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If ${ }_{92} \mathrm{U}^{238}$ emits $8 \alpha$-particles and $6 \beta$-particles, then the resulting nucleus is
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${ }_{82} \mathrm{~Pb}^{206}$
After one $\alpha$-emission, the daughter nucleus reduces in mass number by 4 unit and in atomic number by 2 unit. In $\beta$-emission the atomic number of daughter nucleus increases by 1 unit. The reaction can be written as
${ }_{92} \mathrm{U}^{238} \xrightarrow{-8 \alpha}{ }_{76} X^{206} \xrightarrow{-6 \beta}{ }_{82} Y^{206}$
Thus, the resulting nucleus is ${ }_{82} Y^{206}, i e,{ }_{82} \mathrm{~Pb}^{206}$.
${ }_{92} \mathrm{U}^{238} \xrightarrow{-8 \alpha}{ }_{76} X^{206} \xrightarrow{-6 \beta}{ }_{82} Y^{206}$
Thus, the resulting nucleus is ${ }_{82} Y^{206}, i e,{ }_{82} \mathrm{~Pb}^{206}$.
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