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Take $x(0)=0$. Integrating the velocity solution givesThe initial/noise cross term vanishes under the independence assumption. Evaluating the remaining integral yields the general integrated Ornstein-Uhlenbeck displacementUse the thermal initial velocity and the fluctuation-dissipation relation for a Langevin particle. With diffusion coefficient $D=k_{B}T/\zeta$, this simplifies toFor $t\ll \tau$, expansion gives $⟨x^2⟩=(k_{B}T/m)t^2+O(t^3)$: displacement is initially $v_{0}t$, so the particle is ballistic. For $t\gg \tau$, $⟨x^2⟩=2Dt-2D\tau +o(1)$, hence the leading behavior is one-dimensional diffusion with $D=k_{B}T/(6\pi \mu a)$.

The ballistic conclusion depends on the initial state. For a particle prepared exactly at rest, $v_0=0$, the first expression gives $⟨x^2⟩=A{t}^3/(3m^2)+O(t^4)$ at very short times. A thermal initial velocity, or a nonzero initial velocity variance, is required for the stated $t^2$ mean-square law. It should not be replaced silently by a zero-velocity initial condition.

Model the polymerization Brownian ratchet by a rigid tip and a Brownian load. Let $a$ be the elementary load advance, ${\zeta }_L$ its drag coefficient, and $D=k_{B}T/{\zeta }_L$ its diffusion coefficient. For a spherical load, ${\zeta }_L=6\pi \mu R_L$, but no load radius is specified, so the rate is naturally expressed in terms of $D$. Assume addition is effectively instantaneous once the gap $h$ reaches $a$, and removal is negligible. The gap reflects at zero, first arrival at $a$ causes insertion, and insertion resets it to zero.

With no opposing force, the mean first-passage time satisfies $D{T}^{\prime \prime }=-1$, $T^{\prime }(0)=0$, $T(a)=0$. Integration gives $T(h)=(a^2-h^2)/(2D)$. One advance per passage givesThe advance-event rate is $2D/a^2$, while $V_0$ is a length per time.

A constant resisting force $F\ge 0$ produces drift $-u_F=-DF/(k_{B}T)$. The diffusion-limited Brownian ratchet with constant load now satisfiesAn integrating factor gives $T^{\prime }(h)=[1-e^{u_{F}h/D}]/u_F$. Integrating again from $h$ to $a$ gives $T(0)=D(e^z-1-z)/u_F^2$, where $z=Fa/(k_{B}T)$. ThereforeAt weak load the ratio is $1-z/3+O(z^2)$; at large load $V\sim (D/a)z^2{e}^{-z}$. Replacing the whole force dependence by $e^{-z}$ would discard the first-passage kinetics of this regime.

For two staggered strands that alternately advance the leading tip by half a monomer, $a=\delta /2$ and $V_0=4D/\delta$. Two aligned strands, as schematically drawn, do not automatically have that half-step: an advance may be $a=\delta$, with the other strand filling the same layer afterward. Use the same geometrical step in the passage problem and in converting event rate to velocity. The limiting formula assumes negligible removal, rigid tips, an overdamped load with constant drag, and fast insertion compared with $a^2/D$. With appreciable removal or additional internal strand states, the renewal reset model must be replaced by a coupled gap/reaction process.

The diffusion coefficient becomes $S\sigma (S)=\sqrt{S}$, so the stock is a driftless square-root diffusion, with zero taken as an absorbing boundary. Substitution of $V(t,S)=e^{A(t)S+B(t)}$ into the pricing equation for a local volatility model givesEquating the constant and linear terms and imposing the terminal payoff givesThus $1/A(t)=(1+t)/2$ and $B(t)=0$, and the price isThis is the exponential claim in a driftless square-root model. At the absorbing boundary its value is $1$, matching the payoff at zero.

