Past exam of the physics A course of the University of Cambridge / 2019 / Paper 2

Build a narrow-band wave packet from plane waves with wavenumbers close to $k_0$:The Taylor expansion of the dispersion relation is $\omega (k)\simeq {\omega }_0+{\omega }_0^{\prime }(k-k_0)$, where ${\omega }_0^{\prime }=(d\omega /dk{)}_{k_0}$. Put $\kappa =k-k_0$ to obtainThe carrier's wave phase travels at the phase velocity ${\omega }_0/k_0$, while the envelope has constant shape at constant $x-{\omega }_0^{\prime }t$. Its group velocity is thereforeThe approximation neglects higher derivatives of the dispersion relation. Those derivatives cause the wave packet to spread or distort over sufficiently long propagation times; the result describes the local envelope speed of a sufficiently narrow band.

Assume a massless junction and common tension $T$. The characteristic impedance of a stretched string is $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ is its linear mass density. The displacement reflection coefficient is $r=(Z_1-Z_2)/(Z_1+Z_2)$, so the reflected fraction of power isSet $\mathcal{R}=1/2$. The positive solutions are $z=3\pm 2\sqrt{2}$, which are reciprocals. Thus the string density ratio for prescribed reflected power isThe impedance ratio is about $5.83$, rather than the required density ratio. Reversing which string is on the incident side changes the reflected displacement sign but not the reflected power fraction.

A Michelson interferometer splits a collimated beam at a partially reflecting beam splitter. Two mutually perpendicular arms reflect from mirrors and recombine at the same beam splitter; the superposed returning beams produce interference fringes at the detector. Translate one mirror along its arm while holding the other fixed.

If that mirror moves a measured distance $\Delta x$, the round-trip optical path difference changes by $2\Delta x$ in air, approximated as vacuum. Every change of one wavelength moves one complete bright-dark-bright fringe cycle past a fixed detector point. Count $N$ such cycles to obtainUse a spectral line selected with a suitable optical filter, and count many interference fringes to reduce the fractional uncertainty of the distance and count. Keep the two arms within the source's optical coherence length and maintain overlap and alignment. If the vacuum wavelength is required for an arm filled with air of known refractive index $n$, the corrected relation is ${\lambda }_0=2n\Delta x/N$ at stable $n$.

Take an abrupt equilibrium p-n junction with the p side at $x<0$ and the n side at $x>0$. In the quasi-neutral bulk, electron holes are the majority carriers on the p side and Electrons are the majority carriers on the n side. With ionized dopant densities $N_A,N_D$ and nondegenerate statistics,where $n_i$ is the intrinsic carrier concentration. The semiconductor mass-action law $np=n_i^2$ applies at equilibrium. Both mobile populations are small in the depletion region compared with the dopant densities.

The charge and carrier density of an abrupt p-n junction must distinguish mobile populations from net charge. Diffusion leaves negatively charged ionized acceptor impurities on the p side and positively charged ionized donor impurities on the n side. Under the depletion approximation,The neutral bulks have large carrier densities but nearly zero net charge density. The electric field points from the positive donor region toward the negative acceptor region, balancing diffusion at equilibrium.

The figure uses equal dopant densities only to show a simple symmetric sketch. Generally the more lightly doped side has the wider depleted interval. The carrier profiles are smooth, whereas the step-shaped net charge density is the ideal depletion approximation.

Treat sodium as monovalent, with one mobile Electron per atom. Its mass density is ${\rho }_m=0.97\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}=970\mathrm{kg}{\mathrm{m}}^{-3}$, and its molar mass is $M=0.023\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Using the Avogadro constant,The electron mobility $\mu =e\tau /m_e$ relates the magnitude of drift velocity to electric field. The Drude model then givesUsing $m_e=9.109\times 10^{-31}\mathrm{kg}$,The inferred time is the momentum relaxation time. Identifying it with the mean time between collisions is the intended simple isotropic Drude model assumption; small-angle scattering need not relax directed momentum in a single event.

