Question

The number of $\alpha$ and $\beta$-particles emitted during the transformation of ${}_{90}T{h}^{232}$ to ${}_{82}{P}^{208}$ are respectively

• A. 4, 2
• B. 2, 2
• C. 8, 6
• D. 6, 4

Answer 1

The correct option is D 6, 4

${}_{90}T{h}^{232}{\to }_{82}{P}^{208}{+}_{x_2}H{e}^4{+}_{y_{-1}}{\beta }^0$
Equating mass no.
232 = 208 + 4x + 0 y or 4x = 24 or x = 6
Equating atomic no.
$90=82+2x–y$ or $90=82+2\times 6–y$ or
Hence $6\alpha$ and $4\beta$ particles will be emitted.