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Question

A nucleus with Z = 92 emits the following in a sequence: α, β, β, α, α, α, α, α, β, β, α, β+, β+, α,  The Z of the resultant nucleus is                                                                                                                             [AIEEE-2003] 


A
74
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B
76
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C
78
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D
82
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Solution

The correct option is D 78
From the question, there are 8 α decays, 4 β  decays and  2 β+ decays

Znew=Zold8×2+4×12×1
                           = 92 - 16 + 4 - 2
                           = 78

Physics

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