Question

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

• A. ${}_{82}{X}^{206}$
• B. ${}_{92}{X}^{206}$
• C. ${}_{82}{X}^{210}$
• D. ${}_{82}{X}^{214}$

Answer 1

The correct option is A

${}_{82}{X}^{206}$

${}_{92}{X}^{238}\to {}_Z{X}^A$, say

Now, change in A is only because of $\alpha$ - particles, and given 8 $\alpha -decays$

$\therefore$ A = 238 - 8 $\times$ 4
= 206
and Z = 92 - (8 $\times$ 2) + 6(because of 6 $\beta$ decays)
=82
$\therefore$ ${}_Z{X}^{n}is{}_{82}{X}^{206}$