The payoff in this part is unbounded, unlike the payoff assumed in part (i), so it is useful to verify that the proposed price is a finite conditional expectation. For $h>0$, starting from stock value $s$, defineThen $F_v+\frac{1}{2}x{F}_{xx}=0$, $F(h,x)=e^{-ux}$, and $0\le F\le 1$. By Itô formula, $F(v,S_v)$ is a bounded local martingale, hence a true martingale. It follows thatThis is also the Laplace transform of a sum of $N$ independent variables with exponential distribution of rate $2/h$, where $N$ has Poisson distribution of mean $2s/h$ and is independent of the summands: its transform is $\exp [(2s/h)((2/h)/(2/h+u)-1)]$. Uniqueness of the Laplace transform therefore gives the compound Poisson transition law of a driftless square-root diffusion. The exponential distribution jump has exponential moment $\mathbb{E}[e^Y]=(2/h)/(2/h-1)$ whenever $h<2$. Summing over the Poisson distribution yieldsFor the remaining horizon $h=1-t\le 1$, this is finite and, by the Markov property, is precisely ${\mathbb{E}}_Q[e^{S_1}\mid {\mathcal{F}}_t]=e^{2S_t/(1+t)}$. Thus the displayed process is indeed the risk-neutral pricing value and a true martingale. The replicating strategy from part (ii) has $h_t^S=[2/(1+t)]V(t,S_t)$ and $h_t^B=[1-2S_t/(1+t)]V(t,S_t)/B_t$.

The equation is a conservation law $n_t+J_x=0$ with diffusive flux $J=-D(n_0/n)n_x$. For positive density, insects move down the density gradient, with effective diffusion coefficient $D{n}_0/n$: dispersal is stronger in low-density regions. This is logarithmic diffusion, since the equation is $n_t=D{n}_0(\log n{)}_{xx}$.

Put $\tau =Dt$ and $\xi =x/{\tau }^{\beta }$. A self-similar solution $n=n_0{\tau }^{-\beta }g(\xi )$ conserves total number when ${\int }_{\mathbb{R}}g=1$. Its width shrinks to zero as $t\downarrow 0$; for an integrable nonnegative $g$, changing variables in the integral against a bounded continuous function proves convergence to $n_0$ times the Dirac delta distribution at the origin.

Substitution givesMatching powers for a nonstationary profile forces $\beta =1$. The profile equation integrates to $g^{\prime }/g=-\xi g+C$. For a centred even profile, $C=0$, and integration of $g^{\prime }=-\xi g^2$ gives $g=2/({\xi }^2+a^2)$. Normalization requires $2\pi /a=1$, henceThis is $n_0$ times a Cauchy distribution density with scale $2\pi Dt$. It is positive, has total mass $n_0$, and directly satisfies the nonlinear diffusion equation for $t>0$.

Assume $D_0,n_0,Q>0$. The density-dependent diffusion coefficient makes this a cubic porous medium equation, since its two-dimensional form is $n_t=[D_0/(3n_0^2)]\Delta (n^3)$. The mass-preserving similarity solution is a radial cubic-diffusion source profile. At fixed $r$, the proposed scaled density givesSeparation is possible whenwhere $\gamma$ is a constant. The resulting ordinary differential equation isRecognize the second term as $\gamma ({\xi }^{2}F{)}^{\prime }$. Regularity and zero radial flux at the origin set the integration constant to zero:In the region $F>0$, this reduces to $F{F}^{\prime }=-\gamma \xi$, giving $F^2=1-\gamma {\xi }^2$ from the central normalization. For a nonnegative profile whose front is precisely at $\xi =1$, the first zero fixes $\gamma =1$. Continuity of the density excludes truncating a positive value at the front. HenceHere integrating the scale equation gives ${\lambda }^6=6D_{0}t/r_0^2+\text{constant}$, and the point-source initial condition requires $\lambda (0)=0$.

Use the planar area element, not the one-dimensional area under the drawn profile. Mass conservation givesThus the scale and the complete density arewhere $[u{]}_{+}=\max (u,0)$. The factors of $n_0$ are necessary on dimensional grounds; the length scale is not a function of $Q$ alone.

The front and peak obeyEach radial profile starts with horizontal tangent at the origin, decreases to a square-root edge, and is identically zero beyond $R(t)$. Later profiles are wider and lower. This is finite propagation in porous-medium diffusion, because the diffusion coefficient vanishes at zero density. The sketch uses dimensionless time $\tau =6D_{0}t/r_0^2$; it preserves the radial integral $2\pi \int rndr$, not the unweighted area under each curve.

