_{72}^{180}X xrightarrow{2alpha} xrightarrow{ | vedclass

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Question

(A) The initial nucleus is $_{72}^{180}X$. Emission of $2\alpha$ particles: Each $\alpha$ particle $(_{2}^{4}He)$ reduces the atomic number by $2$ and

Vedclass · Accepted Answer

$69, 172$

Vedclass · Answer

(A) The initial nucleus is $_{72}^{180}X$. Emission of $2\alpha$ particles: Each $\alpha$ particle $(_{2}^{4}He)$ reduces the atomic number by $2$ and mass number by $4$. After $2\alpha$ emission: $Z = 72 - (2 	imes 2) = 68$ and $A = 180 - (2 	imes 4) = 172$. The nucleus becomes $_{68}^{172}P$. Emission of $1\beta$ particle: $\beta$ emission $(_{-1}^{0}e)$ increases the atomic number by $1$ and keeps the mass number constant. After $\beta$ emission: $Z = 68 + 1 = 69$ and $A = 172$. The nucleus becomes $_{69}^{172}Q$. Emission of $\gamma$ radiation: $\gamma$ emission does not change the atomic number or mass number. Final nucleus: $_{69}^{172}X'$. Therefore,$Z = 69$ and $A = 172$.

$_{72}^{180}X$ $\xrightarrow{2\alpha}$ $\xrightarrow{\beta}$ $\xrightarrow{\gamma} _{Z}^{A}X'$. The values of $Z$ and $A$ are:

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