Draw the Fabry–Pérot etalon with parallel mirrors and an incident beam making internal angle $\theta$ with their normal. Every emerging transmitted contribution has undergone one more internal round trip than its predecessor.

The extra internal path is $2d/\cos \theta$, but successive contributions leave at laterally separated points, a distance $2d\tan \theta$ apart. Comparing their phases on the same outgoing wavefront subtracts the path projection $2d\tan \theta \sin \theta$. Their net optical path difference in air isHence, in the reflection-phase convention implicit in the supplied transmission formula,For a gap of refractive index $n$, replace $d$ by $nd$ and use the vacuum wavelength. Coating reflection phases can add a round-trip offset; the expression above assumes that offset is absorbed in the resonance order, or is zero modulo $2\pi$ as in the stated ideal model.

Write $C_R=4R/(1-R{)}^2$, keeping this coefficient distinct from the finesse $\mathcal{F}$. The transmission has maxima at $\delta =2\pi m$. At a displacement $ϵ$ from a maximum, its ratio to peak transmission is $[1+C_R{\sin }^2(ϵ/2){]}^{-1}$. Half maximum therefore requiresThis is the half-width in wave phase; the full width at half maximum is $2{\delta }_{1/2}$. Successive resonances are $2\pi$ apart, so the exact ideal-model result isFor narrow peaks, $(1-R)/(2\sqrt{R})\ll 1$, the small-angle approximation to the inverse sine givesThus the formula requested in the paper is the high-reflectivity approximation, not an exact identity at arbitrary $R$. Ordinary half-maximum peaks exist only when $(1-R)/(2\sqrt{R})\le 1$, or $R\ge 3-2\sqrt{2}$; below that threshold the transmission minimum never reaches half the peak. The prefactor $T^2/(1-R{)}^2$ cancels from this relative-width calculation.

Let ${\lambda }_1=589.0\mathrm{nm}$, ${\lambda }_2=589.6\mathrm{nm}$ and $\Delta \lambda =0.6\mathrm{nm}$. Requiring the separation to be at least $100$ times the measured full width at half maximum means a width no larger than $0.006\mathrm{nm}$, or a resolving power of approximately $589.3/0.006\simeq 9.8\times 10^4$.

The etalon resolving power and order overlap constraints are particularly clear in frequency. Put $\Delta \nu =c|1/{\lambda }_1-1/{\lambda }_2|$. At normal incidence, the free spectral range is ${\nu }_{\mathrm{FSR}}=c/(2d)$ and the instrumental width is $\delta \nu ={\nu }_{\mathrm{FSR}}/\mathcal{F}$. ThusAvoiding ambiguity between the doublet and adjacent interference orders requires a free spectral range exceeding the doublet separation:Together these giveUsing the narrow-peak finesse from the previous part, $\pi \sqrt{R}/(1-R)>100$ gives $\boxed{R\gtrsim 0.9691}$, with a practical margin above that limiting value. At the exact limit, the gap between orders vanishes, so equality is not a sensible operating point.

One feasible ideal design is $R=0.99$, $\mathcal{F}\simeq 313$, and $d=0.14018\mathrm{mm}$. It has an approximate wavelength free spectral range of $1.24\mathrm{nm}$ and instrumental width $0.0040\mathrm{nm}$ near the doublet. The doublet spacing is about $151$ widths. Even the spacing to the nearest replica of the other line exceeds $100$ widths in this example. If that latter separation is also required in general, impose the stronger condition $\min (\Delta \nu ,{\nu }_{\mathrm{FSR}}-\Delta \nu )\ge 100{\nu }_{\mathrm{FSR}}/\mathcal{F}$, rather than just the two basic inequalities.

