JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2014
- Medium
- MCQ
A piece of bone of an animal from a ruin is found to have \(^{14}C\) activity of \(12\) disintegrations per minute per gm of its carbon content. The \(^{14}C\) activity of a living animal is \(16\) disintegrations per minute per gm. How long ago nearly did the animal die? ............\(years\) (Given halflife of \(^{14}C\) is \(t_{1/2} = 5760\,years\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2391\)
Step-by-step Solution
Detailed explanation
Given, for \(^{14} \mathrm{C}\)
\(A_{0}=16\) dis \(\min ^{-1} g^{-1}\)
\(A=12\) dis \(\min ^{-1} g^{-1}\)
\(\mathrm{t}_{1 / 2}=5760\) years
Now, \(\lambda = \frac{{0.693}}{{{t_{1/2}}}}\)
\(\lambda=\frac{0.693}{5760}\) per year
Then, from, \(t=\frac{2.303}{\lambda} \log _{10} \frac{\mathrm{A}_{0}}{\mathrm{A}}\)
\(=\frac{2.303 \times 5760}{0.693} \log _{10} \frac{16}{12}\)
\( = \frac{{2.303 \times 5760}}{{0.693}}{\log _{10}}1.333\)
\( = \frac{{2.303 \times 5760 \times 0.1249}}{{0.693}}\)
\(=2390.81 \approx 2391\) years
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