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${ }_{92}^{238} \mathrm{U}$ atom disintegrates to ${ }_{84}^{214}$ Po with a half life of $4.5 \times 10^{9}$ years by emitting six alpha particles and $\mathrm{n}$ electrons. Here $\mathrm{n}$ is -
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The correct answer is:
4
$\begin{array}{l}
{ }_{92}^{238} \mathrm{U} \rightarrow{ }_{84}^{214} \mathrm{Po}+6 \alpha+\mathrm{ne}^{-} \\
92=84+12-\mathrm{n} \\
\mathrm{n}=4
\end{array}$
{ }_{92}^{238} \mathrm{U} \rightarrow{ }_{84}^{214} \mathrm{Po}+6 \alpha+\mathrm{ne}^{-} \\
92=84+12-\mathrm{n} \\
\mathrm{n}=4
\end{array}$
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