Such mirror reflectivities and submillimetre gaps are plausible for a high-quality Fabry–Pérot etalon. Flatness, parallelism, coating loss, angular spread, mechanical stability and source width limit the achieved finesse. A wavelength width of $0.004\mathrm{nm}$ corresponds to a gap variation of order $d\delta \lambda /\lambda \simeq 1\mathrm{nm}$, so the ideal coating-limited estimate demands stringent uniformity. The paper expressly excludes intrinsic source broadening.

The first panel is the fixed etalon's transmission versus wavelength, a comb of resonances. It does not by itself have one peak at each sodium line. The second panel shows the two peaks obtained by scanning the etalon phase or gap and calibrating the scan to wavelength. A fixed-gap output spectrum is instead the incident spectrum multiplied by the transmission function. This distinction is needed to interpret a doublet measurement correctly.

A high-finesse Fabry–Pérot etalon obtains very narrow transmission peaks by summing many coherent round trips, allowing high resolving power with a compact optical path. It can transmit a substantial fraction of resonant light when coating absorption is small, and works well for precise relative line separations and narrow-line measurements. A diffraction grating may require many illuminated grooves to achieve the same resolving power.

The etalon's free spectral range is limited, and repeated orders cause ambiguity across a broad optical spectrum. It commonly needs scanning and an order-selecting optical filter or another dispersive element. Mirror alignment, flatness and illumination angle strongly affect peak positions and widths, and absorption lowers throughput. A diffraction grating instead separates a broad range of wavelengths spatially for simultaneous detection, although grating orders can also overlap. The preferred instrument depends on whether narrow-band precision or broad spectral coverage is the priority.

Here $x(t)$ is displacement from equilibrium, in metres; $m>0$ is the oscillating mass, in kilograms; $b\ge 0$ is the coefficient of linear viscous damping, in $\mathrm{kg}{\mathrm{s}}^{-1}$; and $k>0$ is the spring constant, in $\mathrm{N}{\mathrm{m}}^{-1}$. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ are respectively the inertial term, the opposing velocity-dependent damping term and the opposing displacement-dependent restoring term when moved to the force side of Newton's second law.

$F_0$ is a possibly complex driving-force complex amplitude, measured in newtons. $\omega$ is the driving angular frequency in $\mathrm{rad}{\mathrm{s}}^{-1}$, and taking the real part of $F_0{e}^{i\omega t}$ gives the physical sinusoidal force. Its frequency in hertz is $f=\omega /(2\pi )$. The natural undamped angular frequency and damping rate areThis damped harmonic oscillator model assumes a linear restoring force and a linear, instantaneous viscous damping law.

Set the drive to zero and try $x=e^{st}$. The characteristic equation isFor underdamping, $\boxed{b^2<4mk}$, the roots are $-\gamma \pm i{\omega }_d$, with ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The displacement oscillates with an exponentially decaying envelope:For critical damping, $\boxed{b^2=4mk}$, there is a repeated root $-{\omega }_0$, giving $x=(C+Dt)e^{-{\omega }_{0}t}$. The motion is nonoscillatory and is the limiting case between oscillatory and overdamped response.

For overdamping, $\boxed{b^2>4mk}$, the roots are distinct negative real numbers and $x=C{e}^{s_{+}t}+D{e}^{s_{-}t}$. Both components decay without sustained oscillation; strong damping gives a particularly slow late-time relaxation. Constants $C,D$ depend on the initial conditions. A nonoscillatory solution can cross equilibrium once for some initial data, so “nonoscillatory” does not mean monotone for every possible initial condition.

Let the initial conditions be $x(0)=0$ and $\dot{x}(0^{+})=v_0>0$. The velocity-impulse response of a damped oscillator isAll curves start at the origin with the same positive slope $v_0$. Underdamping gives a decaying sequence of positive and negative lobes. Critical damping gives one positive lobe followed by return to zero. Overdamping also remains positive for these initial data, but has a longer tail. The initial impulse changes velocity abruptly, not displacement.

