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(B) For a radioactive element to form its own isotope,the atomic number $(Z)$ must remain the same,while the mass number $(A)$ changes.An $\alpha$-par

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$2 \beta$-particles and $1 \alpha$-particle

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(B) For a radioactive element to form its own isotope,the atomic number $(Z)$ must remain the same,while the mass number $(A)$ changes.An $\alpha$-particle emission decreases the atomic number by $2$ and the mass number by $4$ $(_{2}He^{4})$.$A$ $\beta$-particle emission increases the atomic number by $1$ and keeps the mass number unchanged $(_{-1}\beta^{0})$.To keep the atomic number $Z$ constant after $3$ disintegrations,we need to balance the change in $Z$: $\Delta Z = (n_{\alpha} 	imes -2) + (n_{\beta} 	imes 1) = 0$.Given $n_{\alpha} + n_{\beta} = 3$,substituting $n_{\alpha} = 1$ gives $(-2) + (2) = 0$.Therefore,the emission of $1$ $\alpha$-particle and $2$ $\beta$-particles results in the same atomic number,forming an isotope.

$A$ radioactive element forms its own isotope after $3$ consecutive disintegrations. The particles emitted are

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