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The end product of 4n series is a).(82)P...

The end product of `4n` series is a)`._(82)Pb^(208)` b)`._(82)Pb^(207)` c)`._(82)Pb^(209)` d)`._(82)Pb^(204)`

A

`._(82)^(208)Pb`

B

`._(82)^(207)Pb`

C

`._(82)^(209)Pb`

D

`._(83)^(210)Bi`

Text Solution

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The correct Answer is:
a
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