If _a^b X emits a positron,then two alpha-par | vedclass

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(A) The initial nucleus is $_a^b X$. $1$. Positron emission $(_{+1}^0 \beta)$: $_a^b X \rightarrow _{a-1}^b X_1 + _{+1}^0 \beta$. $2$. Emission of two

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$c = b - 12, d = a - 5$

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(A) The initial nucleus is $_a^b X$. $1$. Positron emission $(_{+1}^0 \beta)$: $_a^b X \rightarrow _{a-1}^b X_1 + _{+1}^0 \beta$. $2$. Emission of two $\alpha$-particles $(_{2}^4 \alpha)$: $_{a-1}^b X_1 \rightarrow _{a-1-4}^{b-8} X_2 + 2(_{2}^4 \alpha) = _{a-5}^{b-8} X_2$. $3$. Emission of two $\beta$-particles $(_{-1}^0 \beta)$: $_{a-5}^{b-8} X_2 \rightarrow _{a-5+2}^{b-8} X_3 + 2(_{-1}^0 \beta) = _{a-3}^{b-8} X_3$. $4$. Emission of one $\alpha$-particle $(_{2}^4 \alpha)$: $_{a-3}^{b-8} X_3 \rightarrow _{a-3-2}^{b-8-4} Y = _{a-5}^{b-12} Y$. Comparing this with $_d^c Y$,we get $c = b - 12$ and $d = a - 5$.

If $_a^b X$ emits a positron,then two $\alpha$-particles,two $\beta$-particles,and finally one $\alpha$-particle,it transforms into $_d^c Y$. What is the correct relationship?

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