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Use geometrized units with $\hslash =k_B=1$ initially, and write $f(r)=1-2M/r$, $h(r)=1-Q^2/(Mr)$. The given nonextremality assumption makes $h_H=h(2M)=1-Q^2/(2M^2)>0$. Thus the static Killing vector field has a simple zero of its norm atThe exterior $r>2M$ is static. For radial null geodesics, $dt/dr=\pm 1/f$, independently of $h$. With $v=t+\int dr/f$, the metric iswhich is regular at $r_H$ since $h_H>0$. The outgoing null equation is $dr/dv=f/2$. Inside the future horizon these rays move to smaller radius, while outside they can escape to infinity. Hence the zero is the event horizon. The inner zero of $h$ is not another regular Killing horizon: the Killing norm there diverges rather than vanishes. The strict charge bound matters when making the regular extension.

For the Euclidean black-hole regularity condition, continue $t=i{t}_E$, consistent with the sign of the given quantum time-translation operator. Set $x=r-2M$. To leading order $f=x/(2M)$ and $h=h_H$, so the radial–time part of the Euclidean metric isIntroduce the radial proper length coordinate $\rho$ by $x=h_H{\rho }^2/(8M)$. ThenThe first two terms are a polar plane with angle $t_E/(4M)$. Smoothness at its origin requiresA different period would make a conical singularity. The time coordinate is normalized to proper time at infinity, so no further asymptotic redshift factor is needed. The cancellation of $h_H$ proves the charge independence of the Euclidean horizon period under a regular radial conformal factor; this argument would fail at $h_H=0$.

To justify the thermal interpretation, analytic continuation of the time-evolution operator $U(t)=e^{iHt}$ gives $U(i{t}_E)=e^{-H{t}_E}$. Its matrix element is an imaginary-time propagation kernel. Identifying the initial and final configurations and integrating over them takes its trace, giving the Euclidean path integral $Z=\mathrm{Tr}{e}^{-\beta H}$. In an energy eigenbasis each state has weight $e^{-\beta E}$, exactly the Boltzmann factor of a canonical ensemble. Thus the geometrically required period is the inverse Hawking temperature. With dimensions restored, an imaginary time period $\Delta t_E$ gives $\beta =\Delta t_E/\hslash =1/(k_{B}T)$. ConsequentlyHere $\mathcal{M}$ is mass in ordinary units; for a neutral hole $M=G\mathcal{M}/c^2$ and the physical imaginary-time period is $8\pi G\mathcal{M}/c^3$. Bosonic thermal fields are periodic and fermionic fields antiperiodic over that same Euclidean period. The relation to the canonical density operator, not just dimensional reasoning, identifies the temperature.

For an evaporation estimate take a neutral, isolated Schwarzschild black hole and approximate its radiated power with the Stefan–Boltzmann law. Its area is $A=16\pi G^2{\mathcal{M}}^2/c^4$, and the photon Stefan-Boltzmann constant is ${\sigma }_{SB}={\pi }^2{k}_B^4/(60{\hslash }^3{c}^2)$. Substituting the temperature givesThe black-hole evaporation speeds up as the mass decreases. Integrating, rather than dividing the initial rest energy by the initial power, givesThe corresponding initial Hawking temperature is about $6\times 10^{-8}$K. This is the Stefan-Boltzmann estimate of Schwarzschild evaporation time at the level of accuracy requested. Greybody factors and other emitted species alter the numerical coefficient, but not the leading cubic scaling of a large isolated hole. The estimate assumes negligible accretion or ambient radiation absorption; a stellar-mass hole in the present cosmic radiation bath is not currently losing mass at that isolated-hole rate. Finally the semiclassical formula does not describe the Planck-scale endpoint: extrapolating it to zero mass is an estimate of the macroscopic lifetime, not a proof of what the final quantum object is.

