1.

._(84)Pb^(219) is a member of actinium series. The other member of this series is

Answer»

`._(89)Ac^(225)`
`._(90)Th^(232)`
`._(15)P^(35)`
`._(92)U^(235)`

Solution :The seires `4n + 3` is KNOWN as actinium series. It shows that if mass NUMBER is divided by 4 then the remainder MUST be 3. Hence, the another member of this series is `._(92)U^(235)`
`:' (235)/(4) = 58 +` Remainder 3


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