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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 'D' present after 6930 days are, if initially 10^(20) nuclei of A is taken |
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Answer» `10^(10)` |
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