For the Gaussian critical exponents, use dimensionless statistical-action conventions, with the Boltzmann factor $e^{-H}$. If $H$ is an energy, apply the following to $H/(k_{B}T)$ and absorb that smooth factor into the couplings. Choose the Fourier transform conventionReality gives $\tilde{\varphi }(-p)=\tilde{\varphi }(p{)}^{\ast }$. The Fourier-space statistical Hamiltonian isThe source term is not multiplied by one half. Completing the Gaussian functional integral gives $⟨\tilde{\varphi }(p)⟩=\tilde{J}(p)/({\kappa }^{-1}{p}^2+m^2)$ and the connected Fourier-space two-point correlation functionAssume $m^2>0$ for the stable massive Gaussian field theory. In the continuum long-distance theory, the real-space correlation function is the Green kernel of $-{\kappa }^{-1}{\nabla }^2+m^2$. Away from its source, a radial ansatz $G(R)\sim R^{-(D-1)/2}{e}^{-R/\xi }$ gives, at large $R$, $m^2-{\kappa }^{-1}{\xi }^{-2}=0$. Equivalently the propagator pole is at $p^2=-\kappa m^2$. Thus the exponential correlation length isAt $m=0$ the correlation length diverges. A sharp momentum regulator itself introduces artificial oscillatory long-distance tails; this length describes the physical continuum pole or a local/smoothly regulated model on distances large compared with $a$.

For a momentum-shell renormalization group step with $b>1$, split the Fourier modes into retained modes $|p|<\Lambda /b$ and eliminated modes $\Lambda /b<|p|<\Lambda$. Define ${\tilde{\varphi }}_B(p)$ as the retained field. Integrating the eliminated modes is exact for a Gaussian field theory; a uniform source acts only on the zero mode, so this integration produces a field-independent free-energy term. Rescale $q=bp$ and set ${\tilde{\varphi }}_B(p)=\tilde{Z}{\tilde{\varphi }}^{\prime }(q)$. The retained quadratic action becomesKeeping the gradient coefficient fixed requires $b^{-D-2}{\tilde{Z}}^2=1$. For $J=h$, the source term is $-h{\tilde{\varphi }}_B(0)$, so the complete Gaussian rescaling isThe corresponding real-space field obeys ${\varphi }_B(x)=b^{(2-D)/2}{\varphi }^{\prime }(x/b)$; confusing that real-space factor with $\tilde{Z}$ would change the source exponent incorrectly.

Let $F$ be free energy per original volume, with $t$ proportional to $m^2$. Rescaled volume is $V^{\prime }=V/b^D$, so the Gaussian free-energy scaling relation isA field-normalisation Jacobian can be included in the field-independent shell term. For any finite step on positive-mass modes, that background is analytic near $t=0$. Subtract regular backgrounds and choose $b=t^{-1/2}$ on the massive side. The usual homogeneous singular scaling then givesThese are Gaussian critical exponents for the stipulated Gaussian approximation. A useful check is its source-dependent contribution $-h^2/(2m^2)$: $D/2-2\Delta =-1$ gives precisely that power of $t$.

There is a qualification to a strictly pure-power singular free energy. The Gaussian free-energy logarithm at effective dimension two in $D=2$ contains $t\log t$ after subtracting analytic terms. Thus the displayed exponent assignments remain the formal power indices, but at $D=2$ the advertised homogeneous expression needs this additive logarithmic term. A purely quadratic theory also cannot stabilise the ordered $t<0$ phase or the zero-mass zero mode at nonzero $h$; the scaling calculation is taken from $t>0$, and an ordered-phase continuation requires stabilising interactions.

At a uniaxial Lifshitz point, the quadratic kernel instead isThe vanishing coefficient of $p_{\parallel }^2$ is an additional tuning that distinguishes this Lifshitz point from an ordinary critical point. Use a factorised cutoff, retain $|p_{\perp }|<{\Lambda }_{\perp }/b$ and $|p_{\parallel }|<{\Lambda }_{\parallel }/c$, and integrate its complement. A rectangular or cylindrical retained region is suitable; its precise boundary is not an exponent. This is a uniaxial Lifshitz Gaussian renormalization step.

