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 \[\text{= 80 min}\] 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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