Answer
See the detailed answer below.
Work Step by Step
a) We know that the decay constant is given by
$$\lambda=\dfrac{\ln 2}{T_{\frac{1}{2}}}$$
Plugging the known (for $\rm ^{14}_6C$);
$$\lambda=\dfrac{\ln 2}{5730\times 365.25\times 24\times 60^2 }$$
$$\lambda=\color{red}{\bf 3.833\times10^{-12}}\;\rm s^{-1}$$
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b) We need to find the number of $\rm ^{14}_6C$ atoms per gram of $\rm ^{12}_6C$ in a living organism and we know, in a living organism, that there is about $1.3\times 10 ^{-12}$ $\rm ^{14}_6C$ per carbon atom.
So that
$$N_{^{14}_6C}=\dfrac{N_A}{M} (1.3\times 10 ^{-12} )$$
where $N_A$ is Avogadro's number and $M$ is the molar mass of carbon.
Plugging the known;
$$N_{^{14}_6C}=\dfrac{6.02\times 10^{23}}{12.0107 } (1.3\times 10 ^{-12} ) =6.52\times 10^{10}$$
Therefore, we have $\color{red}{\bf 6.52\times 10^{10}}$ atom of $\rm ^{14}_6C$ per frame of $\rm ^{12}_6C$ in a living organism.
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c) We know that
$$R=N\lambda $$
Plugging from the two final results above;
$$R=(3.833\times10^{-12})(6.52\times 10^{10})$$
$$R=\color{red}{\bf 0.2499}\;\rm \dfrac{decay\cdot s^{-1}}{g\;of\;^{12}_6C}$$
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d) We know that
$$R=R_0e^{-\lambda t}$$
where $R_0$ is the final result we found in part c above.
Solving for $t$, to find how long ago he lived.
$$e^{-\lambda t}=\dfrac{R}{R_0}$$
Hence,
$$ -\lambda t =\ln\left[\dfrac{R}{R_0}\right]$$
$$ t =\dfrac{\ln\left[\dfrac{R}{R_0}\right]}{-\lambda}$$
Plugging the known;
$$ t =\dfrac{\ln\left[\dfrac{0.121}{0.2499}\right]}{- 3.833\times10^{-12} }=1.892\times 10^{11}\;\rm s$$
$$ t =1.892\times 10^{11}\;\rm s\times \dfrac{1\;\rm year}{365.25\times 24\times 60^2\;s}$$
$$t=\rm 5995\;year\approx \color{red}{\bf6000}\;year$$
