A) \[1:4\]
B) \[5:4\]
C) \[1:16\]
D) \[4:1\]
Correct Answer: B
Solution :
For \[{{\operatorname{A}}_{t}}_{^{1/2}}=20\,\,min,\,\,t=80\,\,min\], number of halflifes \[\operatorname{n} = 4\] \[\therefore \,\,\,Nuclei remaining =\frac{{{N}_{o}}}{{{2}^{4}}}\]. Therefore nuclei decayed \[={{N}_{0}} -\frac{{{N}_{o}}}{{{2}^{4}}}\] For \[{{B}_{t}}_{^{1/2}}=40\,\,min,\,\,t=80\,\,min\], number of halflifes \[\operatorname{n} = 2\] \[\therefore \,\,\,Nuclei remaining =\,\,\frac{{{N}_{o}}}{{{2}^{2}}}\] \[\operatorname{Therefore} nuclei decayed ={{N}_{0}}-\frac{{{N}_{0}}}{{{2}^{2}}}\] \[\therefore \,\,\,Required ratio\,\,=\,\,\frac{No-\frac{No}{{{2}^{4}}}}{No-\frac{No}{{{2}^{2}}}}=\frac{1-\frac{1}{16}}{1-\frac{1}{4}}=\frac{15}{16}\times \frac{4}{3}=\frac{5}{4}\]
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