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The half lifeof a radioactivesample._38Sr^90is 28 years. Calculatethe rate ofdisintegrationof 15 mg of this isotope.Given Avogadro number= 6.023xx10^23 |
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Answer» Solution :Given : T=28 yrs w.k.t : `I=lambda = 0.693/T` i.e., `lambda=0.693/28=0.02475 yr^(-1)` but activityof the SAMPLE `A=(dN)/(dt)=lambdaN` 90 g of `Sr^90` has `6.023xx10^(23)` atoms Hence , `15xx10^(-3)g` of `Sr^90`HR .N. atoms. `therefore N=(15xx10^(-3) xx6.023xx10^23)/90` i.e.,`N=1.00xx10^20` Hence , `A=(dN)/(dt)=IN` may be obtained i.e. , `A=0.02475xx1xx10^20` `=2.475xx18` dis. `yr^(-1)` but 1 yr = 365 x 86400 =`3.15xx10^7` s Hence, `A=(2.475xx10^18)/(3.15xx10^7)` `=7.86xx10^10` dis `s^(-1)` but 1 curie = `3.7xx10^10` dis `s^(-1)` So `A=(7.86xx10^10)/(3.7xx10^10)`=2.12 CI Activityof the sample = 2.12 CU. |
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