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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 resulting nucleus is ______.

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Solution

There are 8 α emission, 4 β emission and 2 β+ emission.

Each α emission decreases Z by 2.

Each β emission increases Z by 1.

Each β+ emission decreases Z by 1.

Z=Z8(2)+42=9214

Z=78

Correct answer:78
Why this Question?
This problem involves 3 types of particle emission and gives the better understanding of α,β and β+ decay.

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