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Question

The last product of 4n series is:

A
20882Pb
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B
20782Pb
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C
20982Pb
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D
21083Bi
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Solution

The correct option is A 20882Pb
4n series is the _23290Th decay series.
23290Thα22888Ra....21284Poα20882Po
option (A) is correct.

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