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 the remaining number of A and B nuclei is

• A. $1:16$
• B. $4:1$
• C. $1:4$
• D. $1:1$

Answer 1

Answer: (C)

$1:4$

$N_A\left(t\right)=N_o{e}^{-{\lambda }_{A}t}$
$N_B\left(t\right)=N_o{e}^{-{\lambda }_{B}t}$
$\lambda =\frac{ln2}{t_{1/2}}$
$\frac{N_A}{N_B}=e^{-({\lambda }_A-{\lambda }_B)t}$

At $t=80min$
$\frac{N_A}{N_B}=e^{-ln2(\frac{1}{20}-\frac{1}{40})80}$
$\frac{N_A}{N_B}=e^{-ln2\left(\frac{1}{40}\right)80}$
$\frac{N_A}{N_B}=1:4$