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 . What is the atomic number of the resulting nucleus?

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Correct answer: 78

The nucleus emits 8 &alpha; particles i.e., 8 2 He 4 &there4; Decrease in Z = 8 &times; 2 = 16 Four &beta; - particles are emitted i.e., 4 - 1 &beta; 0 &there4; Increase in Z = 4 &times; 1 = 4 2 positrons are emitted i.e., 2 1 &beta; 0 &there4; Decrease in Z = 2 &times; 1 = 2 &there4; Z of resultant nucleus = 92 - 16 + 4 - 2 = 78.

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 . What is the atomic number of the resulting nucleus?

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