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The number of $\alpha$ and $\beta$-particles emitted during the transformation of ${ }_{90} \mathrm{Th}^{232}$ to ${ }_{82} \mathrm{~Pb}^{208}$ are respectively
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The correct answer is:
$6,4$
${ }_{90}^{232} \mathrm{Th} \longrightarrow{ }_{82}^{208} \mathrm{~Pb}+n \alpha+m \beta$
Number of $\alpha$-particles, $n=\frac{232-208}{4}=6$
Now, for charge balanced
$$
\begin{aligned}
90 &=82+2 n-m \\
m &=4
\end{aligned}
$$
Number of $\beta$-particles, $m=4$
Number of $\alpha$-particles, $n=\frac{232-208}{4}=6$
Now, for charge balanced
$$
\begin{aligned}
90 &=82+2 n-m \\
m &=4
\end{aligned}
$$
Number of $\beta$-particles, $m=4$
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