At the front $F^{\prime }$ diverges, so the profile is a weak solution, not a globally smooth classical solution. Nevertheless $F^2{F}^{\prime }=-\xi \sqrt{1-{\xi }^2}$ tends to zero there. The density and flux both match their zero exterior values, so extending the interior solution by zero creates no spurious boundary source in the conservation law. The behavior at the origin is regular because $F^{\prime }(0)=0$.

Finally, the point release is recovered as a distributional initial condition. For any continuous compactly supported test function $\phi$ on the plane, the total density remains $Q$ and all its support lies in the shrinking disk $r\le R(t)$. ThereforeThus $n(\cdot ,t)\to Q{\delta }^{(2)}$ as $t\downarrow 0$, in the sense of the Dirac delta function.

Use a planar saline Stefan problem with upward coordinate $z$, the initial contact at $z=0$, and phase boundaries $z=a(t)$ and $z=b(t)$ enclosing pure ice, where $a<b$. Fresh liquid occupies $z>b$ and brine occupies $z<a$. We assume negligible bulk flow, equal constant mass density $\rho$, specific heat capacity $c_p$ and thermal diffusivity $\kappa$ in all regions, and zero salt content and salt transport in the ice. Equal densities remove phase-change volume flow; suppressing convection isolates the molecular-transport mechanism. Different material properties would change the numerical coefficients. Set $K=\rho c_p\kappa$ for the common thermal conductivity and $\Delta T=T_m-T_{\infty }>0$. Denote the brine salt diffusion coefficient by $D$.

Use a linear ice liquidus, $T_L(C)=T_m-mC$, with $m>0$ and $C$ the salt mass fraction. Assume the initial brine is a stable liquid, $T_{\infty }>T_L(C_0)$, and all concentrations considered are below the eutectic composition. Thus $R=m{C}_0/\Delta T>1$. These assumptions exclude independent nucleation in a supercooled brine or an eutectic system event, neither of which is specified by the initial data alone. Start with a negligible seed of ice and use local phase equilibrium without interfacial kinetics or curvature. Temperature differences may be in Celsius, since only differences enter this calculation.

WriteThe paper calls $S$ a Stefan number; it is the latent-to-sensible heat ratio, reciprocal to another common Stefan number convention. At the lower phase boundary, define $T_i=T_L(C_i)$ and letThe liquidus relation is $\vartheta =Rc$.

A similarity solution gives an explicit calculation of the two positions. Putand use $u=z/(2\sqrt{\kappa t})$, $v=z/(2\sqrt{Dt})$. The temperatures and salinities areThese error function and complementary error function profiles satisfy the heat equation and the salt diffusion equation, their far-field conditions, and the interfacial temperature conditions. The fresh-water temperature is constant because the interface and the far field are both at $T_m$.

Salt conservation at the moving lower phase boundary requiresFor upward motion this is dilution by melting salt-free ice, rather than salt rejection by freezing brine. Substitution gives the dilution function for a melting saline Stefan front:For $A>0$, $0<c<1$, so the interfacial brine is fresher than the remote liquid.

The Stefan condition at the upper boundary is $\rho L\dot{b}=K{T}_{s,z}(b)$, since the fresh liquid has no temperature gradient. At the lower boundary it isThese signs follow from the jump in enthalpy: the upper front freezes liquid while a positive $\dot{a}$ melts solid. With $E=\mathrm{erf}B-\mathrm{erf}\delta$, they reduce toTogether with $\delta =ϵA$, these equations determine $A,B,\vartheta$, and hence both positions, without discarding the salt-diffusion correction.

For the stated large latent-to-sensible heat ratio, the layer is thin relative to the thermal diffusion length. Expanding the thermal equations for $B,\delta \ll 1$ givesConsequently a useful leading calculation iswhere the positive $A$ is obtained fromThe left side decreases with $A\ge 0$ while the right side increases, and their values at zero and infinity guarantee a unique positive root for large $S$. Thus this leading calculation includes the translation of the lower boundary as well as the increasing thickness.

If the scale separation also obeys $ϵSA\ll 1$, the familiar simpler result iswithThis last simplification needs the logarithmic refinement $ϵS\sqrt{\log (RS)}\ll 1$. The algebraic ordering $ϵ\ll S^{-1}$ alone should not be used to discard an arbitrarily large logarithmic correction; the preceding coupled equations remain the appropriate calculation when that refinement is unavailable. This distinction is captured by large latent heat in a freezing and melting ice layer.

