Question

When ${}_{90}T{h}^{228}$ transforms to ${}_{83}B{i}^{212}$, then the number of the emitted $\alpha$ -and $\beta$ -particles is, respectively

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

Answer 1

The correct option is D $4\alpha ,1\beta$

${}_{Z=90}T{h}^{A=228}{\to }_{Z=83}B{i}^{A^{\prime }=212}$
Number of $\alpha$ -particles emitted $n_{\alpha }=\frac{A-A^{\prime }}{4}=\frac{228-212}{4}=4$
Number of $\beta$ -particles emitted $n_{\beta }=2n_{\alpha }-Z+Z^{\prime }=2\times 4-90+83=1$