At some instant, a radioactive sample `S_(1)` having an activity 5 `muCi` has twice the number of nuclei as another sample `S_(2)` which has an activity of 10 `muCi`. The half lives of `S_(1) and S_(2)` are :

Options

A. 5 years and 20 years, respectively
B. 20 years and 5 years, respectively
C. 20 years and 10 years, respectively
D. 10 years and 20 years, respectively

Answer

B

Explanation

Let us assume number of nuclei for `S_(1)` is `N_(1)` and that for `S_(2)` is `N_(2)`.
`N_(1)=2N_(2)`…….(i)
Activity can be written as `A = lambdaN`
And if t is half-life then: `lambda=(ln(2))/t`
Using above, information we can write the following for two samples:
`A_(1)=lambda_(1)N_(1)=(ln(2))/t_(2) xx N_(1) = 5 muC_(1)`
`A_(2)=lambda_(2)N_(2)=(ln(2))/t_(1) xx N_(2) = 10 muC_(1)`
`t_(1)/t_(2) xx N_(1)/N_(2) =1/2`
From equation (i), `t_(1)/t_(2)=1/4`
Hence 5 years and 20 years satisfy the above relation hence option (a) is correct.