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$\longrightarrow{ }_{82} \mathrm{~Pb}^{208}$. The number of $\alpha$
and $\beta$ -particles emitted during the above reaction is
Options:
and $\beta$ -particles emitted during the above reaction is
Solution:
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Verified Answer
The correct answer is:
$6 \alpha$ and $4 \beta$
No. of $\alpha$ particles $=\frac{232-208}{4}=\frac{24}{4}=6 \alpha$
No. of $\beta$ particles $=2 \times 6-(90-82)$
$$
\begin{array}{l}
=12-8 \\
=4 \beta
\end{array}
$$
Hence, $6 \alpha$ and $4 \beta$ particles are emitted.
No. of $\beta$ particles $=2 \times 6-(90-82)$
$$
\begin{array}{l}
=12-8 \\
=4 \beta
\end{array}
$$
Hence, $6 \alpha$ and $4 \beta$ particles are emitted.
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