• # question_answer                 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

Key-Idea: Total no. of nuclei remained after n half ?lives is $N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}$                 Total time given = 80 min                 Number of half-lives of $A,\,{{n}_{A}}=\frac{80\,\min }{20\,\min }=4$                 Number of half-lives of $B,\,{{n}_{B}}=\frac{80\,\min }{40\,\min }=2$                 Number of nuclei remained undecayed                 $N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}$                 where ${{N}_{0}}$ is initial number of nuclei                 $\therefore \frac{{{N}_{A}}}{{{N}_{B}}}=\frac{{{\left( \frac{1}{2} \right)}^{{{n}_{A}}}}}{{{\left( \frac{1}{2} \right)}^{{{n}_{B}}}}}$                 or            $\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{{{\left( \frac{1}{2} \right)}^{4}}}{{{\left( \frac{1}{2} \right)}^{2}}}=\frac{\left( \frac{1}{16} \right)}{\left( \frac{1}{4} \right)}$                 or            $\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{1}{4}$                 Note:    The graph between number of nuclei decayed with time is shown along side.