Q.Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of 14C averaged about 0.255 Bq per gram of carbon.

[!TLDR] (a) $N_{14}/N_{total}=A/(\lambda N_{total,per\,gram})\approx4$ in $3\times10^{12}$ carbon atoms. (b) Specimen's activity per gram is $0.0967$ Bq/g vs the living value $0.255$ Bq/g; $t=\ln(0.255/0.0967)/\lambda\approx8020$ years. …

(a) The decay constant of $^{14}C$ is $\lambda=0.693/5730\text{ yr}=1.209\times10^{-4}$ yr$^{-1}$, or $3.834\times10^{-12}$ s$^{-1}$. One gram of ordinary carbon contains $N_{total}=6.022\times10^{23}/12\approx5.018\times10^{22}$ atoms (using carbon's molar mass, 12 g/mol). Given activity per gram $A=0.255$ Bq, the number of $^{14}C$ atoms per gram is $N_{14}=A/\lambda=0.255/(3.834\times10^{-12})\approx6.65\times10^{10}$. The fraction is $N_{14}/N_{total}\approx6.65\times10^{10}/5.018\times10^{22}\approx1.33\times10^{-12}$, i.e. about 4 atoms in every $3\times10^{12}$ carbon atoms are $^{14}C$, matching the book's answer. (b) The specimen shows 174 decays per hour from 500 mg = 0.5 g of carbon, i.e. $174/3600=0.04833$ decays/s from 0.5 g, so activity per gram is $0.04833/0.5=0.0967$ Bq/g. Assuming the specimen's activity per gram equalled the living-organism value (0 …

For (a), use $N_{14}=A/\lambda$ per gram and compare with the total carbon-atom count per gram (from Avogadro's number and carbon's molar mass); for (b), scale the specimen's counts to activity-per-gr …

In part (b), using the RAW decay count (174/hour) directly as the specimen's activity per gram, without first dividing by the actu …