Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 1 / Section B

Disperse a very small amount of the dry powder in a clean volatile solvent compatible with aluminium oxide, use brief gentle sonication to separate loose agglomerates, and place a dilute drop on a holey carbon-coated transmission electron microscopy grid. Wick away excess liquid and dry thoroughly. Look first at low magnification and select isolated particles or thin edges over holes, rather than an electron-opaque pile. Record a solvent-only grid to identify contamination and compare several independently prepared areas.

This is quick and preserves individual particles without mechanical thinning. However, spray pyrolysis can make hollow or porous particles that may fracture during sonication; sedimentation and drying can select sizes and aggregate particles. aluminium oxide is insulating, so charging or beam-induced changes can complicate images, and the carbon support can overlap weak contrast. Vary dose and dwell time to test stability. A projected image of one selected thin particle is not an unbiased three-dimensional size or porosity distribution of the whole powder.

This is a conventional transmitted-beam bright-field transmission electron microscopy image containing high-resolution electron phase contrast and visible lattice fringes, rather than an element-specific chemical map. Relative to the scale bar, the smaller particle has an outer projected diameter of roughly $40\mathrm{nm}$. It is approximately rounded and contains a denser outer rim with several lighter internal regions separated by material. A porous or internally hollow particle is a plausible morphological interpretation; the image is not a simple featureless solid sphere.

Greater projected thickness generally removes more intensity from the transmitted beam, producing darker mass–thickness contrast. Different crystallite orientations also scatter intensity differently and give diffraction contrast. Local lattice fringes reveal crystalline material and changes of fringe orientation indicate more than one coherent domain. Bright/dark edge fringes also contain defocus-dependent Fresnel contrast; their width is not a direct shell-thickness measurement. Neither an exact alumina polymorph nor three-dimensional closed porosity can be uniquely established from this projection alone. Use calibrated fringe spacings and electron diffraction for phase identification, tilt-series imaging for void geometry, and a through-focus or dose series to test contrast artefacts.

For a weak phase object, the contrast transfer function gives the signed transfer of an object's spatial Fourier components into intensity contrast. In a common convention,where $u$ is a cyclic spatial frequency, $\Delta f$ the defocus, $C_S$ the spherical aberration coefficient, and $E(u)$ a coherence envelope. A zero suppresses that frequency; crossing a zero reverses contrast. The overall sign depends on the image convention, but the zeros do not. Single-image division by $H$ cannot recover information exactly at its zeros. The microscope manufacturer's discussion explicitly connects the first zero with interpretable resolution and the higher-frequency sign changes.^[ref]

The plotted first nontrivial zero is at approximately $u_0=4.0\mathrm{n}{\mathrm{m}}^{-1}$. Hence the smallest spacing in the first uninterrupted contrast band is about $d_0=1/u_0=0.25\mathrm{nm}$. This axis is cycles per length, not an angular wave number requiring a factor of $2\pi$. Smaller spacings can contribute in later bands, but their contrast has reversed and requires appropriate interpretation or multiple-defocus reconstruction.

There is a verified inconsistency in the printed data: the original PDF says $0.025\mathrm{nm}$, whereas the relativistic de Broglie wavelength at $200\mathrm{kV}$ is approximately $0.00251\mathrm{nm}$:Use consistent length units, $C_S=1.2\times 10^6\mathrm{nm}$, and distinguish literal arithmetic from a physically consistent estimate. The familiar approximate Scherzer defocus is $\Delta f_S\simeq -1.2\sqrt{C_S\lambda }$, or about $-66\mathrm{nm}$ with $\lambda =0.0025\mathrm{nm}$. Slightly different optimum-phase conventions use $\sqrt{4/3}$ in place of $1.2$, giving about $-63\mathrm{nm}$; both are underfocus by a few tens of nanometres. The manufacturer's tabulated $200\mathrm{kV}$ values confirm that scale.^[ref]

Using the question's additional numerical relation exactly as supplied, with $u=1/(0.22\mathrm{nm})$, givesFor the physically appropriate rounded wavelength this is $\Delta f\simeq -67.0\mathrm{nm}$. The CTF itself gives an independent check: its first nonzero zero satisfies $\frac{1}{2}{C}_S{\lambda }^3{u}_0^4+\Delta f\lambda u_0^2=0$, so $u_0\simeq 4.0\mathrm{n}{\mathrm{m}}^{-1}$ implies $\Delta f\simeq -60\mathrm{nm}$. The plotted condition is therefore a little less underfocused than the usual Scherzer setting and has a first-zero spacing near $0.25\mathrm{nm}$. The nominal $0.22\mathrm{nm}$ and approximate optimization formula should not be treated as exactly reproducing every plotted parameter.

If the printed $0.025\mathrm{nm}$ is retained literally, the approximate Scherzer formula instead gives $-208\mathrm{nm}$, the supplied $0.54$ relation gives $-7.75\mu \mathrm{m}$, and the plotted first zero requires $-6.00\mu \mathrm{m}$. These do not describe a consistent $200\mathrm{kV}$ microscope. The physically consistent comparison is approximately $-63$ to $-67\mathrm{nm}$ optimum underfocus versus $-60\mathrm{nm}$ from the plot; the wavelength correction is explicit.

