Question

In the case of thorium ($A=232$ and $Z=90$), we obtain an isotopes of lead $(A=208$ and $Z=82)$ after some radioactive disintegrations. The number of $\alpha$ and $\beta -$ particles emitted are, respectively

• A. $6,3$
• B. $6,4$
• C. $5,5$
• D. $4,6$

Answer 1

The correct option is B $6,4$

${}_{90}T{h}^{232}{\to }_{82}P{b}^{208}+n_1\alpha +n_2{\beta }^{-}$
Mass Balance
(mass of alpha particle =4)
$232=208+4n_1$
$n_1=6$
Now, atomic no./proton balance
$90=82+2n_1-n_2$
$n_2=4$