(-lambda t)` `(A_t)/(A_0) = e^(-lambda t) = 1/(2^n)` `A_t = (A_0)/(2^n) and n = (t_1)/(T_(1//2)) = 40/8 = 5` `:. A_t = (2.368 xx 10^8)/(2^5) = 7.4 xx 10^6 Bq` or `A_t = (7.4 xx 10^6)/(3.7 xx 10^(10)) = 2 xx 10^(-4) Ci = 0.2 mCi`. |
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