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in Physics by (91.5k points)
A nucleus with Z =92 emits the following in a sequence:
`alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(+),alpha`. The Z of the resulting nucleus is
A. 74
B. 76
C. 78
D. 82

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1 Answer

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by (91.5k points)
Correct Answer - C
`Z_("resulting nucleus")=92-8xx2+4xx1-2xx1=78`

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