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 . Then Z of the resulting nucleus is

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 . Then Z of the resulting nucleus is, 76, 78, 82, 74

MARKS App · Accepted Answer

No. of α particles emitted = 8 , No. of β − particles emitted = 4 , No.of β + particles emitted = 2 z = 92 − 2 × 8 + 4 − 2 = 78

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