A nucleus with Z = 92 emits the followi | vedclass

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Question

For each $\alpha$ emission $z$ decreases by 2 ;

For each $\beta^{+}$emission, $z$ decreases by 1 and that for each $\beta^{-}$it increases by 1 .

Vedclass · Accepted Answer

$78$

Vedclass · Answer

For each $\alpha$ emission $z$ decreases by 2 ;

For each $\beta^{+}$emission, $z$ decreases by 1 and that for each $\beta^{-}$it increases by 1 .

No of $\alpha$ emitted $=8$

No of $\beta^{+}$emitted $=2$

No of $\beta^{-}$emitted $=4$

$	herefore$ The resulting nucleus has atomic number $z = z$ (original) $-2(8)-2(1)+1(4)$

$\Rightarrow z=92-16-2+4=78$

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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