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The nucleus ${ }_n^m X$ emits one $\alpha$ particle and $2 \beta$-particles. The resulting nucleus is
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The correct answer is:
${ }_n^{m-4} X$
Emission of $\alpha$-particle decreases the mass number and the atomic number by 4 and 2 respectively. Emission of $\beta$-particle increases the atomic number by 1 while the mass number remains unchanged.
After the emission of the one $\alpha$-particle and two $\beta$-particles
Decreases in mass number $=4-0=4$
Decreases in atomic number $=2-2=0$
$\therefore$ The resulting the nucleus is $m-4, X$.
i.e. the isotope of the given nucleus.
After the emission of the one $\alpha$-particle and two $\beta$-particles
Decreases in mass number $=4-0=4$
Decreases in atomic number $=2-2=0$
$\therefore$ The resulting the nucleus is $m-4, X$.
i.e. the isotope of the given nucleus.
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