Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 63 / 3

Write $n_{56}$ and $n_{54}$ for the number densities of nickel-56 and iron-54. Both nuclei contain 28 neutrons, while their proton numbers differ by two. Dividing the positive-exponential nuclear statistical equilibrium formulas printed in the PDF therefore gives the nickel–iron nuclear equilibrium ratioThe neutron-density factors cancel. $K$ contains the isotope-dependent statistical and translational mass factors suppressed in the supplied proportionality; it is constant in this simplified calculation. The TeX conversion incorrectly reverses the exponential sign, so the PDF sign is essential here.

In the specified restricted composition, equality of total proton and neutron numbers meansConsequentlyDifferentiation, with the suppressed prefactors held constant, givesUsing the paper's supplied signed energy difference producesWith that negative value, $F$ tends to zero at both temperature endpoints, increases below this temperature and decreases above it. At the stationary point $d^2\log F/d{T}^2=-3/T^2<0$, so it is indeed a maximum for the printed toy data.

There is a genuine physical-data qualification: the stated negative energy difference is inconsistent with interpreting $Q$ as the conventional positive total nuclear binding energy of these actual isotopes. Using atomic masses, whose electron masses cancel in this reaction,The masses used are $1.00782503223\mathrm{u}$, $53.93960899\mathrm{u}$ and $55.94212855\mathrm{u}$, from NIST hydrogen masses, NIST iron masses and NIST nickel masses. Small electronic binding corrections do not change the sign. With a positive $\Delta B$, the simplified $F(T)$ above is strictly decreasing for $T>0$ and has no finite maximum. Thus the numerical maximum follows from the supplied negative parameter, not from consistent physical binding energies; changing the TeX sign to recover a maximum would not repair the PDF's data.

Nuclear statistical equilibrium is appropriate when the relevant forward and inverse strong or electromagnetic reactions are fast compared with changes of temperature, density and composition. It is useful in sufficiently hot late silicon-burning material and in the hottest explosive burning regions of massive stars. Some burning zones reach only quasi-equilibrium, and cooling or expansion eventually freezes the abundances out; an equilibrium expression cannot by itself predict the full ashes when these timescale conditions fail. Temperature, density, conserved baryon number and the electron fraction in stellar matter must all be specified. Negligible free-particle contributions in bulk bookkeeping are not the same as exactly zero free densities in the equilibrium formula.

Strong and electromagnetic reactions preserve total charge and baryon number, so they conserve the proton-to-neutron ratio. Weak reactions need not: electron capture and beta-plus decay convert protons into neutrons and lower it. Such conversions occur in hydrogen-burning networks and in the nitrogen-to-neon chain during helium burning. At high density in advanced stellar cores and core collapse, electron capture becomes particularly important; neutrino emission also affects energy and lepton loss. Beta-minus decay changes the ratio in the opposite direction and can compete during later relaxation. Nuclear equilibrium for strong reactions does not automatically imply weak equilibrium or a fixed proton-to-neutron ratio throughout all stellar evolution.

Use $f=X_{22}$ for the neon-22 mass fraction, temporarily keeping it distinct from the earlier metallicity. Let $n_b=\rho /m_u$ be the baryon density, ignoring small nuclear mass defects. Equal masses of carbon and oxygen giveCarbon-12 and oxygen-16 each have equal proton and neutron numbers, while neon-22 has ten protons and twelve neutrons. The total proton and neutron numbers per baryon are consequentlyHenceThe equal carbon/oxygen mass split cancels from this result: any mixture of these two symmetric nuclei has the same electron fraction. Taking the paper's metallicity symbol to mean $f$ gives its requested expression.

The literal pre-helium-burning nitrogen interpretation, however, needs a correction. Each initial nitrogen-14 nucleus receives two alpha particles, including eight additional baryons, before becoming neon-22; the intervening beta-plus decay also emits a neutrino. Thus, if the entire original metallicity is nitrogen-14 with mass fraction $Z_{\mathrm{initial}}$, conservation of the number of seed nuclei givesThis is neon-22 neutron excess from nitrogen-14. The two expressions are compatible only when their arguments are distinguished: the displayed $(11-Z)/(11+Z)$ law uses the final neon mass fraction, not literally the earlier nitrogen mass fraction. Treating those fractions as equal is an extra simplifying approximation, not a consequence of the stated capture chain.

A radioactively powered Type Ia supernova light curve rises to a broad maximum over roughly weeks as energy deposited in expanding ejecta diffuses out, then declines over the following weeks and months. Its power comes predominantly from the nickel-56 decay chain, first nickel to cobalt and then cobalt to iron. Gamma rays and charged decay products deposit heat, which is reprocessed into optical and infrared radiation. At later times, declining cobalt decay power and increasing gamma-ray escape shape the tail. It is not sustained nuclear fusion in the expanding ejecta. The amount of radioactive nickel, photon diffusion time and deposition efficiency all matter, so an uncorrected fixed luminosity is only an idealization.

For the requested two-isotope yield calculation, assume the same total processed iron-group mass, namely $1M_{\odot }$ as inferred from the zero-neon reference. Treat the final freeze-out ashes as nickel-56 and iron-54, neglecting residual light particles and additional weak neutronization. If $w$ is the nickel mass fraction, conservation of electron fraction givesSince $26/54=1/2-1/54$, this reduces to the metallicity dependence of nickel-56 yieldUnder the paper's intended identification $f=Z=1/50$,If instead $Z=1/50$ denotes the earlier nitrogen mass fraction, retaining the actual capture stoichiometry gives $w=1-27Z/7=323/350$ and $M_{56}\simeq 0.92M_{\odot }$. The same temperature and density do not alone fix a total burned mass; equality of that mass is also implicit in the comparison. The two-isotope assumption is a bulk-ash approximation, not a complete equilibrium network with all free particles identically absent.

For otherwise comparable explosions, approximate the peak luminosity by a quantity proportional to the radioactive nickel mass. Let $d$ be the true distance to an extremely low-neon event, and use the solar calibration $L_{\odot }/L_0=523/550$. Its measured flux is $F=L_0/(4\pi d^2)$, but the inferred standard-candle distance isThe source is intrinsically brighter than the calibration, so the distance is underestimated by about $2.5\%$, conventionally described here as about $3\%$. Equivalently, $d/d_{\mathrm{est}}=\sqrt{550/523}\simeq 1.0255$, consistent with the supplied rounded factor $1.03$. Using the literal nitrogen metallicity instead gives about a $4\%$ underestimate. These are conditional composition effects in this simplified model; real Type Ia distance estimation also standardizes light-curve shape and color.