A) 1 : 16
B) 4 : 1
C) 1 : 4
D) 1 : 1
Correct Answer: C
Solution :
Key Idea: Total number 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 remains 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 \[\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{1}{4}\]
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