1.

Half life of a radioactive substance A is two times the half life of another radioactive substance B. Initially, the number of nuclei of A and B are N_(A) and N_(B) respectively. After three half lives of A, number of nuclei of both are equal. Then the ratio (N_(A))/(N _(B)) is

Answer»

`(1)/(3)`
`(1)/(6)`
`(1)/(8)`
`(1)/(4)`

Solution :Three half LIVES of A are equivalent to six half lives of B. As NUMBER of nuclei LEFT are equal in the two cases.
Therefore, `N_(A)((1)/(2))^(3)=N_(B)((1)/(2))^(6)`
`(N_(A))/(N_(B))=((1//2)^(6))/((1//2)^(3))=((1)/(2))^(3)=(1)/(8)`


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