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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 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
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Solution
The correct option is B 78
For each α emission z decreases by 2; For each β+ emission, z decreases by 1 and that for each β− it increases by 1.No of α emitted =8
No of β+ emitted =2
No of β− emitted =4
∴ The resulting nucleus has atomic number z=z(original)−2(8)−2(1)+1(4)
⟹z=92−16−2+4=78
For each α emission z decreases by 2;
For each β+ emission, z decreases by 1 and that for each β− it increases by 1.
No of α emitted =8
No of β+ emitted =2
No of β− emitted =4
∴ The resulting nucleus has atomic number z=z(original)−2(8)−2(1)+1(4)
⟹z=92−16−2+4=78
⟹z=92−16−2+4=78
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