The average moderator temperature is $725{}^{\circ }\mathrm{C}$. First the mean coolant temperature in the lumped thermal model isAt a steady state, the same thermal power crosses both thermal resistances. If $R_1=R_2$, thenThusIn particular $R_1=R_2=(800-650)/(2P)=3.17797\times 10^{-7}\mathrm{K}{\mathrm{W}}^{-1}$ at the specified state.

The coolant mass flow rate is approximately $983\mathrm{kg}{\mathrm{s}}^{-1}$. All thermal power leaves in the steady coolant energy rise:With the stated constant specific heat capacity,The $100{}^{\circ }\mathrm{C}$ difference equals a $100\mathrm{K}$ difference. Thermal power is used directly; no electrical conversion efficiency enters.

Substitute the separated solution into the one-group neutron diffusion equation. The radial and axial Laplacian contributions are $-{\alpha }^2\varphi$ and $-{\beta }^2\varphi$. A nonzero solution therefore requiresThe left side is the geometric buckling; the right side is fixed by the stipulated homogeneous material. Their equality is the model's nuclear criticality condition, expressing the balance of production, absorption and leakage. Both sides have inverse-length-squared units.

The fundamental nonnegative mode is ${\varphi }_0{J}_0(2.405r/R)\cos (\pi z/H)$. Put the origin at the centre of the cylinder, so $-H/2\le z\le H/2$. Neglecting extrapolation imposes zero neutron flux at $r=R$ and $z=\pm H/2$. WriteFor an axisymmetric field, separation of variables in cylindrical coordinates givesChoose the axial separation constant so $Z^{\prime \prime }+{\beta }^{2}Z=0$. Midplane symmetry and the lowest positive axial mode give $Z(z)=\cos (\beta z)$, withThe radial equation isThe solution regular on the axis is the zero-order Bessel function of the first kind, $F(r)\propto J_0(\alpha r)$; the independent second-kind solution is singular there. The first radial zero gives $\alpha R=j_{0,1}\simeq 2.405$. Higher zeros would introduce sign changes and do not describe the fundamental nonnegative neutron flux. Normalize by the central value, using $J_0(0)=1$:The printed $2.405$ is a rounded Bessel function zero, rather than an exact root.

Define ${\gamma }_{SV}$, ${\gamma }_{SL}$ and ${\gamma }_{LV}$ as the solid-vapor, solid-liquid and liquid-vapor interfacial free energies per actual area. For the corresponding planar material, Young's contact-angle equation isConsider advancing the macroscopic liquid footprint over projected area $A$, while retaining the specified wetting only to each sphere's equator. The newly wetted upper hemispheres have actual area $2\varphi A$. The gaps occupy projected area $(1-\varphi )A$, across which a liquid-vapor interface remains instead of solid-liquid contact.

The microscopic free-energy change, apart from the outer macroscopic droplet surface, isEquating this with $-{\gamma }_{LV}\cos {\theta }^{\ast }$, the macroscopic wetting contribution, yieldsThe factor $2$ comes from the curved hemisphere area; the negative gap term is the trapped liquid-vapor interface. This combines Wenzel roughness on the wetted solid fraction with Cassie-type air gaps. It differs from both homogeneous Wenzel wetting and a Cassie formula that treats the solid contacts as flat.

For $\varphi =0.9069$ and $\theta =68^{\circ }$, this gives $\cos {\theta }^{\ast }\simeq 0.5864$ and ${\theta }^{\ast }\simeq 54.1^{\circ }$. The hybrid state is less wettable than the fully penetrated hemispherical model, but still not hydrophobic. The result is conditional on the assumed equatorial meniscus and retained air: pinning or metastability may be required, and allowing further penetration can change the equilibrium geometry.

Let $A$ be projected area. Close-packed circular footprints cover $\varphi A$ and the exposed flat substrate covers $(1-\varphi )A$. A hemisphere's curved area is twice its circular footprint, so the actual wetted solid area isThe Wenzel roughness factor is therefore $r=A_{\mathrm{wet}}/A=1+\varphi =1.9069$, not $\varphi$ or $2$.

Using the Wenzel wetting relation,The nanostructure makes this surface more wettable, because planar PMMA already has $\theta <90^{\circ }$. Under complete Wenzel wetting, roughening amplifies the underlying hydrophilic character; it does not turn this material hydrophobic.

The calculation assumes identical chemistry on spheres and substrate, complete liquid penetration, ideal hemispheres, a macroscopic contact angle and negligible pinning/line-tension effects. Actual dried particles may have different contact geometry, deformation, contamination or trapped air. A hydrophobic coating with planar angle above $90^{\circ }$ would make roughness increase the angle instead. Designing a texture that retains air in a Cassie–Baxter state is another route, but its stability against liquid penetration must be checked.

