In the nuclear decay given below:
$$ _{Z}^{A}X \rightarrow _{Z+1}^{A}Y \rightarrow _{Z-1}^{A-4}B^* \rightarrow

Question

In the nuclear decay given below:
$$ _{Z}^{A}X \rightarrow _{Z+1}^{A}Y \rightarrow _{Z-1}^{A-4}B^* \rightarrow _{Z-1}^{A-4}B $$
The particles emitted in the sequence are:

Accepted Answer

1.  **First Decay ($X ightarrow Y$):** The mass number ($A$) remains the same, while the atomic number ($Z$) increases by 1 ($Z ightarrow Z+1$). This indicates the conversion of a neutron into a proton with the emission of an electron (beta particle). This is **Beta ($\beta^-$) decay**.
2.  **Second Decay ($Y ightarrow B^*$):** The mass number decreases by 4 ($A ightarrow A-4$) and the atomic number decreases by 2 ($(Z+1) - 2 = Z-1$). This corresponds to the emission of a helium nucleus ($_{2}^{4}	ext{He}$). This is **Alpha ($\alpha$) decay**.
3.  **Third Decay ($B^* ightarrow B$):** There is no change in either the mass number or the atomic number, but the nucleus transitions from an excited state (implied by the context of sequential decay ending in the same nucleus) to a lower energy state. This releases energy in the form of a high-energy photon. This is **Gamma ($\gamma$) decay**.
4.  **Sequence:** The order is $\beta, \alpha, \gamma$.

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