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80 minutes = 4 half-lives of A = 2 half-lives of B Let the initial number of nuclei in each sample be N NANA after 80 minutes =N/2^4 => Number of A nuclides decayed =15/16N NBNB after 80 minutes =N/2^4 => Number of B nuclides decayed =3/4N Required ratio =15/16/3/4 =5/4
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N = N'*(1/2)^n  where n is the number of half life passed. N' is initial number of molecules.

for A half life is 20 minutes hence 4 half lefes have passed and for B half life is 40 minutes hence 2 half lifes have passed.

therefore N(A) = N'(1/2)^4 = N'/16 and N(B) = N'(1/2)^2 = N'/4

hence decayed number of A is N'-N(A) = N'-N'/16 = 15 N'/16 and decayed number of B is N'-N(B) = N'-N'/4 = 3 N'/4

so ratio is (15 N'/46)/(3 N'/4) = 5/4 i.e. 5 is to 4

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