For ideal hard spheres, potential energy is zero for nonoverlapping configurations and infinite for overlaps. At fixed temperature and volume, the accessible fluid and crystal configurations therefore have no energetic preference of the ordinary bonding kind: the phase with larger entropy has lower Helmholtz free energy.

Ordering particle centers costs some configurational freedom, but at a high packing fraction it can give each particle more usable cage/free volume than a disordered crowded arrangement. The total number of allowed vibrational and positional configurations can then be greater in a crystal despite its visible long-range order. This is entropy-driven hard-sphere crystallization.

The crystal need not be at close packing. At the stated volume fraction near $0.55$, substantial free space remains for thermal motion. Close packing near $0.74$ is the geometric upper bound for equal spheres, not the fraction at which the crystal first becomes thermodynamically favorable. Treating crystallization as an inevitable decrease in total entropy would confuse visible positional order with the full phase-space count.

The sedimentation scale height simplifies because the same drag coefficient controls both drift and diffusion:Applying the given criterion $a\le z_0$ givesEquivalently, the gravitational Péclet number on the particle-radius scale is $a/z_0=m_{b}ga/(k_{B}T)$ and must be at most one.

At $300\mathrm{K}$ the limiting radius is $a_{\max }\simeq 8.42\times 10^{-7}\mathrm{m}=0.84\mu \mathrm{m}$, corresponding to diameter approximately $1.68\mu \mathrm{m}$. Viscosity cancels from this equilibrium criterion, although it affects the time to approach equilibrium.

For the $100\mathrm{nm}$ radius particle in part (i), $D\simeq 2.20\times 10^{-12}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $z_0\simeq 0.50\mathrm{mm}$, about $5.0\times 10^3$ particle radii. It is therefore firmly diffusion-dominated on its own size scale. This does not mean uniform concentration over an arbitrarily tall vessel: the equilibrium density still decays on the finite scale $z_0$. The supplied size criterion is an approximate gravity-versus-diffusion criterion, not a universal definition of every stable colloid.

At terminal speed the downward buoyant weight balances the Stokes drag:Positive ${\rho }_p-{\rho }_f$ gives downward settling; a lighter particle would rise. With $a=10^{-7}\mathrm{m}$, ${\rho }_p-{\rho }_f=200\mathrm{kg}{\mathrm{m}}^{-3}$, $\eta =10^{-3}\mathrm{Pa}\mathrm{s}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,about $4.4\mathrm{nm}{\mathrm{s}}^{-1}$, or $0.38\mathrm{mm}$ per day.

The very small Reynolds number makes inertial corrections to Stokes flow negligible for an isolated particle. On short particle-scale distances, Brownian motion is much faster than this systematic drift. Nevertheless the nonzero settling speed can produce concentration gradients over sufficiently long times and large sample heights; Brownian motion does not cancel gravity.

The enlarged diamonds reach $|V_{sd}|=5.75\mathrm{mV}$ instead of $5.00\mathrm{mV}$. Subtract the unchanged charging contribution:Thus the two observed closed-orbital gaps are each approximately $0.75\mathrm{meV}$.

Define $A=h^2/(8m^{\ast }{L}_x^2)$ and $B=h^2/(8m^{\ast }{L}_y^2)$, and choose the labels so $A\le B$. With two spins per orbital, the first two closed orbitals correspond to occupancies $N=2$ and $N=4$. The lowest state is $(1,1)$ and the next is $(2,1)$, soThe next orbital is $(1,2)$, yielding the second observed gapThis ordering is self-consistent: $E_{11}=0.75$, $E_{21}=1.50$, $E_{12}=2.25$ and $E_{31}=2.75\mathrm{meV}$, as shown in the original level diagram.

Using $m^{\ast }=0.6m_e$ givesThe axes can be interchanged. Assigning the same $0.75\mathrm{meV}$ directly to both $3A$ and $3B$ would instead give a square box with degenerate $(2,1)$ and $(1,2)$ orbitals. That would require four electrons in that shell and would not give the stated second closed-orbital occupancy $N=4$.

Inside the rectangle the effective-mass Schrödinger equation iswith $\psi =0$ at all four walls because the confining potential is infinite outside. Separate $\psi (x,y)=X(x)Y(y)$. Each factor obeys a one-dimensional particle in a box equation. For $x$,so $X$ is proportional to $\sin (n_x\pi x/L_x)$ with $n_x=1,2,\dots$; the zero and negative integers give no new physical states. The same argument applies in $y$.

The normalized spatial states and their energies areThe product form and energy sum follow from separability. Each spatial orbital can accommodate two opposite-spin electrons under the stated spin degeneracy. These are single-particle confinement energies; the electrostatic charging contribution must still be added when interpreting transport addition energies.

