A) \[_{82}{{U}^{206}}\]
B) \[_{82}P{{b}^{206}}\]
C) \[_{82}{{U}^{210}}\]
D) \[_{82}{{U}^{214}}\]
Correct Answer: B
Solution :
After one \[\alpha \text{-}\]-emission, the daughter nucleus reduces in mass number by 4 unit and in atomic number by 2 unit. In \[\beta \text{-}\]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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