A nucleus with Z = 92 emits alpha, alpha, bet | vedclass

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Question

(B) The initial atomic number is $Z = 92$.An $\alpha$-decay $(_{2}He^{4})$ decreases $Z$ by $2$.$A$ $\beta^-$-decay $(_{-1}e^{0})$ increases $Z$ by $1

Vedclass · Accepted Answer

$78$

Vedclass · Answer

(B) The initial atomic number is $Z = 92$.An $\alpha$-decay $(_{2}He^{4})$ decreases $Z$ by $2$.$A$ $\beta^-$-decay $(_{-1}e^{0})$ increases $Z$ by $1$.$A$ $\beta^+$-decay $(_{+1}e^{0})$ decreases $Z$ by $1$.Counting the particles:- Total $\alpha$ particles: $8$ (emitted at steps $1$,$2$,$5$,$6$,$7$,$8$,$11$,$14$).- Total $\beta^-$ particles: $4$ (emitted at steps $3$,$4$,$9$,$10$).- Total $\beta^+$ particles: $2$ (emitted at steps $12$,$13$).Calculation of final $Z$:$Z_{final} = Z_{initial} - (8 	imes 2) + (4 	imes 1) - (2 	imes 1)$$Z_{final} = 92 - 16 + 4 - 2$$Z_{final} = 78$.

$A$ nucleus with $Z = 92$ emits $\alpha, \alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \alpha, \beta^-, \beta^-, \alpha, \beta^+, \beta^+, \alpha$ particles in sequence. What is the $Z$ of the resulting nucleus?

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