## In This post you will get answer of Q.

The end product of decay of $_{90}Th^{232}$ is $_{82}Pb^{208}$. The number of $alpha$ and $beta$ particles emitted are respectively

Solution:

## $_{90}Th^{232} to _{82}Pb^{208}$

Mass number changes by 24 and hence 6 $alpha$ particles are emitted (as 1 $alpha$ particle emitted decreased mass number by 4) then proton number should decrease by 12, but change in proton number is by 8 and hence 4 $beta$ particle should be emitted so that proton number increased by 4