A) \[_{82}^{208}Pb\]
B) \[_{82}^{206}Pb\]
C) \[_{82}^{207}Pb\]
D) \[_{83}^{210}Bi\]
Correct Answer: B
Solution :
The end product of uranium series or \[(4n+2)\] series is \[_{82}P{{b}^{206}}\]and first member of this series is \[_{92}{{U}^{238}}\] \[_{92}{{U}^{238}}{{\xrightarrow{-\alpha }}_{90}}T{{h}^{234}}{{\xrightarrow{-\beta }}_{91}}P{{a}^{234}}\xrightarrow{-\beta }\]\[_{88}R{{a}^{226}}{{\xrightarrow{-\alpha }}_{86}}R{{n}^{222}}{{\xrightarrow{-\alpha }}_{84}}P{{o}^{218}}\xrightarrow{-\alpha }\] \[_{82}P{{b}^{214}}{{\xrightarrow{-\beta }}_{83}}B{{i}^{210}}{{\xrightarrow{-\beta }}_{84}}P{{o}^{214}}\xrightarrow{-\alpha }\] \[_{82}P{{b}^{210}}{{\xrightarrow{-\beta }}_{83}}B{{i}^{210}}{{\xrightarrow{-\beta }}_{84}}P{{o}^{210}}{{\xrightarrow{-\alpha }}_{82}}P{{o}^{206}}\]
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