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The end product of the decay of ""(90)^(...

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

A

3,4

B

6,4

C

6,0

D

4,6

Text Solution

Verified by Experts

The correct Answer is:
B
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`
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