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The event rate counts unordered pairs of reacting nuclei. For unlike species there are $n_i{n}_j$ pairs per unit-volume normalization; for identical species there are approximately $n_i^2/2$ pairs, since counting ordered pairs counts each collision twice. The Kronecker delta therefore supplies exactly the required divisor $1+{\delta }_{ij}$. It is the event rate, not the destruction rate of identical nuclei: one identical-pair reaction destroys two nuclei.

The temperature dependence combines the high-energy tail of the Maxwell-Boltzmann distribution with quantum tunnelling through the Coulomb barrier. The principal exponential factor in the energy integral is $\exp [-E/(k_{B}T)-b/\sqrt{E}]$. Its stationary point, the Gamow peak, has $E_0\propto T^{2/3}$ and exponent $\eta =3E_0/(k_{B}T)\propto T^{-1/3}$. Integrating around this peak gives a prefactor proportional to $T^{-2/3}$, or ${\eta }^2$, when the nuclear factor varies slowly. Thus the supplied thermonuclear reaction rate is locally proportional to ${\eta }^2{e}^{-\eta }$. The slow weak interaction in the proton-proton reaction also makes its overall coefficient very small; its local thermal slope follows the supplied formula.

At fixed number densities, define the local temperature exponent of a thermonuclear reaction by a logarithmic derivative. Since $d\eta /d\log T=-\eta /3$,For proton-proton reactions, $A=1/2$ and $Z_1=1$. At $T_6=15$, ${\eta }_{11}=13.67$ and $\alpha =3.89$. For two helium-3 nuclei, $A=3/2$ and $Z_3^2{Z}_3^2=16$, giving ${\eta }_{33}=49.68$ and $\beta =15.89$. For helium-3 with helium-4, $A=12/7$ and the same charge product gives ${\eta }_{34}=51.95$ and $\gamma =16.65$. HenceThese are local powers near the specified temperature, not constant exponents over an arbitrarily large range. The greater reduced nuclear mass in the $34$ reaction increases its tunnelling exponent and its thermal sensitivity.

For the effective proton-proton reaction network, denote the event rates byThe short-lived beryllium intermediate can be eliminated, so production through $R_{34}$ feeds $n_7$, representing predominantly lithium-7. This rapid conversion is electron capture. Count the nuclei consumed and produced in each reaction: $11$ consumes two protons and makes one deuterium; $21$ consumes a proton and a deuterium and makes one helium-3; $33$ consumes two helium-3 and makes one helium-4 and two protons; $34$ consumes one of each helium isotope; $17$ consumes a proton and lithium-7 and makes two helium-4. It follows thatFor closure, ${\dot{n}}_7={\lambda }_{34}{n}_3{n}_4-{\lambda }_{17}{n}_1{n}_7$. These nuclear terms conserve $n_1+2n_2+3n_3+4n_4+7n_7$, providing an independent stoichiometric check. They describe a fixed local volume; compression or mixing would add transport terms.

The roughly one-second destruction time of deuterium is negligible compared with the proton-burning and helium-3 evolution times. Its abundance rapidly adjusts to the slowly evolving production rate. Set ${\dot{n}}_2\simeq 0$, giving the deuterium quasi-equilibrium in proton-proton burningSubstitution, without setting ${\dot{n}}_1=0$, givesThe helium-3 relaxation time of about $6\times 10^5$ years is likewise short compared with the central proton-burning time and the age of the Sun. On this relaxation time the proton and helium-4 abundances, and the background temperature, can be treated as fixed. Put $A={\lambda }_{33}$, $B={\lambda }_{34}{n}_4$, $C={\lambda }_{11}{n}_1^2/2$. Then ${\dot{n}}_3=C-A{n}_3^2-B{n}_3$, and its nonnegative equilibrium isThe other root is negative and unphysical. Write $n_3=n_{3e}+x$ and subtract the equilibrium equation. Exactly, for a fixed background,Dropping the quadratic displacement givesThus small displacements decay exponentially. A slowly moving equilibrium introduces an additional $-{\dot{n}}_{3e}$ term, negligible to leading order when $\tau$ is short compared with background evolution.

