• 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

    Correct Answer: C

    Solution :

                    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.                

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