A) 5 : 4
B) 1 : 16
C) 4 : 1
D) 1 : 4
Correct Answer: A
Solution :
\[t=80\min =4{{T}_{A}}=2{{T}_{B}}\] \[\therefore \]no. of nuclei of A decayed \[={{N}_{0}}-\frac{N}{{{2}^{4}}}=\frac{15{{N}_{0}}}{16}\] \[\therefore \]no. of nuclei of B decayed \[={{N}_{0}}-\frac{N}{{{2}^{2}}}=\frac{3{{N}_{0}}}{4}\]required ratio \[=\frac{5}{4}\]
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