To estimate the temperature for helium-3 freeze-out, use the cooler pp-I-dominated region and neglect $B$ compared with $2A{n}_{3e}$. Then $n_{3e}\simeq n_1\sqrt{{\lambda }_{11}/(2{\lambda }_{33})}$ and ${\tau }^{-1}\simeq n_1\sqrt{2{\lambda }_{11}{\lambda }_{33}}$. With the local powers just obtained and fixed proton number density, $n_{3e}\propto T^{-6}$ and $\tau \propto T^{-10}$. Adopting $t_{\odot }\simeq 4.6\times 10^9\mathrm{yr}$ givesThis is an order-of-magnitude extrapolation of local exponents. Keeping the supplied exponential factors gives the helium-3 freeze-out estimate with Gamow factorswhich yields about $6.8\times 10^6\mathrm{K}$ with the same normalization and fixed-density approximation. Thus the robust estimate is several million kelvin. The supplied data do not fix the local density, composition or pp-II fraction away from the centre, so they cannot determine an exact transition temperature or radius. In particular, using $\tau \propto T^{-16}$ alone would omit the temperature dependence of the equilibrium helium-3 abundance.

The solar hydrogen and helium-3 abundance profiles follow from burning, finite equilibration time and convection. Central hydrogen has been consumed, so $X_1$ is lowest in the centre and increases toward its much less processed outer value. In the hot centre, rapid destruction keeps helium-3 small. Moving outward initially raises its equilibrium abundance because ${\lambda }_{11}/{\lambda }_{33}$ rises strongly as temperature falls. Further out, equilibration is no longer reached within the solar age, and still farther out even production is slow. Hence $X_3$ has an off-centre maximum and then declines toward its small outer abundance. The outer convective envelope mixes each abundance toward a constant value.

The sketch shows these qualitative shapes; the helium-3 ordinate is scaled to its own maximum. The illustrated amplitudes, peak radius and envelope boundary are schematic, not a numerical solar-model solution.

The supplied rate law gives a local logarithmic temperature sensitivity, evaluated at fixed number densities. Since $\eta \propto T^{-1/3}$,Thus a power law here describes the tangent logarithmic slope near the chosen temperature, not an exact power law over all temperatures. For the three reactions, the factors $A{Z}_i^2{Z}_j^2$ are $1/2$, $24$ and $192/7$, respectively. At $T_6=15$ the local temperature exponent of a thermonuclear reaction givesHence $\alpha \simeq 4$, $\beta \simeq 16$, $\gamma \simeq 16$ and $\gamma >\beta$. The larger reduced mass for reaction 34 produces the last inequality. Composition changes contribute separately to an evolving reaction rate; they are held fixed for this derivative.

For the effective proton-proton reaction network, let the event rates beThe factors of one half count identical pairs once. Event 11 consumes two protons and makes one deuteron; event 21 consumes a deuteron and a proton and makes helium-3; event 33 consumes two helium-3 nuclei and makes one helium-4 nucleus and two protons. Under the stipulated fast-capture approximation, event 34 consumes helium-3 and helium-4 and supplies one mass-seven nucleus, and event 17 consumes that lithium-7 nucleus and a proton and makes two helium-4 nuclei. ThereforeFor closure, ${\dot{n}}_7={\lambda }_{34}{n}_3{n}_4-{\lambda }_{17}{n}_1{n}_7$. The stoichiometry conserves $n_1+2n_2+3n_3+4n_4+7n_7$, the baryon number density at fixed volume. In particular the mass-seven capture produces two helium-4 nuclei, not one.

