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In the reaction given below
${ }_{86} A^{222} \longrightarrow{ }_{84} B^{210}$
how many $\alpha$ and $\beta$-particles are emitted?
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${ }_{86} A^{222} \longrightarrow{ }_{84} B^{210}$
how many $\alpha$ and $\beta$-particles are emitted?
Solution:
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Verified Answer
The correct answer is:
$3 \alpha, 4 \beta$
${ }_{86} A^{222} \longrightarrow{ }_{84} B^{210}$
$\begin{aligned} \text { Decrease in mass number } & =222-210 \\ & =12\end{aligned}$
$\therefore$ Number of $\alpha$-particles emitted
$\begin{aligned} & =\frac{12}{4}=3 \\ { }_{86} A^{222} \longrightarrow{ }_{80} X^{210} & \longrightarrow 84 B^{210}\end{aligned}$
Increase in atomic number $=84-80=4$
$\therefore$ Number of $\beta$-particles emitted $=4$
$\begin{aligned} \text { Decrease in mass number } & =222-210 \\ & =12\end{aligned}$
$\therefore$ Number of $\alpha$-particles emitted
$\begin{aligned} & =\frac{12}{4}=3 \\ { }_{86} A^{222} \longrightarrow{ }_{80} X^{210} & \longrightarrow 84 B^{210}\end{aligned}$
Increase in atomic number $=84-80=4$
$\therefore$ Number of $\beta$-particles emitted $=4$
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