The illustration uses $S=40$, $ϵ=2.5\times 10^{-4}$ and $R=2$. Solving the full equations gives $A=1.73707$, $B=0.0143053$ and $\vartheta =0.0158766$. It shows the ice layer between freshwater and cold brine, with both fronts advancing upward and a strongly diluted lower liquid boundary.

The phase diagram shows the spatial path from remote brine to the lower interface: temperature first changes over $\sqrt{\kappa t}$ at almost unchanged salinity, then salinity changes over its much smaller diffusion scale at nearly constant temperature. The liquid path ends on the liquidus at $(C_i,T_i)$; the solid has $C=0$ and its temperature rises from $T_i$ to $T_m$. The eutectic position is schematic, with $C_E=2C_0$ in the illustration, and is not used in the calculation. A point initially in the fresh layer freezes when $b$ reaches it, cools in the ice, and later melts when $a$ reaches it. The inset indicates this temporal path in the opposite direction through the solid branch and into the liquid.

The mechanism is simultaneous upper freezing and lower melting, with net growth of the ice thickness. Cold brine accepts most of the latent heat released by upper freezing. Salt reaching the lower contact lowers its equilibrium melting temperature, and melting freshens that liquid until its liquidus is close to $T_m$. The heat delivered through the ice supplies the much smaller melting demand there. The remote brine remains liquid even though it is below the pure-water melting temperature, because it lies above its own saline liquidus. Thus “cold” does not by itself decide which phase is stable. These conclusions apply while the layers remain effectively deep and the stated no-convection, pure-ice and phase-equilibrium assumptions hold.

Write $W(\Sigma )=\bar{\nu }\Sigma$. In a small radial patch $r=r_c+\chi$ with disturbance scale much smaller than $r_c$, the leading disk diffusion equation is ${\partial }_t\Sigma \simeq 3{\partial }_{\chi }^{2}W$. Taking the linearization around a uniform surface density ${\Sigma }_c$ givesFor a local Fourier mode $\delta \Sigma \propto e^{st+ik\chi }$,because $W\propto {\Sigma }^{-1}$ has $W^{\prime }<0$. Thus the effective density diffusion coefficient is negative and the disk has a viscous instability of an accretion disk.

A localized overdensity is a maximum of $\Sigma$ but a minimum of $W$. At its center ${\partial }_{\chi }^{2}W>0$, so ${\partial }_t\Sigma >0$: it grows. The local radial mass flux is proportional to $-{\partial }_{\chi }W$, directing matter toward that minimum from both sides. This is the physical feedback: an overdense patch has a smaller viscous torque in an accretion disk and attracts further accumulation. The arbitrarily rapid $k\to \infty$ growth in this ideal negative-diffusion model is not a resolved prediction below the validity scale of the thin-disk and local viscous closure.

Fix $M_{\star },{\rho }_p,t_{\star }$ and an initial-orbit ratio $\kappa =a_0/a_p$; use $\kappa =1$ for the figure. The qualitative finite-age planetary scattering regimes have two useful boundaries.

First, the order-unity escape-to-orbital-speed boundary $\chi =1$ isIt slopes down with logarithmic slope $-3/2$. Above it a planet can give large fractional binding-energy changes, favouring planetary ejection of a comet; below it weak kicks and repeated collision opportunities favour accretion.

Second, equating the characteristic comet energy diffusion time to the age givesThis boundary slopes up with logarithmic slope $3/4$. Above it the many-encounter diffusion estimate fits within the available age; below it that estimate exceeds the age.

Above both boundaries, rapid ejection is favoured. Above the age boundary but below the escape-speed boundary, repeated encounters can act during the age while individual kicks remain weak, so collision or accretion is commonly favoured. Below the age boundary, the diffusion model predicts incomplete ejection; the escape-speed boundary still distinguishes strong from weak individual kicks.

These labels are statistical expectations under the stated encounter model. In particular, the given diffusion coefficient supplies no collision rate, so it does not prove that every object in the low-$\chi$ region is accreted within $t_{\star }$. Nor does a long ejection time exclude faster collisions or other loss mechanisms. Very massive planets with $N_{\mathrm{ej}}\lesssim 1$ require a few-encounter treatment rather than an extrapolation of the diffusion formula.

