Question

If ${}_{92}{U}^{238}$ undergoes successively $8\alpha -decays$ and $6\beta -decays$, then the resulting nucleus is

• A. ${}_{82}{U}^{214}$
• B. ${}_{82}{U}^{210}$
• C. ${}_{82}{U}^{206}$
• D. ${}_{82}{X}^{206}$

Answer 1

The correct option is D ${}_{82}{X}^{206}$

$\alpha$ particles decreases $A$ by $4$ and $Z$ by $2$.

$\beta$ particles increases $Z$ by $1$ and $A$ remains same.

Hence,

${92}^{u^{238}}\to {8a}_{u_{92}}-{(8\times 2)}^{238-(6\times 4)}={76}^{X^{206}}$

and ${76}^{X^{206}}\to {6B}_{76}+{(6\times 1)}^{X^{206}}={82}^{X^{206}}$