Question

The number of $\alpha$ and $\beta$ -particles emitted in the conversion of ${}_{90}T{h}^{232}$ to ${}_{82}P{b}^{208}$ are :

• A. $6,4$
• B. $4,6$
• C. $8,6$
• D. $6,8$

Answer 1

The correct option is A $6,4$

${}_{90}^{232}Tn{\to }_{82}^{208}Pb+n\alpha +m\beta$

The change in mass no. is completely decided by no. of $\alpha -$particles.

No. of $\alpha -$particles $=\frac{232-208}{4}=6=n$

Now for charge balance.

$90=82+2n-m$

$\Rightarrow m=4$