Explanation

(c) Given $\ _{90}Th^{232} \rightarrow \ _{82}Pb^{208}$ …(i)
Change in mass number
$\ \ \ \ \ \ \ \ = 232 - 208 = 24$
No. of$\ \alpha$-particles emitted = $\frac{24}{4}$ =6
Now. Eq. (i) becomes
$\ _{90}Th^{232}\ \ \overset{\ \ \ - 6\alpha\ \ \ }{\rightarrow}\ _{78}A^{208}\overset{\ \ \ - n\beta\ \ }{\rightarrow}\ _{82}Pb^{208}$
Further change in atomic number is $82 - 78 = 4$
It means atomic no. 78 is increased by 4 to make the atomic no 82.
Therefore $6\alpha$-particles and 4$\beta - particles\ $will be emitted.