AP EAMCET · PHYSICS · Nuclear Physics
A ancient discovery found a sample, where \(75 \%\) of the original carbon \(\left(\mathrm{C}^{14}\right)\) remains. Then the age of the sample is
\(\left(\begin{array}{r}T_{\frac{1}{2}}\left(\mathrm{C}^{14}\right)=5730 \text { years, } \ln 0.5=-0.7 \\ \ln (0.75)=-0.3\end{array}\right)\)
- A 2300 years
- B 2456 years
- C 2546 years
- D 3456 years
Answer & Solution
Correct Answer
(B) 2456 years
Step-by-step Solution
Detailed explanation
For a carbon sample \(\left(\mathrm{C}^{14}\right)\), \(T_{1 / 2}=5730\) year Decay constant, \(K=\frac{0.693}{T_{1 / 2}}\) \(\begin{aligned} & =\frac{0.693}{5730} \\ & =1.209 \times 10^{-4} / \text { year }\end{aligned}\) The rate of counts is proportional to the number of…
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