A ancient discovery found a sample, where $75 \%$ of the original carbon $\left(\mathrm{C}^{14}\right)$…
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
$\left(\begin{array}{r}T_{\frac{1}{2}}\left(\mathrm{C}^{14}\right)=5730 \text { years, } \ln 0.5=-0.7 \\ \ln (0.75)=-0.3\end{array}\right)$
2300 years
2456 years
2546 years
3456 years
Solution
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 $\mathrm{C}^{14}$ atom in the sample
$N_0=100, N=75$
The age of the sample is given as,
$t=\frac{2.303}{K} \log \frac{N_0}{N}=\frac{2.303}{K} \log \frac{1}{0.75}$
$\begin{aligned} & =\frac{2303}{1.209 \times 10^{-4}} \log \frac{100}{75} \\ & =2456 \times 10^3 \text { years } \\ & =2456 \text { years }\end{aligned}$