Set $q_{\perp }=b{p}_{\perp }$, $q_{\parallel }=c{p}_{\parallel }$, and ${\tilde{\varphi }}_B=\tilde{Z}{\tilde{\varphi }}^{\prime }$. The momentum measure acquires $b^{-(D-1)}{c}^{-1}$. Keeping both kinetic coefficients fixed requiresTherefore the anisotropic blocking and source factors areThe anisotropic effective dimension is $d_{\mathrm{eff}}=D-1+1/2=D-1/2$: one parallel coordinate contributes half the scaling weight of a perpendicular coordinate. The rescaled-volume factor is $b^{-d_{\mathrm{eff}}}$, soChoosing $b=t^{-1/2}$ gives the requested Lifshitz Gaussian exponents:Again $(D-1/2)/2-2\Delta =-1$, checking the uniform-source susceptibility. The associated length powers are ${\xi }_{\perp }\sim t^{-1/2}$ and ${\xi }_{\parallel }\sim t^{-1/4}$. A Gaussian determinant has the analogous logarithmic exception if its effective dimension equals 2. Interactions can alter these exponents; dimensional rescaling here solves the specified Gaussian model.

Take the couplings in the Hamiltonian to include inverse temperature, so the Boltzmann factor is $e^{-H}$. If physical energies are used instead, first replace $J,L,D,K$ by their products with $1/(k_{B}T)$. Split the on-site term equally between its two bonds. For spin states $s,t\in \{1,0,-1\}$ the spin-chain transfer matrix has entriesUsing the specified coupling coordinates, in the order $(1,0,-1)$ this isFor example, $e^{D/2}=x/z$ and $e^{-J+L+D}=y/z$. This is the nearest-neighbour Blume–Emery–Griffiths model with an additive constant and the paper's sign convention for $D$.

Summing the periodic spin chain gives the partition function $Z_N=\mathrm{tr}{W}^N={\sum }_{a=1}^3{\lambda }_a^N$. For finite real couplings all entries of $W$ are strictly positive; the Perron–Frobenius theorem gives a unique positive dominant eigenvalue with ${\lambda }_1>|{\lambda }_a|$ for $a\ne 1$. Since $W$ is a symmetric matrix, all its eigenvalues are real. Thus in the thermodynamic limit,The positivity condition is stronger and more useful than mere ordering by signed value: subdominant eigenvalues can be negative. Also, the printed strict ordering between the other two is not guaranteed for all couplings. For instance, $J=L=D=0$ makes $W$ the positive constant matrix $e^{K}11^T$ with two equal zero eigenvalues. Degeneracy there does not affect the largest-eigenvalue limit; no explicit generic eigenvalues are needed.

For spin magnetization, introduce a dimensionless field $h$ through $-h{\sum }_i{\sigma }_i$, or insert $S=\mathrm{diag}(1,0,-1)$ into the trace. Thenwhere $v_1$ is a normalized eigenvector of the largest eigenvalue and $W_{st}(h)=e^{h(s+t)/2}{W}_{st}(0)$. Spin inversion symmetry gives $v_{1,+}=v_{1,-}$, since the positive eigenvector of the largest eigenvalue is unique. Thereforeat zero field, both at finite $N$ and in the finite-coupling thermodynamic limit. The finite-$N$ result follows directly by pairing each configuration with its spin-reversed partner. Eigenvalues at a single fixed field do not determine a general observable: a field derivative of the largest eigenvalue, or its eigenvector, is required. Positivity and the real analytic dependence on the couplings exclude a finite-temperature spontaneous symmetry breaking transition in this one-dimensional finite-range chain; singular zero-temperature coupling limits require separate treatment.

For even $N$, spin decimation on alternate sites sums the middle spin of each two-bond segment, hence the coarse spin-chain transfer matrix is $W^{\prime }=W^2$. DefineDirect multiplication gives $W^2=e^{2K}{z}^{-2}\left(\begin{array}{ccc}A& B& C\\ B& E& B\\ C& B& A\end{array}\right)$. Matching its entry ratios to the original parameterization yields the spin-1 chain decimation recursionThe remaining overall positive factor is absorbed into $K^{\prime }$. Keeping that factor preserves the free energy as well as normalized spin probabilities. Indeed $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$, exactly; on an odd ring an unmatched boundary segment needs separate handling rather than assuming a uniform two-site block decomposition.

