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The end product of the decay of ""_(90)^(232)Th is ""_(82)^(208)Pb. The number of alpha and beta particle emitted are, respectively. |
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Answer» 3,4 Then, `""_(90)^(232)Th to ""_(90)^(208)Pb+n(""_(2)^(4)He)+m(""_(-1)^(0)beta)` `rArr 90=82+2n-m` and `232=208+4n` `rArr n=6, m=7` |
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