For underdamping, the general response is a free transient plus a particular driven response. Use the paper's $e^{i\omega t}$ convention. The steady displacement complex amplitude isThusHere $\gamma =b/(2m)$, ${\omega }_d=\sqrt{k/m-{\gamma }^2}$ and the real constants $C,D$ enforce the initial conditions. For a real positive $F_0$, the steady term is $|x_0|\cos (\omega t-\varphi )$, whereFor light damping, ${\omega }_d\simeq {\omega }_0$ and the transient decays slowly compared with an oscillation. The two-argument phase convention makes the displacement lag evolve from approximately zero below resonance through $\pi /2$ near ${\omega }_0$ to approximately $\pi$ far above it.

The response is quasistatic at low frequency, resonantly enhanced when the spring and inertia nearly cancel, and suppressed as $1/{\omega }^2$ at high frequency. The original graph gives approximate readings $x_s=|x_0(0)|\simeq 0.10\mathrm{mm}$, $x_p\simeq 1.0\mathrm{mm}$ and $f_r\simeq 150\mathrm{Hz}$. These are graph estimates, not exact supplied measurements.

Use oscillator parameters from static and resonant displacement. The static spring constant isFor light damping, ${\omega }_0\simeq {\omega }_r=2\pi (150)\simeq 9.42\times 10^2{\mathrm{s}}^{-1}$, givingNear resonance, $x_p\simeq F_0/(b{\omega }_0)$, soIndependently, the resonance enhancement $x_p/x_s\simeq 10$ estimates the same quality factor. The half-power displacement level is $x_p/\sqrt{2}\simeq 0.71\mathrm{mm}$; its intersections are approximately $142$ and $157\mathrm{Hz}$, giving bandwidth $\Delta f\simeq 15\mathrm{Hz}$ and $Q\simeq f_0/\Delta f\simeq 10$.

The exact displacement-peak frequency is ${\omega }_r^2={\omega }_0^2-b^2/(2m^2)$. At $Q\simeq 10$ its difference from ${\omega }_0$ is only about $0.25\%$, well below the precision justified by this printed curve. Greater numerical precision for the inferred parameters would be misleading.

In Fresnel diffraction, the propagation wave phase varies quadratically across the aperture and that curvature must be retained. For a transverse scale $w$ at propagation distance $L$, its size is of order $\pi w^2/(\lambda L)$. In Fraunhofer diffraction, that quadratic term is negligible, approximately $w^2/(\lambda L)\ll 1$, while the linear phase encoding the observation angle remains. A lens can also form the far-field angular pattern in its focal plane.

For a normally incident plane wave and aperture transmission $h(x,y)$, the far-field complex amplitude is its Fourier transform:The detected light intensity is proportional to $|\psi {|}^2$. For identical parallel slits, write the transmission as a sum of shifted copies of one slit. The Fourier shift theorem gives one single-slit transform multiplied by the sum of positional phase factors:The single-slit diffraction envelope sets the broad scale, while coherent interference between slit centres produces the finer array pattern. One sums complex amplitudes before squaring, rather than adding independent slit intensities.

Put the slit centres at $x_j=\pm D/2,\pm 3D/2$. Each slit has the same width $a$ and illumination. Its contribution to the Fraunhofer diffraction complex amplitude isThe four phase factors pair into cosines:Therefore the four-slit Fraunhofer diffraction field isThe common factor $2a$ can be absorbed in the proportionality constant, giving the requested form. In particular $\psi (0)\propto 4a$, as expected from four equal in-phase slits. Normalizing by the central intensity givesThe last form follows from $\cos x+\cos 3x=2\cos 2x\cos x$ and avoids apparent singularities in the usual sine-ratio array formula.

Let $X$ be horizontal position on the screen, $L=0.50\mathrm{m}$, and use $\sin \theta =X/\sqrt{L^2+X^2}$. The plotted interval is $-50\mathrm{mm}\le X\le 50\mathrm{mm}$. With $a=10\mu \mathrm{m}$, $D=20\mu \mathrm{m}$ and $\lambda =0.500\mu \mathrm{m}$, the four-slit Fraunhofer diffraction intensity is the previous normalized expression evaluated at this angle. The total aperture width is $3D+a=70\mu \mathrm{m}$, giving a Fresnel number of order $0.020$, consistent with the far-field approximation.

