Q.Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be $3.839 \times 10^{-12}$ per second.

[!TLDR] $t=\frac{\ln(15.3/12.3)}{\lambda}=\frac{0.2182}{1.211\times10^{-4}/\text{yr}}\approx1 …

Carbon-14 dating uses $A(t)=A_0e^{-\lambda t}$, comparing the CURRENT (old-sample) activity to the activity of a LIVING sample (taken as $A_0$, since a living organism continuously exchanges carbon with the atmosphere and maintains a roughly constant $^{14}C$ fraction). Here $A_0=15.3$ decays/g/min (living) and $A(t)=12.3$ decays/g/min (old sample). From $A(t)=A_0e^{-\lambda t}$, $t=\frac{1}{\lambda}\ln\left(\frac{A_0}{A(t)}\right)=\frac{1}{\lambda}\ln(15.3/12.3)=\frac{\ln(1.2439)}{\lambda}=\frac{0.2182}{\lambda}$. Converting the given $\lambda=3.839\times10^{-12}$ s$^{-1}$ to per-year units ($\times3.156\tim …

Use $A(t)=A_0e^{-\lambda t}$ with the living-sample rate as $A_0$ and the old-sample rate as $A(t)$, solving for $t=\frac{1}{\lambda}\l …

Forgetting to convert the given decay constant from per-second to per-year units before dividing, which would give an …