Question

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

• A. 1:16
• B. 4:01
• C. 1:04
• D. 1:01

Answer 1

Answer: (C)

1:04

Answer 2

Total time given= 80 min
Number of half-lives of A, $n_A = \frac{80 \, min}{20 \, min} = 4$
Number of half-lives of A, $n_B = \frac{80 \, min}{40 \, min} = 2$
Number of nuclei remains undecayed
$N = N_0 \big(\frac{1}{2}\big)^n$
where $N_0$ is initial number of nuclei.
$\therefore \frac{N_A}{N_B} = \frac{\big(\frac{1}{2}\big)^{n_A}}{\big(\frac{1}{2}\big)^{n_B}}$
or $\\\frac{N_A}{N_B} = \frac{\big(\frac{1}{2}\big)^4}{\big(\frac{1}{2}\big)^2} = \frac{\big(\frac{1}{16}\big)}{\big(\frac{1}{4}\big)}$
or $\frac{N_A}{N_B} = \frac{1}{4}$