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.