In chemical equilibrium, the reaction $p+e\leftrightarrow H+\gamma$ obeys chemical-potential balance for ionizationbecause a photon gas in equilibrium has photon chemical potential ${\mu }_{\gamma }=0$. Applying the given nonrelativistic Maxwell-Boltzmann distribution to each massive species givesThe Hydrogen binding energy is $I=m_e+m_p-m_H$ in units with $c=1$. Thus Saha's equation isFor ground-state hydrogen, $g_e=g_p=2$ and $g_H=4$, so the degeneracy factor is one. Since $m_p\simeq m_H$, the translational factor is commonly written $(2\pi m_e{k}_{B}T/h^2{)}^{3/2}$.

Charge neutrality gives $n_e=n_p$. Write $n_B=n_e+n_H\simeq \eta n_{\gamma }$. Thenand henceCombining this with the stated photon density produces

In our universe the baryon-to-photon ratio is tiny, so there are roughly $10^9$ photons per baryon. Even when the mean photon energy is far below $I$, the high-energy blackbody radiation tail contains enough ionizing photons to suppress neutral hydrogen. The recombination temperature is reached only after the Boltzmann factor overwhelms this large photon-to-baryon ratio, at $k_{B}T\simeq 0.3\mathrm{eV}$.

For a universe with $\eta =1$, take the degeneracy factor as one, $m_H\simeq m_p$, and define $x=I/(k_{B}T)$. Neutral formation occurs when the boxed ratio is of order one, giving approximatelyWith $I=13.6\mathrm{eV}$ and $m_e=5.11\times 10^5\mathrm{eV}$, this gives $x\simeq 19$ and thereforeto order unity. This is much hotter than recombination in our universe because there is no enormous excess of photons, but it remains below $I$ because the electron's large translational phase space still favours the ionized state.

A subcritical droplet must make a rare thermal fluctuation from the metastable basin over the free-energy barrier $F^{\ast }$. Its probability carries the Boltzmann factor $e^{-F^{\ast }/(k_{B}T)}$, so the nucleation rate has the Arrhenius formFor a volume containing order one candidate subcritical droplet, the waiting time for a supercritical droplet and subsequent macroscopic growth is thereforeup to an algebraic kinetic prefactor. This is the Arrhenius nucleation time.

If a physical tension $f$ acts on a state of extension $l_i$, its mechanical energy is $-f{l}_i$. The canonical ensemble therefore assigns that state the Boltzmann factorComparison with the statistical weight $e^{\tau l_i}$ givesEquivalently, $f$ is the thermodynamic conjugate variable to $L$ in the Gibbs free energy, just as pressure is conjugate to volume.

For a ferromagnetic nearest-neighbour Ising model on a chain let $s_i=\pm 1$, $K=J/(k_{B}T)>0$ and $h=H/(k_{B}T)$. The dimensionless Hamiltonian is $-K{\sum }_i{s}_i{s}_{i+1}-h{\sum }_i{s}_i$. In one dimension, decimating every other spin closes exactly within the nearest-neighbour family, including its additive constant. This is the exact chain building block of the Migdal-Kadanoff approach; bond moving becomes an additional approximation in higher-dimensional lattices.

For exact zero-field Ising-chain decimation, sum over an intermediate spin $\sigma$ between two retained spins:Parallel and antiparallel end spins give $A{e}^{K^{\prime }}=2\cosh 2K$ and $A{e}^{-K^{\prime }}=2$. HenceFor an even periodic chain, $Z_N(K)=A^{N/2}{Z}_{N/2}(K^{\prime })$. Rescaling the lattice spacing $2a$ back to $a$ completes the renormalization step. Decimation of $b$ successive links gives $\tanh K^{\prime }=(\tanh K{)}^b$, as follows by multiplying the transfer matrix $b$ times.

