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Question

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

A
76
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B
78
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C
82
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D
74
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Solution

The correct option is C 78
Given A nucleons with z=92 Sequence 2α,2β1,4α,2β1,α,2β+,α Total =8α particles, 4β1,2β+particles emitted.

we know - * 1α particle decreases z by (2) * 1β+decay decreases z by (1) * 1β decay increases z by (1)zfinal =928αdecay 2β+decay +4betadecay =92162+4zf=78

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