Question

Atomic mass number of an element is $232$ and its atomic number of $90$. The end product of this radioactive element is an isotopes of lead (atomic mass $208$ and atomic number $82$). The number of $\alpha$ and $\beta -$ particles emitted are :

• A. $\alpha =3,\beta =3$
• B. $\alpha =6,\beta =4$
• C. $\alpha =6,\beta =0$
• D. $\alpha =4,\beta =6$

Answer 1

The correct option is B $\alpha =6,\beta =4$

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