In the radioactive decay of X into Y as shown | vedclass

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(C) In a $\beta$-decay process,the mass number $(m)$ remains constant while the atomic number $(Z)$ increases by $1$ for each $\beta$-particle emitted

Vedclass · Accepted Answer

$_{9}{Y^{14}}$

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(C) In a $\beta$-decay process,the mass number $(m)$ remains constant while the atomic number $(Z)$ increases by $1$ for each $\beta$-particle emitted.The initial nucleus is $_{6}{X^{14}}$.After the emission of $3\beta$-particles:$1$. The mass number remains $14$,so $m = 14$.$2$. The atomic number changes as $Z = 6 + 3 = 9$.Therefore,the resulting nucleus is $_{9}{Y^{14}}$.

In the radioactive decay of $X$ into $Y$ as shown below,what is the value of $_Z{Y^m}$?
$_{6}{X^{14}} \xrightarrow{-3\beta} {_{Z}}{Y^{m}}$

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In the radioactive decay of $X$ into $Y$ as shown below,what is the value of $_Z{Y^m}$?$_{6}{X^{14}} \xrightarrow{-3\beta} {_{Z}}{Y^{m}}$