Half-life of a radioactive substance` A` and `B` are, respectively, `20 min` and `40min`. Initially, the samples of `A` and `B` have equal number of nuclei. After `80 min`, the ratio of the ramaining number of `A` and `B` nuclei is

Options

A. `1:16`
B. `4:1`
C. `1:4`
D. `1:1`

Answer

c

Explanation

We know that
`N=N_(0)((1)/(2))^(n_A)`
For `A`,
`N=N_(0)((1)/(2))^(n_A) = N_(0)((1)/(2))^(4) = (N_(0))/(16)`
`[:' n_(A)=(t)/(T_(A))=(80)/(20)=4]`
For `B`,
`N_(B)=N_(0)((1)/(2))^(n_B) = N_(0)((1)/(2))^(2) =(N_(0))/(4)`
`:. (N_(A))/(N_(B))=(1)/(4)` or `N_(A):N_(B)=1:4`.