Write $x=1/a$ for the comet's inverse semi-major axis. Its specific orbital energy is $-G{M}_{\star }x/2$, so escape corresponds to $x=0$. The printed quantityhas dimensions of inverse length. It cannot, literally, be a standard variance-per-time diffusion coefficient. Interpret it as the characteristic root mean square step in $x$ per periapsis passage, as in a discrete comet energy diffusion model.

If successive kicks are unbiased and uncorrelated, after $N$ passages the root mean square displacement is $s_x\sqrt{N}$. Starting at $a_0$ gives $x_0=1/a_0$, hence $N_{\mathrm{ej}}\sim (x_0/s_x{)}^2$. Taking the characteristic interval to be the initial orbital period $P(a_0)$ givesHere $P_p=2\pi \sqrt{a_p^3/(G{M}_{\star })}$. In particular, for $a_0\sim a_p$ this is $P_p(M_{\star }/M_p{)}^2/100$. Without specifying the initial semi-major axis and the time per statistically independent encounter, the PDF cannot determine a unique time.

With the convention $⟨(\Delta x{)}^2⟩=2D_{x}t$, the actual diffusion coefficient would be $D_x=s_x^2/[2P(a_0)]$. The estimate requires $N_{\mathrm{ej}}\gg 1$, orbit crossing, and sufficient phase decorrelation. Near escape the orbital period grows, and resonant or secular correlations invalidate the simple constant-step clock. This is a characteristic diffusion scale, not an exact mean first-passage time: even an unbiased Brownian motion on an unbounded half-line has an infinite mean time to reach its absorbing endpoint.

Let $\tau =|t-t^{\prime }|$. Integrating the equation for $x$ and using independence of the two noises givesWith the supplied integral,For $\alpha \tau \ll 1$, the active contribution is ballistic, $f_0^2{\tau }^2+O({\tau }^3)$, in addition to the Brownian term. For $\alpha \tau \gg 1$,so the long-time motion is diffusive with an enhanced diffusion coefficient.

Put $D={\sigma }^2>0$ and $\xi =x-x_0$. Logarithmic differentiation of the positive Gaussian function givesDifferentiating once more in space givesMultiplying by $D$ provesThe assumption $\sigma \ne 0$ ensures both the normalization and exponential denominator are defined. This is the heat kernel, with diffusion coefficient $D$ and centre $x_0$.

An exciton is a Coulomb-correlated electron-hole pair, electrically neutral overall. Three useful properties are:

In many molecular organic semiconductors, limited screening and localization make exciton binding important, so absorption need not immediately produce collectable free charges.

A useful microscopic screening estimate compares thermal energy with the buoyant gravitational potential energy over one particle radius. Write $\Delta \rho =50\mathrm{kg}{\mathrm{m}}^{-3}$ and $m_b=(4\pi /3)a^3\Delta \rho$. The sedimentation length is ${\ell }_g=k_{B}T/(m_{b}g)$, and the gravitational Peclet number isUsing $T=298\mathrm{K}$ and requiring ${\mathrm{Pe}}_g\lesssim 1$ givesEquivalently, combining Stokes law speed $v_s=2\Delta \rho g{a}^2/(9\eta )$ and diffusion coefficient $D=k_{B}T/(6\pi \eta a)$ gives $v_{s}a/D={\mathrm{Pe}}_g$. Buoyancy, not the full polystyrene density, is essential to the estimate. Taking a diameter rather than radius as the comparison length changes the numerical cutoff by an order-one factor.

There is no universal largest particle that stays uniformly suspended indefinitely. If uniformity over a vessel height $H$ is required, use ${\ell }_g\gtrsim H$, not merely ${\ell }_g\gtrsim a$. For example $H=1\mathrm{cm}$ gives a much smaller radius, about 58 nm. Observation time, density matching, convection, particle polydispersity and shape also matter. Independent of sedimentation, electrostatic/steric barriers must prevent aggregation; low ionic strength, suitable surface chemistry and clean solvent can be as important as size.

