Question

In the given radioactive disintegration series,
${}_{92}^{235}U{\to }_{82}^{207}Pb$
Calculate the difference between the number of $\alpha$ and $\beta$ particles emitted in this series.

Answer 1

${}_{92}^{235}U{\to }_{82}^{207}Pb$

As the mass is reduced from 235 to 207, there must be 7 $\alpha$ particles released as the mass is reduced by 28.

4 $\beta$ particles must also be released as the atomic number reduces only by 10 not by 14.

${}_{92}^{235}U{\to }_{82}^{207}Pb+7{}_2^4\alpha +4{}_{-1}^0\beta$

Hence the diffrence is:
7 - 4 = 3