Nominal principal array directions satisfy $D\sin {\theta }_m=m\lambda$. Their screen positions areThus the visible nonzero orders are near $\boxed{X=\pm 12.50\mathrm{mm}(m=\pm 1),X=\pm 37.61\mathrm{mm}(m=\pm 3)}$, in addition to the central maximum. The even nonzero orders are missing diffraction orders: at $m=2$ the first envelope zero occurs at $X=\pm 25.03\mathrm{mm}$. The next envelope zero and $m=4$ direction lie at about $\pm 50.25\mathrm{mm}$, just beyond the screen edges.

At the nominal first and third array directions, $I/I(0)=4/{\pi }^2\simeq 0.405$ and $4/(9{\pi }^2)\simeq 0.0450$, respectively. The closest central minima satisfy $D\sin \theta =\pm \lambda /4$, hence $\boxed{X\simeq \pm 3.125\mathrm{mm}}$. Two small subsidiary array maxima occur between adjacent principal directions. The single-slit envelope slightly shifts the true off-axis maxima relative to the nominal principal array angles.

Model the two openings as identical axis-aligned rectangles of width $w$ in $x$, height $h$ in $y$, and centre separation $s$ along $x$. Their Fraunhofer diffraction field has the formThe original image shows first broad envelope zeros near $q_x=\pm 5\mu {\mathrm{m}}^{-1}$ and $q_y=\pm 1\mu {\mathrm{m}}^{-1}$. The horizontal interference period is about $\Delta q_x=2.5\mu {\mathrm{m}}^{-1}$; the first interference minima near $\pm 1.25\mu {\mathrm{m}}^{-1}$ distinguish that fine scale from the envelope. The rectangular-aperture dimensions from diffraction zeros are thereforeThe edge-to-edge gap is $s-w\simeq 1.26\mu \mathrm{m}$. The long direction is vertical. These dimensions also explain the suppressed alternate principal array maxima: $s/w\simeq 2$.

The factors $2\pi$ are required by the stated $q=2\pi \sin \theta /\lambda$ convention; treating the plotted coordinates as cycles per micrometre would give different dimensions. The finite printed figure supports approximate dimensions. An overall translation and inversion of the aperture pair leave its Fourier intensity unchanged, so their absolute placement is not recoverable; the diagram chooses a convenient centred representative.

A waveguide confines the transverse structure of a wave while permitting propagation along a chosen route. It can deliver energy to a distant receiver without the same geometric spreading as an unguided beam, protect a signal from unwanted coupling, and select transverse normal modes. It also makes propagation paths and coupling to devices more controllable. Confinement does not eliminate wave absorption or scattering loss, and it can introduce wave dispersion and cutoff frequencies of a membrane waveguide or corresponding cutoffs for other wave types.

Two examples are an optical fiber, which guides light by its core-and-cladding refractive index structure, and a hollow electromagnetic waveguide used to carry microwaves between a source and an antenna. A duct for acoustic waves is a further example. Each uses different material and boundary conditions, but the common idea is controlled transverse confinement.

For the trial displacement $z=\sin (qy)e^{i(kx-\omega t)}$, direct differentiation givesThe two-dimensional wave equation is therefore satisfied ifHere $k$ is the longitudinal wavenumber, $q$ the transverse wavenumber, and $c_m$ the speed of an unconstrained membrane wave. The physical displacement is the real part, multiplied by an arbitrary displacement amplitude. $T$ has units $\mathrm{N}{\mathrm{m}}^{-1}$ and $\sigma$ is surface mass density in $\mathrm{kg}{\mathrm{m}}^{-2}$, so $T/\sigma$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$ as required.

