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 N0 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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