If `._92 U^238` undergoes successively `8 alpha-`decays and `6 beta-`decays, then resulting nucleus is.

Options

A. `._82 U^206`
B. `._82 Pb^206`
C. `._82 U^210`
D. `._82 U^214`

Answer

B

Explanation

(b) `._((Z = 92))U^((A = 238)) overset (((8 alpha, 6 beta)))rarr ._(Z')X^(A')`
so `A' = A - 4 n_alpha = 238 - 4 xx 8 = 206`
and `Z' = n_beta - 2n_alpha + z = 6 -2 xx 8 + 92 = 82`.