Half-lives of two radioactive substances A and B are respectively 20 min and 40 min . Initially the samples of A and B have equal number of nuclei. After 80 min the ratio of remaining number of A and B nuclei is

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Correct answer: C. 1: 4

Key Idea : Total number of nuclei remained after n half-lives is N=N_0((1)/(2))^n . Total time given =80 ~min Number of half-lives of A, n_A= 80 ~min 20 ~min =4 Number of half-lives of B, n_B= 80 ~min 40 ~min =2 Number of nuclei remains undecayed N=N_0 ((1)/(2))^n where N_0 is initial number of nuclei. array ll & (N_A)/(N_B)= ((1)/(2))^ n_A ((1)/(2))^ n_B or & (N_A)/(N_B)=(((1)/(2))^4)/(((1)/(2))^2)=(((1)/(16)))/(((1)/(4))) or & (N_A)/(N_B)=(1)/(4) array

Half-lives of two radioactive substances A and B are respectively...

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