Let $v=\tanh K$. Then $v^{\prime }=v^2$ on $0\le v\le 1$. The infinite-temperature renormalization-group fixed point $v=0$ attracts every $0\le v<1$; the zero-temperature renormalization-group fixed point $v=1$ is unstable to heating, with slope two. There is no intermediate ferromagnetic renormalization-group fixed point. Equivalently $K^{\prime }\simeq K^2$ for small $K$, while $K^{\prime }\simeq K-\frac{1}{2}\log 2$ for large finite $K$. Thus there is no finite-temperature ordered thermodynamic phase in the short-range chain.

The domain-wall scaling coordinate of an Ising chain near $T=0$ is the domain-wall Boltzmann factor $w=e^{-2K}$. Its recursion and the zero-field correlation length areSince $\xi (K^{\prime })/a=[\xi (K)/a]/2$, the recursion has the correct length scaling. Its exponent is one with respect to $w$; it is not a finite power of physical temperature about $T_c=0$. In the thermodynamic limit, a bond is a domain wall with probability $(1-\tanh K)/2\sim e^{-2K}$, giving the same exponentially large length scale. Their finite energy allows entropy to proliferate walls at any positive temperature.

The Ising-chain decimation constant matters for thermodynamics. With dimensionless free energy per spin $f=-\lim N^{-1}\log Z_N$,The second expression follows independently from transfer eigenvalues $2\cosh K$ and $2\sinh K$ and satisfies the recursion. Multiplying it by $k_{B}T$ restores the physical free energy. Dropping $A$ would preserve normalized retained-spin probabilities but lose the eliminated-spin entropy.

Finite-field Ising-chain decimation is also exact. Use the symmetric transfer matrixand read the new couplings fromHere $W_{++}=e^{2K+2h}+e^{-2K}$, $W_{--}=e^{2K-2h}+e^{-2K}$ and $W_{+-}=2\cosh h$. The latter off-diagonal entry, the field convention split between adjacent bonds and the constant must all be tracked consistently. Near the zero-temperature zero-field renormalization-group fixed point $h^{\prime }\simeq 2h$, while at $K=0$ the independent-spin field is unchanged.

In the higher-dimensional Migdal-Kadanoff approximation, moving bonds creates chains that can be decimated. With bond moving first on a $d$-dimensional hypercubic lattice and length factor $b$, the strengthened links have $b^{d-1}K$, giving the conventionFor $d=1$ no transverse bonds move, so this reduces to the exact chain result. In higher dimensions it discards geometric correlations and is generally approximate on the original lattice; moving and decimating in another order yields another approximation, not a contradictory exact chain formula.^[ref]

As an estimate, model the confinement by an infinite square well of width $L$ and use the stipulated electron mass. Its energy eigenvalues are $E_j=j^2{\pi }^2{\hslash }^2/(2m{L}^2)$, $j=1,2,\dots$, so the gap controlling thermal occupation of a quantum well isThus $T\ll 130\mathrm{K}$ suppresses excitation: $p_2/p_1=e^{-\Delta E/(k_{B}T)}$. For example $T\lesssim 28\mathrm{K}$ makes this ratio below one percent in this model. There is no sharp cutoff. The ground state energy itself corresponds to about $44\mathrm{K}$, but it is the excitation gap that enters the Boltzmann factor. Finite barrier height and a different effective mass would change the numerical scale; neither is supplied.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

The expression is the canonical partition function for a system in thermal equilibrium with a large heat bath at fixed temperature $T$. The system exchanges energy with the bath, while its particle number and external mechanical parameters, such as volume, are fixed. The bath is large enough that its temperature is not appreciably changed by this exchange.

The system has allowed energy levels ${ϵ}_j$. The index $j$ labels distinct energy values, and the integer degeneracy factor $g_j$ counts distinct quantum states with that same energy. $k_B$ is the Boltzmann constant, and $\exp [-{ϵ}_j/(k_{B}T)]$ is a state's Boltzmann factor. The dimensionless sum $Z$ normalizes the probabilities:Thus the degeneracy multiplies the probability of occupying the level, not the probability assigned to each individual state within it.

