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A nucleus ${ }_{Z}^{A} X$ emits an $\alpha$-particle. The resultant nucleus emits a $\beta^{+}-$particle. The respective atomic and mass number of final nucleus will be
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The correct answer is:
Z-3, A-4
\begin{array}{l}
{ }_{Z}^{A} X \rightarrow_{2}^{4} H e+_{Z-2}^{A-2} Y \\
{ }_{Z-2}^{A-4} Y \rightarrow e^{+}+{ }_{Z-3}^{A-4} Y^{\prime}
\end{array}
During $\beta^{+}$ emission.
$$
{ }_{1} \mathrm{p}^{1} \rightarrow{ }_{0}^{n}{ }^{1}+\beta^{+}
$$
The proton changes into neutron. So, charge number decreases by 1 but mass number remains unchanged.
{ }_{Z}^{A} X \rightarrow_{2}^{4} H e+_{Z-2}^{A-2} Y \\
{ }_{Z-2}^{A-4} Y \rightarrow e^{+}+{ }_{Z-3}^{A-4} Y^{\prime}
\end{array}
During $\beta^{+}$ emission.
$$
{ }_{1} \mathrm{p}^{1} \rightarrow{ }_{0}^{n}{ }^{1}+\beta^{+}
$$
The proton changes into neutron. So, charge number decreases by 1 but mass number remains unchanged.
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