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

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Question

(B) The change in atomic number $Z$ due to radioactive decay is as follows:$1$. An $\alpha$ particle emission decreases $Z$ by $2$.$2$. $A$ $\beta^{-}

Vedclass · Accepted Answer

$78$

Vedclass · Answer

(B) The change in atomic number $Z$ due to radioactive decay is as follows:$1$. An $\alpha$ particle emission decreases $Z$ by $2$.$2$. $A$ $\beta^{-}$ particle emission increases $Z$ by $1$.$3$. $A$ $\beta^{+}$ particle emission decreases $Z$ by $1$.Counting the particles emitted:- Number of $\alpha$ particles $= 8$- Number of $\beta^{-}$ particles $= 4$- Number of $\beta^{+}$ particles $= 2$The final atomic number $Z'$ is given by:$Z' = Z_{initial} - (8 	imes 2) + (4 	imes 1) - (2 	imes 1)$$Z' = 92 - 16 + 4 - 2$$Z' = 78$Therefore,the atomic number of the resulting nucleus is $78$.

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

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