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

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(C) For each $\alpha$ decay,the atomic number $Z$ decreases by $2$.For each $\beta^-$ decay,the atomic number $Z$ increases by $1$.Counting the emissi

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$82$

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(C) For each $\alpha$ decay,the atomic number $Z$ decreases by $2$.For each $\beta^-$ decay,the atomic number $Z$ increases by $1$.Counting the emissions from the given sequence:Total number of $\alpha$ particles emitted $= 8$.Total number of $\beta^-$ particles emitted $= 6$.Initial atomic number $Z_i = 92$.The final atomic number $Z_f$ is given by:$Z_f = Z_i - 2(	ext{number of } \alpha) + 1(	ext{number of } \beta^-)$$Z_f = 92 - 2(8) + 1(6)$$Z_f = 92 - 16 + 6$$Z_f = 82$.

$A$ nucleus with $Z = 92$ emits the following in a sequence: $\alpha, \alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \alpha, \beta^-, \beta^-, \alpha, \beta^-, \beta^-, \alpha$. The $Z$ of the resulting nucleus is

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