For the stipulated effective-sphere polymer drag, the radius is the end-to-end RMS distance, $R=2a\sqrt{N}$, rather than the single-monomer radius. Its Stokes law and translational diffusion coefficient areUnder the same three-dimensional RMS convention as used above,Thus for sufficiently long chains ($N>4$), this model gives a larger displacement than independent monomer drag: solvent motion couples the parts of a coil, reducing its effective friction relative to $N$ isolated moving beads. The $D\propto N^{-1/2}$ theta-solvent scaling is consistent with the dilute-coil hydrodynamic scaling underlying the Zimm model, but treating a permeable fluctuating coil as a solid sphere with radius exactly $R_{\mathrm{rms}}$ gives only an order-one hydrodynamic approximation. The hydrodynamic radius and end-to-end radius are not generally equal. Neither model by itself restores the long-time constraints of a truly entangled solution.

If the one-dimensional contour statistic in part (iii) is used literally, the comparison of this three-dimensional $d_2$ with that contour RMS is $\sqrt{3/2}{N}^{1/4}$; it compares different coordinates. Using mean radial displacements in both three-dimensional models instead changes both prefactors equally and leaves their ratio unchanged.

For a dilute suspended sphere in Stokes flow, the translational friction coefficient is $f_h=6\pi \eta R$, with dynamic viscosity $\eta$ and hydrodynamic radius $R$. To derive the diffusion-friction relation, put particles in a weak conservative force $F=-U^{\prime }$. Their drift velocity is $F/f_h$, while the diffusive flux is $-Ddc/dx$. At equilibrium the Boltzmann distribution gives $c^{\prime }=-c{U}^{\prime }/(k_{B}T)=cF/(k_{B}T)$. Zero total flux therefore gives $cF/f_h-DcF/(k_{B}T)=0$, hence the Stokes–Einstein relationHere $k_B$ is the Boltzmann constant, $T$ absolute temperature and $D$ the translational diffusion coefficient. The sphere, no-slip continuum-fluid and dilute thermal-equilibrium assumptions enter through the drag law and flux balance.

At $T=293.15\mathrm{K}$ and the supplied viscosity, the predicted product is $DR=2.1472\times 10^{-19}{\mathrm{m}}^3{\mathrm{s}}^{-1}$. The derived predictions and radii inferred from the measured diffusion coefficients are:

The last three agree closely at the precision of the supplied data. The ribonuclease diffusion coefficient is about $16\%$ below the spherical prediction, corresponding to a larger frictional radius than the stated radius. Hydration, shape, a different operational radius and coarse measurement precision can matter; molar mass is not an independent parameter in the spherical diffusion law. A compact constant-density globule would have $R\propto M^{1/3}$, whereas collagen's large frictional radius relative to its mass is consistent with an extended nonspherical molecule. The inferred Stokes radius is a friction-equivalent radius, not automatically a geometric half-length.

In analytical ultracentrifugation, scan the radial protein concentration while a rotor spins at known angular speed $\omega$. Optical absorbance or refractive-index interference records a moving sedimentation boundary and its diffusive broadening.^[ref] With solvent density ${\rho }_s$, partial specific volume $\bar{v}$ and molar mass $M$ in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, the buoyant mass is $m_b=M(1-\bar{v}{\rho }_s)/N_A$. Force balance gives $f_{h}v=m_b{\omega }^{2}r$, so the sedimentation coefficient is $s=v/({\omega }^{2}r)=m_b/f_h$. A boundary trajectory gives $s={\omega }^{-2}d\ln r/dt$; diffusion gives $f_h=k_{B}T/D$.

Combining them gives the molar mass from sedimentation and diffusion:Known mass and buoyancy can instead determine friction from $s$ alone, and comparison with a compact sphere tests shape or hydration. In sedimentation equilibrium, the stationary radial profile obeys $d\ln c/d(r^2)=m_b{\omega }^2/(2k_{B}T)$, providing buoyant mass without a friction model. This is a sedimentation-diffusion balance in a uniform solvent, rather than density-gradient banding at an isopycnic point.

