Question

How many $\alpha$ and $\beta$ particles will be emitted by an elements ${}_{84}{A}^{218}$ in changing to a stable isotope of ${}_{82}{B}^{206}$?

• A. $3\alpha ,4\beta$
• B. $8\alpha ,6\beta$
• C. $6\alpha ,8\beta$
• D. $3\alpha ,3\beta$

Answer 1

Answer: (A)

$3\alpha ,4\beta$

Solution:- (A) $3\alpha ,4\beta$

${}_{82}{A}^{218}⟶{}_{84}{B}^{206}+\alpha {}_2{He}^4+\beta {}_{-1}{e}^0$

Comparing mass number from both sides, we have

$218=206+4\alpha +(\beta \times 0)$

$4\alpha =218-206$

$\Rightarrow \alpha =\frac{12}{4}=3$

Comparing atomic numbers from both sides, we have

$84=82+2\times \alpha +\beta (-1)$

$\Rightarrow 2\alpha -\beta =2$

$\Rightarrow b=2\left(3\right)-2=4$

Therefore,

Number of $\alpha$-particle emitted $=3$

Number of $\beta$-particles emitted $=4$