Question 1 of 7: Binding Energy and Nuclear Stability

Question 1: Binding Energy and Nuclear Stability (12 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. Three atomic masses — deuterium (²H, $Z{=}1,N{=}1$), calcium-41 (⁴¹Ca, $Z{=}20,N{=}21$) and uranium-238 (²³⁸U, $Z{=}92,N{=}146$) — plus the proton mass, neutron mass and the mass–energy conversion $1\text{ u}=931.4943$ MeV.

Find. (a) binding energy per nucleon of ${}^{2}$H, ${}^{41}$Ca and ${}^{238}$U; (b) physical explanations, grounded in those numbers, for deuterium burning, ${}^{238}$U's fertility, and ${}^{41}$Ca's fission/fusion inertness.

Approach. Build each nuclide's mass defect against $Z$ free protons plus $N$ free neutrons using the two masses the question actually supplies, convert to MeV, then read the three qualitative sub-parts of (b) directly off the binding-energy-per-nucleon curve these three points sit on.

1. Part (a) — binding energy per nucleon. The question supplies the bare proton mass (not the hydrogen-atom mass), so the mass-defect formula used literally as given is $$\text{BE} = \big[Z\,m_p + N\,m_n - M\big]c^2$$ For ${}^{2}$H ($Z{=}1,N{=}1$): $$\Delta m = (1.00727647+1.008665012) - 2.0141018 = 1.83968\times10^{-3}\text{ u}$$ $$\boxed{\text{BE}({}^{2}\text{H}) = 1.83968\times10^{-3}\times931.4943 = 1.714\text{ MeV}, \quad \text{BE}/A = 1.714/2 = 0.857\text{ MeV/nucleon}}$$ For ${}^{41}$Ca ($Z{=}20,N{=}21$): $$\Delta m = \big[20(1.00727647)+21(1.008665012)\big]-40.9622783 = 0.36522\text{ u}$$ $$\boxed{\text{BE}({}^{41}\text{Ca}) = 0.36522\times931.4943 = 340.20\text{ MeV}, \quad \text{BE}/A = 340.20/41 = 8.297\text{ MeV/nucleon}}$$ For ${}^{238}$U ($Z{=}92,N{=}146$): $$\Delta m = \big[92(1.00727647)+146(1.008665012)\big]-238.0507826 = 1.88376\text{ u}$$ $$\boxed{\text{BE}({}^{238}\text{U}) = 1.88376\times931.4943 = 1754.70\text{ MeV}, \quad \text{BE}/A = 1754.70/238 = 7.373\text{ MeV/nucleon}}$$
2. Part (b)(i) — why deuterium burning is possible. ${}^{2}$H's computed binding energy per nucleon (0.857 MeV) is far below the value for a mass-3 nucleus such as helium-3 (${}^{3}\text{He}$ binding energy per nucleon is about 2.6 MeV) — the deuteron is only barely bound. Fusing it with a proton, $${}^{2}\text{H} + p \longrightarrow {}^{3}\text{He} + \gamma$$ moves two nucleons from a loosely-bound A=2 system to a much more tightly-bound A=3 system, so the reaction is strongly exothermic. Because both reactants carry only a single proton, the Coulomb barrier to fuse them is the lowest of any charged-particle fusion reaction, which is why deuterium burning ignites at relatively modest stellar-core temperatures (well below what hydrogen-hydrogen fusion itself needs), consistent with it occurring even in substellar objects (brown dwarfs) that never reach true main-sequence core temperatures.
3. Part (b)(ii) — why ${}^{238}$U is fertile rather than fissile. ${}^{238}$U's binding energy per nucleon (7.373 MeV) sits high on the curve for a heavy nucleus — splitting it into two mid-mass fragments (each landing nearer the curve's peak, around 8.5 MeV/nucleon) is energetically favourable, which is why fission of uranium releases energy at all. But ${}^{238}$U is even-$N$ (even-even overall), so absorbing a neutron to form ${}^{239}$U pairs the odd neutron and releases relatively little extra binding energy on capture — not enough to push the compound nucleus over its fission barrier with a slow (thermal) neutron, so ${}^{238}$U does not fission promptly. Instead the captured neutron survives, and ${}^{239}$U beta-decays ($T_{1/2}\approx 23$ min) to ${}^{239}$Np and then to ${}^{239}$Pu, which is fissile. "Fertile" means exactly this: not itself thermal-fissile, but convertible by neutron capture into a fissile isotope.
4. Part (b)(iii) — why ${}^{41}$Ca suits neither fission nor fusion. ${}^{41}$Ca's binding energy per nucleon (8.297 MeV) already sits close to the peak of the binding-energy curve (the peak, near iron/nickel, is about 8.5–8.8 MeV/nucleon). Splitting ${}^{41}$Ca would produce fragments that are lighter still and therefore sit further from, not closer to, the peak — fission is endothermic here, not exothermic. Fusing ${}^{41}$Ca with anything produces a heavier nucleus that has moved past the peak, which is also energetically unfavourable. Only nuclei well below the peak (light nuclei, which gain binding energy per nucleon by fusing up toward iron) or well above it (heavy nuclei, which gain binding energy per nucleon by splitting down toward iron) release energy this way — mid-mass nuclei already near the peak, like ${}^{41}$Ca, do neither.