Question

If $92U^{238}$ undergoes successively 8 α-decays and 6 β-decays, then resulting nucleus is.

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

Answer 1

Answer: (B)

$82Pb^{206}$

Answer 2

$(Z = 92)U^{(A = 238)}{\overset{\quad(8\alpha,6\beta)\quad}{\rightarrow}}_{Z^{'}} ⥂ X^{A^{'}}$

So $A^{'} = A - 4n_{\alpha}$= 238 – 4 × 8 = 206

and $Z^{'} = n_{\beta} - 2n_{\alpha} + z$ = 6 – 2 × 8 + 92 = 82.