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 : 1
• C. 1 : 4
• D. 1 : 1

Answer 1

The correct option is C 1 : 4

$\begin{array}{cc}{t}_A=20min& t_B=40min\\ n_A=\frac{T}{t_A}=\frac{80}{20}=4& n_B=\frac{80}{40}=2\\ N_A=N{\left(\frac{1}{2}\right)}^{n_A}& n_B=N{\left(\frac{1}{2}\right)}^{n_B}\\ N_A=N_0{\left(\frac{1}{2}\right)}^4& N_B=N_0{\left(\frac{1}{2}\right)}^2\\ \frac{N_A}{N_B}=\frac{1}{4}& \end{array}$