A nucleus \( {}^{Q}_{P}X \) emits one alpha \( (\alpha) \) and two beta \( (\beta^{-}) \) particles. The resulting nucleus will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( {}^{Q-4}_{P}X \)
In radioactive decay the mass number (A) and charge number (Z) are conserved. Here the nucleus has A = Q and Z = P.
\( \alpha \text{ decay } ({}^{4}_{2}\mathrm{He} \text{ emitted}): \ A \rightarrow A - 4,\ \ Z \rightarrow Z - 2 \)
\( \beta^{-} \text{ decay } ({}^{0}_{-1}e \text{ emitted}): \ A \text{ unchanged},\ \ Z \rightarrow Z + 1 \)
One alpha and two \( \beta^{-} \) emissions:
\( A = Q - 4 \)
\( Z = P - 2 + 2(1) = P \)
\( {}^{Q}_{P}X \rightarrow {}^{Q-4}_{P}X + {}^{4}_{2}\mathrm{He} + 2\,{}^{0}_{-1}e \)
The resulting nucleus has the same atomic number as the parent, so it is an isotope of X.