The fixed edge at $y=0$ is automatically satisfied by $\sin (qy)$. At $y=a$, the displacement must vanish at every $x,t$, so $\sin (qa)=0$. Thus the allowed transverse normal modes haveThe case $n=0$ gives zero displacement for this sine solution and is not a physical propagating mode. Negative $n$ only reverses the amplitude sign and does not create a distinct transverse mode. This discrete transverse spectrum is a consequence of the Dirichlet boundary conditions, rather than quantization of the longitudinal wavenumber in an indefinitely long guide.

For the $n$th transverse mode, substitute $q_n=n\pi /a$ into the membrane waveguide dispersion relation:Longitudinal propagation requires real nonzero $k$, henceBelow the cutoff frequency of a membrane waveguide, $k=i\kappa$, with $\kappa =\sqrt{(n\pi /a{)}^2-\sigma {\omega }^2/T}$, and a bounded excitation decays as $e^{-\kappa x}$ rather than carrying a travelling wave along an infinite guide. At equality $k=0$, the displacement has no longitudinal phase advance and the group velocity is zero; it is a cutoff oscillation, not a propagating mode carrying finite longitudinal power.

For positive angular frequency, the first three branches areEach branch is even in $k$, has minimum ${\omega }_n(0)=n\pi c_m/a$, and starts with zero slope. At large $|k|$, all approach the straight asymptote $\omega =c_m|k|$. Their group velocities are $v_{g,n}=c_m^{2}k/{\omega }_n$, whereas their longitudinal phase velocities are $v_{p,n}={\omega }_n/k$ for $k\ne 0$, giving $v_{p,n}{v}_{g,n}=c_m^2$.

Resistance to bending adds a restoring contribution that grows more rapidly with spatial curvature than the tension contribution. Short-wavelength modes are therefore stiffened especially strongly, raising their frequencies and changing the large-wavenumber behaviour.

For example, a thin plate of flexural rigidity $D_b$ has $\sigma z_{tt}=T{\nabla }^{2}z-D_b{\nabla }^{4}z$. When its edge conditions admit the same sine transverse mode, bending stiffness in a membrane waveguide givesThus the cutoffs increase, and bending-dominated propagation has $\omega \propto k^2$ at large $|k|$, rather than approaching $c_m|k|$. Its phase velocity and group velocity then grow with $|k|$. If “rigidly fixed” means a genuinely clamped plate edge, the slope must also vanish and the mode shapes change; the qualitative stiffening and stronger wave dispersion remain, but simple $q_n=n\pi /a$ need not remain exact.

A Bravais lattice is the set of positions generated by integer combinations of independent primitive translation vectors, $\mathbf{R}=n_1{\mathbf{a}}_1+n_2{\mathbf{a}}_2+n_3{\mathbf{a}}_3$. Every lattice point has the same translational environment. A crystal may additionally have a basis of several atoms attached to each lattice point; those atoms are not automatically separate points of its Bravais lattice.

A unit cell is a region whose translated copies fill the crystal without gaps or overlaps. It can be chosen for convenient symmetry, and a conventional cell may contain several lattice points. A primitive unit cell is a smallest such repeating cell, containing exactly one lattice point when boundary sharing is counted. Its shape is not unique, but its volume $\Omega =|{\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)|$ is fixed for a given lattice. For a monatomic Bravais crystal, each primitive cell contains one atom.

Interpret the sphere as an atom of iron, and assume neighbouring atoms touch without overlap. Its radius is $r=0.124\mathrm{nm}=1.24\times 10^{-10}\mathrm{m}$, and its mass is $m_{\mathrm{Fe}}=56u\simeq 9.30\times 10^{-26}\mathrm{kg}$, using atomic mass unit $u=1.66054\times 10^{-27}\mathrm{kg}$.

