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

Question

Dr. Abubacker · Accepted Answer

(b) Since, 8$\alpha - particles\ \ 4\beta$$^{—}$particles are emitted, and $2\beta$$^{+}$ particles are emitted so new atomic number.
$Z’ = \ Z - 8\  	imes \ 2\  + \ 4\  	imes 1\  - \ 2 	imes 1$
$= 92 - 16\  + \ 4 - 2$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 92 - 14$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 78$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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