1.

Half-lives of two radioactive substances `A` and `B` are respectively `20` minutes and `40` minutes. Initially, he sample of `A` and `B` have equal number of nuclei. After `80` minutes the ratio of the remaining number of `A` and `B` nuclei is :A. `1 : 16`B. `4 : 1`C. `1 : 4`D. `1 : 1`

Answer» Correct Answer - C
( c) 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)`
`[because 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`.


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