Q103 / 180 · AIIMS 2006 · Physics

The fossil bone has a ${ }^{14} \mathrm{C}:{ }^{12} \mathrm{C}$ ratio, which is $[1 / 16]$ of that in a living animal bone. If the halflife time of ${ }^{14} \mathrm{C}$ is 5730 years, then the age of the fossil bone is

  • A
    11460 years
  • B
    17190 years
  • Correct answer
    C
    22920 years
  • D
    45840 years.

Solution

After 5730 years, the number of $\mathrm{C}^{14}$ remaining is $\frac{1}{2}$ th original. To have $(1 / 16)$ original value, it takes $(1 / 2)^4, 4$ half lives. Therefore, the bone is $4 \times 5730$ years $=22920$ years old.

Hence, the correct answer is option: 'C'.

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