The canonical model is not restricted to a classical ideal gas. It applies to a quantum rotor, for example, provided the sum counts all of that system's allowed states with their proper degeneracies. For a many-particle system, ${ϵ}_j$ would be the energy of a many-particle state, with any quantum-statistical restrictions already built into the state count.

Write $y=g{\mu }_{B}B/(k_{B}T)$. Summing the Boltzmann factors gives the one-atom partition functionSince $⟨m⟩=-{\partial }_y\log Z_1$, the Brillouin magnetization formula isHere $M$ is the total magnetic moment used by the question; the magnetization per unit volume is $M/V$. For independent distinguishable atoms, $Z_N=Z_1^N$, with no indistinguishability factor.

Putting $J=1/2$ yieldsThe apparent singularities of the hyperbolic cotangents at $B=0$ cancel; the finite sum makes the zero-field limit explicit.^[ref]

Nondegenerate carrier product. Assume three-dimensional parabolic energy bands near their extrema, spin degeneracy two, and no additional uncounted valley multiplicity. Near the conduction edge, a Boltzmann approximation to the Fermi-Dirac distribution givesAngular integration and the Gaussian radial integral yieldFor valence electron holes, the empty-state probability is $1-f(E)\simeq e^{(E-\mu )/(k_{B}T)}$. Measuring electron hole kinetic energy downwards from $E_v$ similarly givesThe factors two count spin once in each band. Multiplying cancels the chemical potential and proves the semiconductor mass-action law:In an intrinsic neutral semiconductor, $n=p=n_i$. In a doped equilibrium semiconductor the same product holds under the stated nondegenerate assumptions, even when $n\ne p$. Extra valleys or several valence branches must be included in the effective density-of-states factors rather than counted inconsistently.

Equilibrium junction. Put the p region on the left and the n region on the right. Before contact their majority carriers are electron holes and electrons, respectively. Diffusion initially sends electrons from n to p and electron holes from p to n. This leaves negative ionized acceptors on the p side and positive ionized donors on the n side, creating a depletion region and an electric field directed from n to p. At equilibrium this field balances diffusion, so the net electric current vanishes.

For a homojunction the gap remains constant while the conduction band and valence band edges bend downwards from p to n by $e{V}_{\mathrm{bi}}$. The equilibrium chemical potential is flat. Since $E_{c,v}(x)=E_{c,v}^{\mathrm{ref}}-e\varphi (x)$, the electrostatic potential rises from p to n. The carrier profiles satisfyWith fully ionized nondegenerate dopants, the neutral p side has $p_p\simeq N_A$, $n_p\simeq n_i^2/N_A$, while the neutral n side has $n_n\simeq N_D$, $p_n\simeq n_i^2/N_D$. Thus the electron concentration rises across the junction and the electron hole concentration falls, with both small relative to the far-side majority densities in much of the depleted region. The ideal depletion approximation neglects these mobile populations when calculating the fixed space electric charge, but the actual equilibrium profiles are smooth.

The built-in potential follows from these Boltzmann factors:There is no applied terminal bias; the internal electrostatic potential variation does not contradict a constant electrochemical potential.

Opposing carrier fluxes under reverse bias. Use $V>0$ for the magnitude of reverse bias and $I>0$ for reverse electric current, as in the PDF. The junction barrier increases by $eV$. The Shockley diode equation in that orientation can be writtenThe reverse-directed contribution is supplied by minority electrons on the p side and minority electron holes on the n side. They reach the depletion edge by diffusion and are swept across by the built-in field. The opposing contribution is associated with majority carriers thermally crossing the barrier in the forward direction; reverse bias suppresses it by the Boltzmann factor. At zero bias detailed balance makes the two equal, while at large enough reverse bias the opposing flux becomes negligible.

