InterviewSolution
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InterviewSolution
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The isotope 92U238 successfully undergoes three α – decays and two β – decays. What is the resulting isotope |
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Answer» Number and nature of decays = 3α and 2β Atomic number of isotope is Z = 92 Mass number of isotope is A = 238 Resulting isotope: After the emission of 3α particles atomic number Z’ = Z = (3 × 2) = Z – 6 = 92 – 6 Z’ = 86 After the emission of 2β particles atomic number Z” = Z – 2 (-1) = 86 + 2 = 88 Mass number after the emission of 3α particles A’ = A – 3α = 238 – 3(4) = 238 – 12 = 226 Mass number after the emission of 2β particles A” = A’ – 2(0) = 226 ∴ Resulting isotope is 88Ra266 |
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