A) \[{{\,}_{82}}{{U}^{206}}\]
B) \[{{\,}_{82}}P{{b}^{206}}\]
C) \[{{\,}_{82}}{{U}^{210}}\]
D) \[{{\,}_{82}}{{U}^{214}}\]
Correct Answer: B
Solution :
After one \[\alpha -\]emission, the daughter nucleus reduces in mass number by 4 unit and in atomic number by 2 unit. In \[\beta -\]emission the atomic number of daughter nucleus increases by 1 unit. The reaction can be written as \[{{\,}_{92}}{{U}^{238}}\xrightarrow{-8\alpha }{{\,}_{76}}{{X}^{206}}\xrightarrow{-6\beta }{{\,}_{82}}{{Y}^{206}}\] Thus, the resulting nucleus is \[{{\,}_{82}}{{Y}^{206}},\]i.e., \[{{\,}_{82}}P{{b}^{206}},\]
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