The beryllium-to-lithium step is electron capture. Eliminating beryllium assumes that its capture flux tracks its production on the slow evolutionary timescale. More generally one retains ${\dot{n}}_{\mathrm{Be}}=R_{34}-{\lambda }_e{n}_{\mathrm{Be}}$ and ${\dot{n}}_7={\lambda }_e{n}_{\mathrm{Be}}-R_{17}$; setting the former to zero gives the effective equation used above. Fast capture relative to slow evolution alone would not prove that lithium exceeds beryllium: if both reach steady state their ratio is $n_{\mathrm{Be}}/n_7={\lambda }_{17}{n}_1/{\lambda }_e$. The mass-seven abundance assertion is thus part of the stipulated schematic limit, not a consequence to impose on every detailed solar model.

The enormous separation between the one-second deuterium destruction time and the other stated timescales justifies deuterium quasi-equilibrium in proton-proton burning:Substitution givesThe approximation is to the rapidly adjusting intermediate abundance, not to the slow proton abundance.

Near the centre the helium-3 relaxation time is also short compared with solar age and with the timescale of significant hydrogen evolution. Put $A={\lambda }_{33}$, $B={\lambda }_{34}{n}_4$ and $C={\lambda }_{11}{n}_1^2/2$. For slowly varying background quantities, the stable positive root of $A{n}_3^2+B{n}_3-C=0$ gives the helium-3 equilibrium abundanceThe other root is negative and unphysical. The production-minus-destruction function decreases strictly with positive $n_3$, so this is the unique attracting equilibrium. For $n_3=n_{3e}+x$, with the background held fixed during relaxation,Dropping the quadratic perturbation term gives the helium-3 relaxation timeIf the equilibrium itself evolves slowly, an additional forcing term $-{\dot{n}}_{3e}$ appears; tracking is accurate when this drift is small over one relaxation time. The given central value $6\times 10^5$ years is much shorter than the roughly $4.6\times 10^9$-year age of the Sun.

To estimate the temperature for helium-3 freeze-out, take the cooler pp-I-dominated regime and keep the background proton mass density approximately fixed for this order-of-magnitude scaling. ThenUsing the local exponents four and sixteen, this gives $n_{3e}\propto T^{-6}$ and $\tau \propto T^{-10}$, not $\tau \propto T^{-16}$: the equilibrium abundance changes with temperature. Normalizing at the central temperature givesSetting this equal to solar age yieldsThis is the requested power-law estimate. The exponents were evaluated locally at the central temperature, so extrapolation over this large range is approximate. Keeping the supplied full exponential rate factors in the same fixed-density pp-I estimate giveswhich gives about $6.8\times 10^6\mathrm{K}$. Thus the robust scale is a few million kelvin, roughly six to seven million in these estimates. A unique precise solar transition temperature cannot be obtained from the given numbers without the mass density, composition and pp-II contribution as functions of radius.

Finally, use mass fractions $X_i\simeq i{m}_u{n}_i/\rho$. The hydrogen mass fraction is depleted most strongly in the central burning region, so $X_1$ increases outward and approaches the nearly unprocessed envelope value. Convective mixing makes the outer envelope composition approximately uniform.

In the hot core helium-3 is quickly destroyed and remains close to its small equilibrium abundance. Moving outward, its destruction rates fall much faster than its production rate, so the equilibrium ratio $n_{3e}/n_1\propto T^{-6}$ rises. Where the relaxation time becomes comparable with solar age the abundance ceases to follow that rising equilibrium. Farther out, production itself becomes too slow to accumulate much helium-3, so $X_3$ falls again toward the envelope value. The result is a broad off-centre maximum, rather than a central maximum. These are the solar hydrogen and helium-3 abundance profiles requested by the sketch.

hydrogen rises from a depleted centre to an almost uniform envelope; helium-3 has a small central abundance and an off-centre peak before declining outward. The figure shows qualitative shapes with independently normalized vertical axes, not a fitted or computed solar model.