Question

${}_{92}^{235}U$ decays with emission of $\alpha$ and $\beta -$ particles to form ultimately ${}_{82}^{207}Pb$. How many $\alpha -$ and $\beta -$particles are emitted per atom of $Pb$ produced?

Answer 1

${}_{92}^{235}U{⟶}_{82}^{207}Pb+7\left({}_2^{4}He\right)+4\left(\beta \right)$

$\alpha \to 7$

$\beta \to 4$