A radioactive nucleus _{92}X^{235} decays to | vedclass

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(A) The decay process can be analyzed by looking at the change in atomic number $(Z)$ and mass number $(A)$.Initial nucleus: $_{92}X^{235}$.Final nucl

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One alpha and one electron

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(A) The decay process can be analyzed by looking at the change in atomic number $(Z)$ and mass number $(A)$.Initial nucleus: $_{92}X^{235}$.Final nucleus: $_{91}Y^{231}$.Change in mass number $\Delta A = 235 - 231 = 4$.Change in atomic number $\Delta Z = 92 - 91 = 1$.An alpha particle $(_{2}He^{4})$ emission reduces the mass number by $4$ and atomic number by $2$. After alpha emission,the nucleus becomes $_{90}X^{231}$.To reach $_{91}Y^{231}$ from $_{90}X^{231}$,the atomic number must increase by $1$ while the mass number remains constant. This corresponds to the emission of a beta particle (electron,$_{-1}e^{0}$).Thus,the process is: $_{92}X^{235} \rightarrow _{90}X^{231} + _{2}He^{4} \rightarrow _{91}Y^{231} + _{-1}e^{0} + _{2}He^{4}$.

$A$ radioactive nucleus $_{92}X^{235}$ decays to $_{91}Y^{231}$. Which of the following particles are emitted?

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