A nucleus _n{X^m} emits one alpha particle an | vedclass

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(D) When an alpha particle $({}_2^4He)$ is emitted,the mass number decreases by $4$ and the atomic number decreases by $2$. When a beta particle $(\be

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$_{n}{Z^{m - 4}}$

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(D) When an alpha particle $({}_2^4He)$ is emitted,the mass number decreases by $4$ and the atomic number decreases by $2$. When a beta particle $(\beta^-)$ is emitted,the atomic number increases by $1$ and the mass number remains unchanged. Starting with $_n^mX$: $1$. After emitting one $\alpha$ particle: $_n^mX \rightarrow {}_{n-2}^{m-4}Y + {}_2^4He$. $2$. After emitting two $\beta$ particles: ${}_{n-2}^{m-4}Y \rightarrow {}_{n-2+2}^{m-4}Z + 2_{-1}^0e = {}_n^{m-4}Z$. Thus,the resulting nucleus is $_n^{m-4}Z$.

$A$ nucleus $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is

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