A) \[6000\text{ }yr\]
B) \[3000\text{ }yr\]
C) \[9000\text{ }yr\]
D) \[12000\text{ }yr\]
Correct Answer: D
Solution :
\[K=\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{6000}=1.155\times {{10}^{-4}}\] \[t=\frac{2.303}{k}\log \frac{{{N}_{0}}}{N}\] \[=\frac{2.303}{1.55\times {{10}^{-4}}}\log \frac{100}{25}\] \[=12000yr\]You need to login to perform this action.
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