The zero-bias gate period in the first plot is $\Delta V_g=20\mathrm{mV}$. In the constant interaction model, changing the gate-induced island charge by one electron givesThe tip is at $|V_{sd}|=5.0\mathrm{mV}$ above or below zero; use this half-height, not the full $10\mathrm{mV}$ vertical extent. The addition energy is $E_{\mathrm{add}}=e(5.0\mathrm{mV})=e^2/C_{\Sigma }$, giving

To distinguish the junction capacitances, explicitly adopt the circuit convention $V_s=V_{sd}$ and $V_d=0$. The diamond with zero-bias endpoints $20$ and $40\mathrm{mV}$ has its positive tip near $(V_g,V_{sd})=(35,5)\mathrm{mV}$. Its two edge slopes are therefore $+1/3$ and $-1$.

Writing the island electrochemical potential as a charge-dependent constant minus $e(C_s{V}_{sd}+C_g{V}_g)/C_{\Sigma }$, alignment with the drain gives slope $-C_g/C_s$, while alignment with the source gives $C_g/(C_{\Sigma }-C_s)$. ThusThe source/drain names are exchanged if the opposite lead is biased under the voltage convention. The figure alone establishes the pair $\{8,16\}\mathrm{aF}$ unless that wiring convention is specified. Symmetric-looking diamond heights do not imply equal junction capacitances.

The pattern is characteristic of a single-electron transistor: a small conducting island is connected to source and drain by two tunnel junctions, and to a gate through a capacitance. Inside a Coulomb diamond, changing the island's integer charge costs more energy than the leads can supply, so sequential tunnelling is blocked. Outside it, an island electrochemical potential lies in the transport window and current can flow.

For clear Coulomb blockade, the island charging scale must exceed thermal broadening, and the tunnel junctions must be sufficiently resistive to preserve well-defined charge states:in the simplest circuit. Stray capacitances should be included in $C_{\Sigma }$ if appreciable. The gate changes electrostatic energy but should not provide a conducting leakage path. Lifetime broadening and cotunnelling should be small enough not to wash out the blocked regions.

The almost identical unconfined diamonds are consistent with a constant interaction model having negligible or unresolved changes in single-particle level spacing.

In catalytic chemical vapor deposition, a metal nanoparticle activates a carbon-containing feedstock and supplies mobile carbon to the growing rim. Carbon can diffuse over the particle surface and, depending on the catalyst state, through its bulk. Once a curved carbon cap nucleates, continued incorporation extends the tube. Strong catalyst-support adhesion favors base growth; weaker adhesion can allow the particle to lift with the tube tip. A liquid catalyst is not universally required.

Particle size constrains tube diameter, but semiconducting character requires a particular wrapping family: $n-m\not\equiv 0(\mathrm{mod}3)$. Merely making particles smaller does not select that congruence class. Routes toward predominantly semiconducting carbon nanotubes include controlling catalyst composition, exposed facets and size distribution to bias cap/rim nucleation and growth, or adjusting carbon chemical potential and growth/etching conditions so that metallic tubes are less likely to survive.

These are mechanistic strategies rather than a universal recipe. Experiments with catalyst conditioning found that high semiconducting selectivity depended sensitively on oxidizing/reducing gas conditions and came with a strong yield penalty, motivating a catalyst-deactivation model.^[ref]

The useful distinction is between selecting a structure at nucleation, selecting its growth rate, and selectively terminating or removing unwanted tubes. Each can enrich the semiconducting fraction, but they do not establish the same growth mechanism.

Write the reaction using the species actually supplied by solution. For either halide, a bulk contribution per converted formula unit isThe stated room-temperature feasibility means that the overall reaction free energy is negative under those conditions; it does not say that each microscopic step is downhill. Equal nominal $50\mathrm{mM}$ concentrations also do not make iodide and bromide activities or solvation free energies identical.

At the nanoscale the balance additionally contains changing interfaces:Here $N_c$ is the number of converted formula units, and $j$ includes solvent-facing surfaces and contact with the titanium-dioxide support. Surface terms scale approximately as size squared whereas bulk terms scale as size cubed. Particle size, wetting, expansion during conversion and new grain interfaces can therefore shift both the conversion driving force and the nucleation barrier.

Microscopically, conversion requires desolvation, uptake of the organic cation and halide, redistribution through the precursor, and reorganization of the lead-halide framework. A study using these conversion conditions found energetically costly dispersed-ion intercalation for both compounds, with a much larger calculated penalty for MABr in PbBr${}_2$ than for MAI in PbI${}_2$. The iodide's layered framework accommodates the cation more favorably; the bromide's undulated framework gives poorer coordination and greater local distortion. Stabilization on forming intermediate compounds and the final perovskite can nevertheless make the complete reaction favorable. The calculations are approximate energetic comparisons, not measured finite-temperature activation free energies.^[ref]

Thus both reactions can be spontaneous yet proceed at very different rates: the same study observed faster MAPbI${}_3$ formation. One must distinguish net reaction free energy from intermediate penalties and diffusion barriers. Spontaneity alone supplies no quantitative ranking of the complete iodide and bromide reaction free energies.