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The half-life of ""_(38)^(90)Sr is 28 years. What is the disintegration rate of 15 mg of this isotope? |
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Answer» Solution :`T_(1//2) = 28 "YEARS" = 28 XX 365 xx 24 xx 60 xx 60 s = 8.83 xx 10^8 s` `(dN)/(DT) = lambda N = (0.693)/(T_(1//2)) xx N = (0.693 xx 15 xx 6.023 xx 10^(23))/(8.83 xx 10^8 xx 90)` `= 7.877 xx 10^(10) BQ = (7.877 xx 10^(10))/(3.7 xx 10^(10)) = 2.1289 Ci`. |
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