A) 0
B) 12.5
C) 8
D) 25
Correct Answer: D
Solution :
We know that\[{{C}_{n}}=\frac{{{C}^{o}}}{{{2}^{n}}}\] \[{{C}_{n}}=\]concentration after n half-lives \[n=\]number of half-lives \[{{C}^{o}}=\]initial concentration \[{{C}_{n}}=\frac{1}{{{2}^{2}}}=\frac{1}{4}=0.25\]or \[25%\]
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