Question 1 of 7: Atomic and Nuclear Structure, Isotopes, Radioactivity

Question 1: Atomic and Nuclear Structure, Isotopes, Radioactivity (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Atomic masses and particle masses below; ¹²C has the exact reference mass 12.000000 u by definition of the unified atomic mass unit. ¹⁴C decays by β⁻ emission with a half-life of 5,730 years. The mummy's tissue carries a ¹⁴C/¹²C ratio 0.000795 times today's living ratio.

Find. (a) binding energy per nucleon of ${}^{12}$C and ${}^{14}$C (and of ${}^{13}$C, whose mass is also given); (b) the energy released adding a neutron to ${}^{12}$C; (c) the nuclear reactions that produce and destroy atmospheric ${}^{14}$C; (d) the age of the mummy from its measured isotope ratio.

Approach. Build each isotope's binding energy from the mass defect against $Z$ free hydrogen atoms plus $N$ free neutrons, apply the same mass-defect logic to the single-neutron-capture reaction in (b), write the two nuclear reactions literally for (c), then use the exponential decay law directly on the isotope ratio for (d), since the stable ${}^{12}\text{C}$ denominator does not itself decay.

1. Part (a) — binding energy per nucleon. For an atom with $Z$ protons and $N$ neutrons, the binding energy compares the atom's mass to $Z$ free hydrogen atoms (proton + electron) plus $N$ free neutrons: $$\text{BE} = \big[Z\,m_{\text{H}} + N\,m_n - M\big]c^2, \qquad m_{\text{H}} = m_p + m_e = 1.00782558\text{ u}$$ For ${}^{12}$C ($Z=6,\,N=6$): $$\Delta m = 6(1.00782558) + 6(1.008665) - 12.000000 = 0.098943\text{ u}$$ $$\boxed{\text{BE}({}^{12}\text{C}) = 0.098943 \times 931.494 = 92.17\text{ MeV}, \quad \text{BE}/A = 92.17/12 = 7.680\text{ MeV/nucleon}}$$ For ${}^{14}$C ($Z=6,\,N=8$): $$\Delta m = 6(1.00782558) + 8(1.008665) - 14.003241 = 0.113032\text{ u}$$ $$\boxed{\text{BE}({}^{14}\text{C}) = 0.113032 \times 931.494 = 105.29\text{ MeV}, \quad \text{BE}/A = 105.29/14 = 7.521\text{ MeV/nucleon}}$$ For completeness, ${}^{13}$C ($Z=6,\,N=7$), whose mass is also supplied: $$\Delta m = 6(1.00782558) + 7(1.008665) - 13.00335 = 0.104258\text{ u} \;\Rightarrow\; \text{BE} = 97.12\text{ MeV},\quad \text{BE}/A = 7.470\text{ MeV/nucleon}$$ Both heavier isotopes are slightly less tightly bound per nucleon than ${}^{12}\text{C}$ — their extra neutrons take them off the $N=Z$ line that makes ${}^{12}$C (an alpha-cluster nucleus) especially stable.
2. Part (b) — neutron-capture energy. The same mass-defect logic applies to the single reaction ${}^{12}\text{C} + n \rightarrow {}^{13}\text{C}$: the energy released equals the mass lost, $$\Delta m = \big[M({}^{12}\text{C}) + m_n\big] - M({}^{13}\text{C}) = (12.000000 + 1.008665) - 13.00335 = 0.005315\text{ u}$$ $$\boxed{S_n = 0.005315 \times 931.494 = 4.95\text{ MeV}}$$ This is the neutron separation energy of ${}^{13}\text{C}$: because ${}^{13}\text{C}$ is more tightly bound than ${}^{12}\text{C}$ plus a free neutron, the reaction is exothermic and releases 4.95 MeV rather than requiring an energy input — the phrase "energy needed to add a neutron" is best read as the energy that must be supplied to reverse the reaction (strip the neutron back out), which is numerically the same 4.95 MeV.
3. Part (c) — production and decay relationships. ${}^{14}$C is produced when a cosmic-ray-spallation neutron is captured by atmospheric nitrogen, ejecting a proton (an $(n,p)$ reaction): $$n + {}^{14}_{\ 7}\text{N} \longrightarrow {}^{14}_{\ 6}\text{C} + p$$ It decays back to nitrogen by $\beta^-$ emission, converting a neutron into a proton and emitting an electron and antineutrino: $${}^{14}_{\ 6}\text{C} \longrightarrow {}^{14}_{\ 7}\text{N} + \beta^- + \bar\nu_e, \qquad T_{1/2} = 5{,}730\text{ years}$$ These two reactions set up the steady-state atmospheric ${}^{14}\text{C}/^{12}\text{C}$ ratio that dating in part (d) compares against.
4. Part (d) — age from the isotope ratio. While an organism is alive it exchanges carbon with the atmosphere and keeps the living ratio; once it dies, ${}^{14}$C decays with no replenishment while the ${}^{12}$C denominator stays fixed (it is stable), so the ratio itself decays exponentially: $$\frac{R(t)}{R_0} = e^{-\lambda t}, \qquad \lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{5{,}730} = 1.2097\times10^{-4}\text{ yr}^{-1}$$ Solving for $t$ with $R(t)/R_0 = 0.000795$, $$t = -\frac{\ln(0.000795)}{\lambda} = \frac{7.137}{1.2097\times10^{-4}}$$ $$\boxed{t \approx 5.90\times10^4\text{ years} \approx 59{,}000\text{ years}}$$