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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2014
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  • 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 \(t_{1/2} = 5760\,years\) )

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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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