Physics - Modern Physics Question with Solution | TestHub

${}_{92}^{238}\mathrm{U}{\to }_{82}^{206}\mathrm{Pb}+n_{\mathit{1}}\mathrm{\alpha }+n_{\mathit{2}}\mathrm{\beta }$

From conservation of mass number:
$238=206+4n_1$
$32=4n_1$
$\Rightarrow n_1=8$

From conservation of charge:
$92=82+2n_1-n_2$
$92=82+2\left(8\right)-n_2$
$\Rightarrow n_2=6$