Question

The number of $\alpha$ and $\beta$-particles emitted in the nuclear reaction are:
${}_{90}^{228}Th⟶$ ${}_{83}^{212}Bi$

• A. $4\alpha$ and $1\beta$
• B. $3\alpha$ and $7\beta$
• C. $8\alpha$ and $1\beta$
• D. $4\alpha$ and $7\beta$

Answer 1

Answer: (A)

$4\alpha$ and $1\beta$

${}_{90}^{228}Th{⟶}_{83}^{212}Bi$

$\therefore \Delta$ mass$=16$

$\Delta$ nuclear$=7$

$\therefore 4\alpha$ particle and $1\beta$ particle is needed to create a nuclear decay of 16 a.m.u and 7 nucleides.