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A radioactive substance 'A' converts to stable nuclei D by following series of reaction : AtoBtoCtoD Given : t_(1//2)"for"'A'=0.0693 days t_(1//2)"for"'B'=6930 days t_(1//2)"for"'C'=6.93 days Number of nuclei of 'C' formed in the 10 days are, if initially 10^(20) nuclei of A is taken |
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Answer» `10^(18)` `"SINCE "lamda_(1)gtgtlamda_(2) ltltlamda_(3)` we can assume that all the 'A' has been converetd into 'B' in small duration Number of moles of C formed = number of number of moles of 'B' dissociated `DeltaB=lamda_(2)"N t"` `=(ln2)/(6930)xx10^(20)xx10=10^(17)]` |
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