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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)Pb^(208)`

B

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

C

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

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
A
`underset("Initial")(""_(90)Th^(232)) underset((-4)beta- "particle")overset((-6) alpha-"particle")to underset("final")(""_(82)Pb^(208))`
All the nuclei from the initial element to the final stable element constitute a series known as disintegration series. 4n-series , (4n+1) series , (4n +2) series , (4n+3) series
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