A) 11460 years
B) 17190 years
C) 22920 years
D) 45840 years
Correct Answer: C
Solution :
After n half-lives the number of nuclides left un decayed is,\[N=\frac{{{N}_{0}}}{{{2}^{n}}}\]where\[{{N}_{0}}\] is initial number of nuclides. Here in present case the activity decreases to \[1/16={{\left( \frac{1}{2} \right)}^{4}}\]times of initial one i.e. 4 half-life is the age of fossil bone.
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