In the face-centered cubic lattice, atoms touch along a face diagonal: $\sqrt{2}{a}_F=4r$, so $a_F=2\sqrt{2}r=0.3507\mathrm{nm}$. Four atoms occupy a conventional cube. The hard-sphere atomic volume in cubic lattices therefore givesIn the body-centered cubic lattice, atoms touch along the body diagonal: $\sqrt{3}{a}_B=4r$, so $a_B=4r/\sqrt{3}=0.2864\mathrm{nm}$. There are two atoms per conventional cube, givingThese are primitive unit cell volumes, not the volumes of the conventional cubes or of the atomic spheres. The denser FCC contact packing gives the higher mass density in this fixed-radius model. Actual hot phases have temperature-dependent dimensions and bonding, so these numbers should not be treated as exact experimental phase densities.

The reciprocal lattice is the set of wavevectors $\mathbf{G}$ satisfying $e^{i\mathbf{G}\cdot \mathbf{R}}=1$ for every direct Bravais lattice translation $\mathbf{R}$. Equivalently, $\mathbf{G}\cdot \mathbf{R}$ is an integer multiple of $2\pi$. It describes the spatial periodicities that can produce coherent Bragg scattering.

For direct primitive vectors ${\mathbf{a}}_i$ and oriented cell volume $\Omega ={\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)$, the primitive reciprocal vectors areThey obey ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$, and every reciprocal vector is an integer combination of them. Their units are inverse length. The reciprocal primitive cell volume is $(2\pi {)}^3/|\Omega |$, so expanding a direct lattice contracts its reciprocal scale.

The direct body-centered cubic lattice has a face-centred cubic reciprocal lattice, whereas the direct face-centered cubic lattice has a body-centred cubic reciprocal lattice. Thus the reciprocal lattice change in a structural phase transition reverses the two cubic types: the reciprocal lattice changes from FCC to BCC when the real material changes from BCC to FCC. If the direct conventional side is $a$, the reciprocal conventional side is $4\pi /a$ in either case, with the appropriate phase-specific $a$.

Within one phase, ordinary thermal expansion increases direct spacings and decreases reciprocal spacings. At the structural phase transition, reciprocal points acquire the new symmetry and allowed positions. A first-order transformation may show both phases simultaneously and may create differently oriented domains; it need not continuously preserve one perfect reciprocal array throughout.

Using the fixed-radius estimates from part (b), ${\Omega }_F/{\Omega }_B\simeq 0.919$, so the reciprocal primitive cell volume would increase by about $8.9\%$. That comparison belongs to the ideal contact model. The question supplies neither the actual thermal expansion nor a crystallographic orientation relationship, so it does not determine a unique trajectory of individual reciprocal points during the transition.

Use temperature-dependent X-ray diffraction, with the crystal in a controlled heating stage and a detector measuring diffracted directions. A reflection occurs when the scattering wavevector matches a reciprocal lattice vector, or equivalently when Bragg's law $2d_{hkl}\sin \theta =\lambda$ is satisfied. Record peak positions and intensities while heating to infer changes in lattice spacings and structure.

A monatomic conventional BCC cell has nonzero crystal basis structure factor only when $h+k+l$ is even. For FCC the indices must be all even or all odd. The change in reflection selection and peak positions therefore distinguishes the phases. For instance, BCC starts with a $(110)$ family whereas FCC starts with $(111)$, when indexed in their own conventional cells.

The transformation can create multiple orientation variants, grains, strain and phase coexistence. A reciprocal point that formerly satisfied the diffraction geometry may disappear or move off a fixed detector; a loss of one spot alone does not identify the new phase. Thermal expansion shifts peaks, thermal motion weakens coherent intensities, and hot iron may oxidize without a protective atmosphere.

After transformation, scan crystal angle and scattering angle or use an area detector to locate the new reflections. If single-crystal order has been lost, powder X-ray diffraction or a sufficiently broad orientation survey collects rings or many grain reflections instead of relying on the original alignment. Index several new peaks using the FCC selection rule and fit their common lattice parameter; this probes the new structure even when the original crystal orientation is no longer usable.

