= Solution
\b[Dating principle]. Atmospheric <cosmic rays> produce secondary <neutrons>, which can generate <carbon-14> through $^{14}\mathrm N(n,p){}^{14}\mathrm C$. <Carbon> then enters atmospheric <carbon dioxide> and living organisms. While an organism exchanges carbon with its environment, its <carbon-14> to <carbon-12> ratio approximately tracks that reservoir. After exchange stops, surviving <carbon-14> undergoes <beta-minus decay>, whereas the stable <carbon> remains. On the closed-sample, constant-initial-ratio model, <radioactive decay> gives
$$
N_{14}(t)=N_{14}(0)e^{-\lambda t},\qquad
\mathcal A(t)=\lambda N_{14}(t)=\mathcal A_0e^{-\lambda t}.
$$
Thus <radiocarbon dating> infers elapsed time from the loss of an isotope ratio or of <radioactive activity> per fixed carbon mass. It dates cessation of carbon exchange, which need not equal the date an object was manufactured or buried. Actual calendar dating also requires <radiocarbon calibration> and control of contamination and reservoir differences.
\b[Mean lifetime]. For <exponential decay>, the <half-life> obeys $e^{-\lambda T_{1/2}}=1/2$, while the <mean lifetime> is $\tau=1/\lambda$. Hence
$$
\boxed{\tau=\frac{5730}{\ln2}\,\mathrm{yr}=8.26664\times10^3\,\mathrm{yr}.}
$$
Using $1\,\mathrm{yr}=365.25\times86400\,\mathrm s$ gives $\lambda=3.83325\times10^{-12}\,\mathrm s^{-1}$ and $\tau=2.60875\times10^{11}\,\mathrm s$.
\b[The mass minimum and beta stability]. Treat the supplied <masses> as nuclear <rest energies> in $\mathrm{MeV}$, and omit <nuclear pairing> as instructed. At fixed <mass number> $A$, the volume and surface terms of the <semi-empirical mass formula> are constant. Expanding the remaining terms gives a convex <nuclear mass parabola>,
$$
M(A,Z)=C(A)+\left(\frac{a_C}{A^{1/3}}+\frac{4a_A}{A}\right)Z^2
+\left(m_p-m_n-4a_A\right)Z.
$$
The continuous minimum in <atomic number> is therefore
$$
\frac{\partial M}{\partial Z}
=m_p-m_n+\frac{2a_CZ}{A^{1/3}}-\frac{4a_A(A-2Z)}{A}=0,
\qquad
Z_* =\frac{4a_A+m_n-m_p}{2a_CA^{-1/3}+8a_A/A}.
$$
At $A=14$, the quadratic coefficient is $7.01302\,\mathrm{MeV}$ and $Z_*=6.79450$. The nearest integer is $Z=7$; the <nuclear isobar> of minimum <mass> is <nitrogen-14>. Maximizing <nuclear binding energy> alone would omit the neutron-proton rest-mass difference; its continuous optimum is slightly different, but also selects integer $Z=7$ here.
The values most useful for checking neighboring decay thresholds are
$$
\begin{array}{c|rrrr}
Z&5&6&7&8\\ \hline
M(14,Z)\;(\mathrm{MeV})&13065.584&13047.427&13043.296&13053.192.
\end{array}
$$
Consequently <carbon-14> can lower its nuclear <mass> by changing a <neutron> into a <proton>:
$$
\boxed{{}^{14}_6\mathrm C\longrightarrow{}^{14}_7\mathrm N+e^-+\bar\nu_e
\quad\text{(beta-minus decay).}}
$$
For bare nuclear <masses>, the <Q value> must include the emitted <electron mass>. Taking $m_ec^2=0.511\,\mathrm{MeV}$,
$$
Q_{\beta^-}^{\mathrm{model}}=M(14,6)-M(14,7)-m_ec^2
=4.13070-0.511=3.61970\,\mathrm{MeV}>0.
$$
The opposite proton-to-neutron conversion would lead to $Z=5$, whose nuclear <mass> is $18.1567\,\mathrm{MeV}$ higher than that of <carbon-14>. Neither <beta-plus decay> nor <electron capture> can bridge that gap by supplying an electron-mass correction. For the <nitrogen-14> daughter, both neighboring nuclear <masses> are higher: $Z=6$ by $4.13070\,\mathrm{MeV}$ and $Z=8$ by $9.89535\,\mathrm{MeV}$. Even <electron capture> to $Z=6$ has $Q=0.511-4.13070<0$. Thus the daughter has \b[no further energetically allowed single beta decay in this model]; physical <nitrogen-14> is stable.
These threshold numbers belong to the supplied smooth <semi-empirical mass formula>, not to an accurate prediction of the <carbon-14> electron spectrum or its long <half-life>. <Semi-empirical mass errors can change beta-decay thresholds>. Also, the instruction to omit <nuclear pairing> is a model simplification: even $A=14$ does not make physical pairing vanish, and even-even carbon and odd-odd nitrogen need not have the same pairing correction. No omitted pairing coefficient is guessed here.
\b[Activity of modern carbon]. Neglect the small contribution of other stable <carbon> <isotopes> when counting atoms in a gram. The <Avogadro constant> gives
$$
N_{12}\simeq\frac{1\,\mathrm g}{12\,\mathrm{g\,mol^{-1}}}
(6.02214076\times10^{23}\,\mathrm{mol^{-1}})
=5.01845\times10^{22},
\qquad
N_{14}=1.3\times10^{-12}N_{12}=6.52399\times10^{10}.
$$
The expected <radioactive activity> of that one gram is
$$
\boxed{\mathcal A_0=\lambda N_{14}=0.25008\,\mathrm{Bq}
\simeq0.25\,\mathrm{Bq},}
$$
or approximately $15.0$ decays per minute per gram of carbon. A <becquerel> is one decay per second; the normalization is per gram of carbon, rather than per gram of the entire organic sample.
\b[Artefact age]. Its remaining activity fraction is $0.13/0.25008=0.51983$. Inverting the <exponential decay> law,
$$
\boxed{t=\frac{1}{\lambda}\ln\frac{\mathcal A_0}{\mathcal A}
=8266.64\,\mathrm{yr}\times\ln\frac{0.25008}{0.13}
=5408.44\,\mathrm{yr}\simeq5.4\times10^3\,\mathrm{yr}.}
$$
This is the <radiocarbon age> on the supplied constant-initial-activity, $5730$-year <half-life> model. It is not a calibrated calendar date; the sample's historical initial ratio and any later carbon exchange are not supplied.
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