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

Answer

A

Explanation

The number of radioactive atoms, that reamain undecayed is given by
`N = N_0 (1/2)^n = N_0 (1/2)^(t/T)`
For A, n ` = t/T = 80/20 = 4`
`therefore N_A = N_0 (1/2)^4 = (N_0)/(16)`
For `B, n = 80/40 = 2 therefore N_B = N_0 (1/2)^2 = (N_0)/(4)`
`therefore N_A/N_B = (N_0)/(16) xx (4)/(N_0) = 1/4`