The quantum free electron model treats conduction Electrons as independent particles moving in an approximately constant potential, with parabolic dispersion relation $E={\hslash }^2{k}^2/(2m)$. Fermi-Dirac statistics successfully explains important features such as a Fermi energy and a small electronic heat capacity, but a constant potential gives no band gaps and cannot by itself explain why different band fillings produce metals, semiconductors and band insulators. It also misses lattice-dependent effective masses, detailed directional responses and the topology of real Fermi surfaces. Scattering and finite electrical resistivity require additional physics.

The nearly-free electron model introduces a weak periodic lattice potential. Bloch's theorem replaces individual free plane waves by Bloch states, and Fourier components of the potential mix plane waves whose wavevectors differ by a reciprocal lattice vector. Near degeneracy at a Brillouin zone boundary, this coupling opens a band gap; for a two-wave crossing its size is $2|V_G|$. Away from the crossing, energies remain close to free-particle values. Band curvature produces band effective masses, and band filling determines whether the Fermi energy lies inside an allowed band or a gap.

The improvement is the inclusion of lattice periodicity, rather than a switch from classical to quantum statistics: the quantum free electron model already uses Fermi-Dirac statistics. The strong delta barriers in the later numerical example are treated exactly by the Kronig-Penney model, rather than by assuming that their strength is perturbatively small.

A Bloch state on the loop has $\psi (x)=u(x)e^{i\varphi x}$, with $u(x+a)=u(x)$. Its circumference is $L=Na$, so $u(x+Na)=u(x)$. Single-valuedness after one complete circuit requires the periodic boundary conditionThereforeThe allowed Bloch wavevectors are spaced by $2\pi /(Na)$. Values of $\varphi$ differing by $2\pi /a$ label the same crystal momentum modulo a reciprocal lattice vector, so there are $N$ inequivalent choices in one Brillouin zone. In this notation $\varphi$ is the Bloch wavenumber, not a dimensionless phase; $\varphi a$ is dimensionless. Spin gives additional states but does not alter this condition.

Between delta barriers, $V=0$ and the Schrödinger equation gives ${\psi }^{\prime \prime }+k^2\psi =0$, with $k^2=2mE/{\hslash }^2$ for positive energy. Write the state and derivative as a two-component vector $\mathbf{v}=(\psi ,{\psi }^{\prime }{)}^T$. Free propagation across one interval of length $a$ isThe wavefunction is continuous at a delta barrier, while its derivative jumps by $g\psi$, with $g=2m\alpha /{\hslash }^2$. Thus the barrier matrix isBloch's theorem requires $\mathbf{v}(a^{+})=e^{i\varphi a}\mathbf{v}(0^{+})$. Both $D$ and $P$ have determinant one, so the two eigenvalues of $M$ are $e^{i\varphi a}$ and $e^{-i\varphi a}$. Taking its trace givesTherefore the Floquet discriminant of the delta-comb Kronig-Penney model yieldsThis trace argument is the one-cell transfer method for the Kronig-Penney model. The loop imposes the additional discrete values of $\varphi$ derived in part (b). Real $\varphi$ requires the right-hand side to lie in $[-1,1]$, giving allowed energy bands; values outside this range are band gaps. The strength $\alpha$ has units of energy times length, because a Dirac delta distribution has units of inverse length.

With length measured in units of $a=1$, plot the real functionOnly the portions with $|f(k)|\le 1$ can equal $\cos \varphi$, so the strip between the horizontal lines $+1$ and $-1$ identifies allowed energy bands. The band edges of a repulsive delta-comb potential have the following approximate first intervals:Each interval corresponds to energy $E={\hslash }^2{k}^2/(2m)$; the intervening intervals are forbidden. The exact upper endpoints follow from $f(j\pi )=(-1{)}^j$. The lower endpoints solve $f(k)=(-1{)}^{j-1}$ inside $((j-1)\pi ,j\pi )$. The divergence suggested by $\sin k/k$ at the origin is removable, and $f(0)=11>1$, so there is no zero-energy state in the repulsive model.