A) 4 : 1
B) 1 : 16
C) 1 : 1
D) 1 : 4
Correct Answer: D
Solution :
From the relation\[\frac{N}{{{N}_{0}}}=\frac{1}{{{2}^{n}}}\] so \[N\propto \frac{1}{{{2}^{n}}}\] Hence, \[\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{1/{{2}^{4}}}{1/{{2}^{2}}}=\frac{1}{{{2}^{4}}}\times \frac{{{2}^{2}}}{1}=\frac{1}{4}\]
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