Question

In the nuclear reaction ${}_{90}T{h}^{232}{\to }_{82}P{b}^{208}$. The number of $\alpha$ and $\beta$ particles emitted are:

• A. $1\alpha ,4\beta$
• B. $2\alpha ,2\beta$
• C. $6\alpha ,4\beta$
• D. $8\alpha ,4\beta$

Answer 1

The correct option is C $6\alpha ,4\beta$

$6\alpha ,4\beta$ are emitted in the reaction

${}_{90}T{h}^{232}{\to }_{82}P{b}^{208}$

The number of $\alpha$ particles $=\frac{232-208}{4}=\frac{24}{4}=6$

The number of $\beta$ particles $=\frac{2\times 6-(90-82)}{1}=12-8=4$

Note:
When an alpha particle is lost, the atomic number decreases by 2 and the mass number decreases by 4. When a beta particle is lost, the atomic number increases by 1 and the mass number remains unchanged.