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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 :Here half-life period `T_(1/2)`=28 YEARS` = 28 XX 3.154 xx 1^(7)`s, and mass of sample m=15 mg = 0.015 g `therefore` Number of NUCLEI in given sample of `" "_(38)^(90)Sr N =(0.015xx6.023xx10^(23))/90` `therefore` Disintegration rate `abs((dN)/(DT))=lambda N=(0.6931 N)/T_(1/2)=(0.6931xx0.015xx6.023xx10^(23))/(90xx28xx3.154xx10^(7))=7.878xx10^(10)Bq =(7.878xx10^(10))/(3.7xx10^(10))C_(i)=2.13C_(i)`. |
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