1.

The end product of the decay of ""_(90)^(232)Th is ""_(82)^(208)Pb. The number of alpha and beta particle emitted are, respectively.

Answer»

3,4
6,4
6,0
4,6

Solution :Let the number of `ALPHA-` particle be N and the number of `BETA`- PARTICLES be m
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`


Discussion

No Comment Found

Related InterviewSolutions