A nucleus with Z = 92 emits the following in | vedclass

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(C) The initial atomic number is $Z = 92$.Each $\alpha$ decay decreases $Z$ by $2$.Each $\beta^-$ decay increases $Z$ by $1$.Each $\beta^+$ decay decr

Vedclass · Accepted Answer

$78$

Vedclass · Answer

(C) The initial atomic number is $Z = 92$.Each $\alpha$ decay decreases $Z$ by $2$.Each $\beta^-$ decay increases $Z$ by $1$.Each $\beta^+$ decay decreases $Z$ by $1$.Counting the emissions:- $\alpha$ particles: $8$ (emitted at positions $1, 4, 5, 6, 7, 8, 11, 14$)- $\beta^-$ particles: $4$ (emitted at positions $2, 3, 9, 10$)- $\beta^+$ particles: $2$ (emitted at positions $12, 13$)Calculation: $Z_{final} = 92 - (8 	imes 2) + (4 	imes 1) - (2 	imes 1) = 92 - 16 + 4 - 2 = 78$.

$A$ 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

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