Q. If $\ _{92}U^{238}$ undergoes successively 8 $\alpha$-decays and 6 $\beta$-decays, then resulting nucleus is
A) $\ _{82}U^{206}$
B) $\ _{82}{Pb}^{206}$
C) $\ _{82}U^{210}$
D) $\ _{82}U^{214}$
Explanation
(b) $\ _{(Z = 92)}U^{(A = 238)}\overset{\ \ (8\alpha,6\beta)\ \ }{\rightarrow}\ _{Z^{'}}X^{A^{'}}$
So $A^{'} = A - 4n_{\alpha} = 238 - 4 \times 8 = 206$
and $Z^{'} = n_{\beta} - 2n_{\sigma} + z = 6 - 2 \times 8 + 92 = 82$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.