A nucleus with Z = 92 emits the following in a sequence: α, α, β^-, β^-, α, α, α, α, β^-, β^-, α, β^+, β^+, α . The Z of the resulting nucleus is: - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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A nucleus with Z = 92 emits the following in a sequence: α, α, β^{-}, β^{-}, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α . The Z of the resulting nucleus is:

Solution & Step-by-step Explanation

Changes in Atomic Number Z : α -emission: Z \to Z - 2 . β^{-} -emission: Z \to Z + 1 . β^{+} -emission: Z \to Z - 1 .Sequence check:Total α : 1, 2, 5, 6, 7, 8, 11, 14 ⟹ 8 α . Loss = 8 \times 2 = 16 .Total β^{-} : 3, 4, 9, 10 ⟹ 4 β^{-} . Gain = 4 \times 1 = 4 .Total β^{+} : 12, 13 ⟹ 2 β^{+} . Loss = 2 \times 1 = 2 .Final Z = 92 - 16 + 4 - 2 = 78 .

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A nucleus with Z = 92 emits the following in a sequence: α, α, β^{-}, β^{-}, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α . The Z of the resulting nucleus is:

76

78

82

74

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