Question

When ${{}_{90}T}^{228}$ transforms to ${{}_{83}Bi}^{212}$, then the number to 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 $4\alpha ,1\beta$

${{}_{Z=90}T}^{A=228}⟶{{}_{Z^{\prime }=83}Bi}^{A^{\prime }=212}$

Number of $\alpha$ particles emitted

$n_{\alpha }=\frac{A-A^{\prime }}{4}=\frac{226-212}{4}=4$

Number of $\beta$ particles emitted

$n_{\beta }=2n_{\alpha }-Z+Z^{\prime }=2\times 4-90+83=1$