The Stokes–Einstein relation gives the translational diffusion coefficientat $T=298\mathrm{K}$. For isotropic three-dimensional Brownian motion, the mean squared displacement is $⟨|\Delta \mathbf{r}{|}^2⟩=6Dt$. Thus after one hour the root mean square displacement isThe phrase “on average” needs a convention: the mean displacement vector is zero; the mean displacement magnitude of three-dimensional Brownian motion is $⟨|\Delta \mathbf{r}|⟩=4\sqrt{Dt/\pi }=1.22\mathrm{mm}$, obtained by integrating the three-dimensional Gaussian distribution. These are displacements, not the total microscopic path length. The estimate assumes dilute unconfined water, a spherical hydrodynamic radius and constant temperature.

Equilibrium junction. Put the p-type side on the left and n-type side on the right. When they are joined, electrons diffuse from n to p and electron holes diffuse from p to n. This leaves negative ionized acceptors on the p-side and positive ionized donors on the n-side of the depletion region. The resulting built-in electric field points from n to p, opposing further diffusion. At equilibrium there is one constant Fermi level, and drift and diffusion currents cancel for each carrier species.

In the nondegenerate, fully ionized limit, the neutral-region carrier densities areThe conduction band and valence band lie higher on the p-side and bend down towards the n-side. The gap between them stays fixed for a homojunction. If $\psi$ increases towards the n-side, both edges shift by $-e\psi$, whileThus $n$ rises and $p$ falls continuously across the junction; neither mobile population is an exact step function. Their depletion exposes fixed dopant charge density, rather than an absence of every kind of charge. The built-in voltage is

Light emission. Forward bias lowers the barrier and injects minority carriers across the p-n junction. In a light-emitting diode, electron-hole recombination converts some of their energy into photons. A direct band gap permits efficient radiative transitions with nearly conserved crystal momentum; nonradiative channels compete. Near-edge emission has $\hslash \omega \simeq E_g$ and $\lambda \simeq hc/E_g$. Material composition and confinement select the emission colour. Reverse-bias saturation is not the ordinary LED operating mode.

Reverse current. Take $V=V_R>0$ and $I_R>0$ to mean reverse voltage and reverse current magnitudes. The Shockley diode equation in this convention isThe bias-independent $I_S$ term is the supply of thermally excited minority carriers from the neutral regions, swept across the junction by its electric field. The opposing injection current is reduced by the extra barrier energy $e{V}_R$ and is $I_S{e}^{-e{V}_R/(k_{B}T)}$. Their equality gives zero net current at zero bias; for $e{V}_R\gg k_{B}T$, the reverse current approaches $I_S$ until leakage or breakdown invalidates this ideal law. The collecting field need not create the minority carriers; their thermal supply limits the diffusion-dominated current.

For an ideal long junction of area $\mathcal{A}$, steady diffusion with recombination gives the diode minority-carrier diffusion currentIt depends on doping, geometry, carrier diffusion coefficients and lifetimes, and especially the intrinsic carrier concentration. Depletion-region generation current and surface leakage can add nonideal contributions to the measured reverse current.

Band-gap estimate from the data. Read a nearly flat region well above the small-bias knee and below the high-bias upturn, rather than estimating $I_S$ in the breakdown region. For a diffusion-dominated junction with nearly temperature-independent gap and effective masses,so $I_S=C(T)T^3{e}^{-E_g/(k_{B}T)}$, where $C(T)$ includes transport and doping factors. Extracting a semiconductor band gap from diode saturation current amounts to measuring the slope of $\ln [I_S/(C(T)T^3)]$ against $1/T$: it is $-E_g/k_B$. If the prefactor is taken approximately constant over a short interval,Approximate readings from the original PDF plateaus areUsing the two warmer curves and $k_B=8.6173\times 10^{-5}eV/K$ givesKeeping the $T^3$ factor but taking the residual transport factor constant changes this estimate to approximately $0.57\mathrm{eV}$. The justified graphical estimate is thereforeThe colder curve is a genuine limitation of the supplied data: the $0$ to $-45{}^{\circ }\mathrm{C}$ pair gives only about $0.32\mathrm{eV}$ with a constant prefactor, and the endpoint pair gives about $0.42\mathrm{eV}$. Thus all three cannot support one precise diffusion-only activation slope. A depletion-region generation current, approximately proportional to $n_i$ rather than $n_i^2$, can have a roughly half-gap activation energy; other leakage and temperature-dependent transport can also alter the slope. The PDF alone does not identify their relative contributions. A reliable precision gap measurement would use more temperatures and verify the transport regime instead of forcing one fit through these three readings.