In the ideal neutral-region diffusion model,Here $D_n,D_p$ are minority-carrier diffusion coefficients, $L_n,L_p$ their diffusion lengths, and $\mathcal{A}$ is junction area. The two terms inside this positive prefactor are the electron and electron hole contributions; both flow in the same reverse-current direction. This separates the two carrier-species contributions from the two opposing fluxes in the Shockley diode equation. Calling the saturation term a drift current refers to its passage across the depleted region, and does not eliminate the minority-carrier diffusion needed to supply it.

Reading the measured curves. At $eV\ll k_{B}T$, the ideal equation gives $I\simeq I_{\mathrm{sat}}eV/(k_{B}T)$, initially proportional to reverse electric potential difference. At $eV\gg k_{B}T$, it approaches a plateau $I\simeq I_{\mathrm{sat}}$. The thermal electric potential differences at the three temperatures are about $21$–$26\mathrm{mV}$, consistent with a rise at small reverse biases and near saturation at larger ones. Higher temperature produces a much larger plateau because thermal electron-hole excitation raises $n_i^2$ exponentially.

The lower-temperature curve has a noticeable residual electric potential difference dependence rather than a perfectly flat plateau. Depletion-region generation, surface leakage or other departures from the ideal diffusion model can contribute. The sharp upturn at the largest reverse electric potential differences is reverse breakdown, outside the unextended Shockley diode equation; the plot alone does not identify a unique microscopic breakdown mechanism. The original horizontal axis duplicates the final decade label, so a precise breakdown electric potential difference cannot be read reliably. Neither issue prevents estimating the gap from the intermediate-bias saturation region.

Band-gap estimate. Use plateau electric currents at a common intermediate reverse bias, away from both the small-bias rise and the breakdown upturn. Approximate readings are $I_{\mathrm{sat}}(298\mathrm{K})\sim 2.5\times 10^{-6}\mathrm{A}$, $I_{\mathrm{sat}}(273\mathrm{K})\sim 1.5$–$2\times 10^{-7}\mathrm{A}$, and $I_{\mathrm{sat}}(248\mathrm{K})\sim 10^{-8}\mathrm{A}$. Kelvin temperatures are required in the activation law.

For fixed effective masses, $n_i^2\propto T^3{e}^{-E_g/(k_{B}T)}$. If the remaining transport prefactor is approximately constant, extracting a semiconductor band gap from diode saturation current therefore usesUsing the widest temperature separation and the rounded readings givesNeglecting all algebraic prefactors and fitting $\ln I_{\mathrm{sat}}$ instead gives about $0.70\mathrm{eV}$. These differ by roughly $0.07\mathrm{eV}$, a material correction on the graph's precision scale. Thus an honest graph-based conclusion isThe nonflat low-temperature plateau, rough electric current readings, unprovided temperature dependence of $D/L$, and possible gap variation prevent a more precise extraction. If depletion-region generation dominated rather than the specified $I_{\mathrm{sat}}\propto n_i^2$ mechanism, its activation slope could involve a different energy scale; the displayed estimate explicitly uses the diffusion-dominated saturation model requested here.

Assume a dilute nondegenerate hydrogen gas in thermal and chemical equilibrium, with Maxwell-Boltzmann translational partition functions, ground-state neutral atoms, single ionization and charge neutrality $n_p=n_e$. For $\mathrm{H}\rightleftharpoons p+e$, chemical equilibrium gives ${\mu }_H={\mu }_p+{\mu }_e$. The ideal-gas relation ${\mu }_i=E_{i0}+k_{B}T\log (n_i{\lambda }_i^3/g_i)$, with $E_{H0}=-R$, givesNeglecting the electron mass in $m_H/m_p$ and using the ground-state electronic statistical weights, $g_e=2$ and $g_H=2$ with cancelling nuclear-spin factors, this becomesExcited neutral states have an energy gap of at least $3R/4$ above the ground state. At the temperatures relevant to the plotted maximum their Boltzmann factors are very small despite the finite degeneracy increase. Highly excited levels also require a physical plasma cutoff; the untruncated infinite isolated-atom partition sum is not a valid finite-density model. Ground-state dominance is an approximation in this regime, not a theorem at all temperatures.