Question

If ${}_{92}{U}^{206}$ undergoes successively 8 $\alpha$ - decays and 6 $\beta$ - decays, then resulting nucleus is
[CBSE PMT 2002]

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

Answer 1

The correct option is B

${}_{92}P{b}^{206}$

${}_{92}{U}^{238}\to {}_Z{X}^A$, 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(cos of 6 $\beta$ decays)
=82
$\because$ the Z is changed the element changes from U to Pb

$\therefore$ ${}_Z{X}^{n}is{}_{82}P{b}^{206}$

(Tip: You could have directly selected this option as all others have U as the unchanged element!)