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Half life of a radioactive substance A i...

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

A

`(1)/(3)`

B

`(1)/(6)`

C

`(1)/(8)`

D

`(1)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
C

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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Knowledge Check

  • Half-life of a radioactive substance A is two times the half-life of another radioactive substance B . Initially, the number 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 .

    A
    `1//4`
    B
    `1//8`
    C
    `1//3
    D
    `1//6`
  • Half life of a radioactive substance with progress of time

    A
    increase
    B
    decrease
    C
    remain same
    D
    none
  • The half-life of a radioactive substance is 10 days. This means that

    A
    the substance completely disintergrates in 20 days
    B
    the substance completely disintergrates in 40 days
    C
    `1//8` part of the mass of the substance will be ledt intact at the end of 40 days
    D
    `7//8` part of the mass of the substance disintegrates in 30 days
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