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Q. What is the age of an ancient wooden piece if it is known that the specific activity of $C^{14}$ nuclide in its amounts is $3/5$ of that in freshly grown trees? Given the half of $C$ nuclide is $5570\, yr$.

VITEEEVITEEE 2011

Solution:

$N_{0}=\frac{3}{5}\,N_{0}e^{-\lambda t}$
$e\lambda^{t}=\frac{5}{3}$
$\therefore log\,e^{\lambda t}=log_{e} \frac{5}{3}$ or $\lambda t=log_{e} \frac{5}{3}$
or $t=\frac{1}{\lambda}\,log_{e} \frac{5}{3}$
$=\frac{T}{0.693}\times0.5\,\left(\because T=\frac{0.693}{\lambda}\right)$
$=\frac{5570\times0.5}{0.693}yr$
$=4018.7\,yrs$
$= 4000\,yr$