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`._(38)^(90)Sr` की अर्द्ध आयु 28 वर्ष है। इस समस्थानिक के 15 mg के विघटन की दर क्या है?

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दिया है
`T_(1//2)=28` वर्ष `=28xx365x24xx60xx60`
`=8.83xx10^(8)s`
`lamda=0.6931/(T_(1//2))=0.6931/(8.83xx10^(8))=0.0785xx10^(-8)`
`=7.85xx10^(-10)s^(-1)`
Sr के एक मोल अर्थात 90g में `6.023xx10^(23)` परमाणु होते हैं।
अतः `15mg=15xx10^(-3)g` में परमाणुओं की संख्या
`N=(6.023xx10^(23))/90xx15xx10^(-3)`
`=1.0038xx10^(20)`
`:.` विघटन की दर `R=lamdaN`
`=7.85xx10^(-10)xx1.0038xx10^(20)`
`=7.88xx10^(10)Bq`
`=(7.88xx10^(10))/(3.7xx10^(10))Ci=2.13Ci`

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