Question

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

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

Answer 1

Answer: (D)

$4\alpha ,1\beta$

$\alpha$-decay

${}_Z^{A}X\underrightarrow{\alpha }{}_{Z-2}^{A-4}Y$

$\beta$-decay

${}_Z^{A}X\underrightarrow{\beta }{}_{Z+1}^{A}Y$

Decrease in atomic mass will be caused only by

$\alpha$-decay

${}_{90}^{228}Th\underrightarrow{x\alpha ,y\beta }{}_{83}^{212}Bi$

No. of alpha particles emitted (x) = $\frac{228-212}{4}$

x = 4

No. of beta particles emitted (y) = 2x-(90-83)

= 8-7 = 1.

y = 1

Hence, 4$\alpha$, 1$\beta$ are emitted in the number reaction

${}_{90}^{228}Th\to {}_{83}^{212}Bi$