Chemistry - Chemical Kinetics Question with Solution | TestHub

The reaction is:

${}_{90}{\text{Th}}^{232}{\to }_{82}{\text{Pb}}^{208}+6{}_2{\text{He}}^4+4{}_{-1}{\text{e}}^0$

The number of$\mathrm{\alpha }$and$\mathrm{\beta }$particles emitted in the conversion of${}_{90}\mathrm{T}{\mathrm{h}}^{232}$to${}_{82}\mathrm{P}{\mathrm{b}}^{208}$are 6 and 4 respectively.