From the illuminated side downward, the p-type gallium arsenide cap provides a low-resistance route to the front metal contact. It should remain beneath the contact grid but be removed from the optical aperture to reduce parasitic absorption. The wider-band gap p-type aluminium gallium arsenide window transmits much of the useful light and reduces front-surface recombination of carriers in the adjacent absorber. The optical-window and reduced-recombination rationale has a primary experimental basis in early heteroface devices.^[ref]

The p-type gallium arsenide emitter and n-type gallium arsenide base form the p-n junction. Absorbed photons generate electrons and electron holes; diffusion and the junction's electric field direct them toward the appropriate contacts. The n-type aluminium gallium arsenide back-surface-field layer provides a barrier to minority holes moving toward the substrate and reduces rear recombination. The n-type gallium arsenide substrate supports the epitaxial layers and conducts majority carriers to the rear contact.

The front metal must collect current with limited shading and the back metal provides the return contact. Interpreting the patterned hatched region as the optical antireflection coating, it reduces reflected light over the exposed window. Carrier selectivity, low recombination, low series resistance and low shading must work together; the equilibrium junction field alone is not a source of electrical energy.

One workable route uses two resist coatings and a metal-defined optical aperture. The following are eleven fabrication operations; exposure/development is grouped as one lithographic operation, and rear metallization includes its contact activation if required. Detailed cleaning, rinsing and inspection occur between compatible operations.

1. Spin-coat a first photoresist on the front surface to define a mesa mask.
2. Expose and develop that mask, retaining resist over the complete active-cell mesa, including its future front contacts.
3. Etch the exposed semiconductor through the p-type layers and slightly into the n-type base. This removes the unwanted junction around the cell and produces the lateral mesa step.
4. Strip the first photoresist, leaving a clean mesa with its cap intact.
5. Spin-coat a second photoresist suitable for the lift-off process.
6. Expose and develop openings at the intended front grid or contact pads, positioned safely inside the mesa edge.
7. Evaporate a p-contact metal stack onto the exposed gallium arsenide and resist. Au/Zn-based contact metallization is a conventional candidate, subject to doping and contact-resistance qualification.
8. Dissolve the second photoresist to lift off unwanted metal, leaving the front contacts on the cap.
9. Selectively remove the cap from the light-receiving aperture, using the front metal as a hard mask and the aluminium gallium arsenide window as the etch stop.
10. Deposit the rear n-contact metallization, for example a qualified AuGe/Ni/Au system, with an appropriate contact-activation anneal if required and compatible with the front contact.
11. Deposit an antireflection dielectric over the optical aperture through a mechanical shadow mask that keeps the contact pads accessible. Its refractive index and thickness are chosen for the target optical band.

The essential material changes are junction isolation at the mesa edge, metal retained on the front cap, cap removed only from the optical aperture, and a conducting rear contact. The metal systems and anneal are a process design, not compositions specified by the drawing. A workable design must also qualify the passivation of etched sidewalls and the front-contact surface preparation.

Use optical photolithography for the two masks and a resist profile compatible with the selected process. A controlled wet mesa etch or a calibrated reactive-ion etching process must pass completely through the p–n interface, with limited sidewall damage. Use profilometry or cross-section controls to verify depth and a test device to check leakage. For the front lift-off process, directional evaporation and an undercut resist profile discourage a continuous metal fence bridging the mask sidewall.

The cap-removal step requires a separately qualified selective etchant: its selectivity between gallium arsenide and the particular aluminium gallium arsenide alloy depends on alloy composition, chemistry and processing conditions. Establish etch rate and endpoint on a witness stack; a universal etch time cannot be inferred from a diagram without thicknesses. The front metal must resist that etchant. Back metallization and activation must establish an ohmic contact rather than an unintended rectifying junction. Verify specific contact resistance on test structures and avoid an anneal that degrades the p–n profile. The final dielectric can be deposited by a low-damage evaporation or other compatible thin-film method.

Mesa definition comes first so that the two front contacts can be aligned to an already isolated active area and can avoid damaged mesa edges. Keeping the cap until after metallization provides the intended contact surface and protects the window during the contact process. lift-off process must precede selective cap removal if the retained metal is to act as that etch mask. The cap etch must stop above the emitter rather than destroy the junction.

Rear metallization after the wet-processing stages avoids unnecessarily exposing the rear contact to their etchants. Contact activation precedes the optical coating where possible, because the anneal can alter dielectric properties and optical thickness. The antireflection coating goes last and leaves the pads uncoated for electrical access. The key ordering constraints are clean semiconductor before contacts, electrical isolation without severing the intended current path, selective aperture opening, and a thermal budget compatible with every retained layer. If a different metal or etchant requires a changed order, it must preserve these same functions.

For identical spherical drops, total oil volume and interfacial area satisfyTaking $10\mathrm{vol}\%$ to mean the oil fraction of the final emulsion, $V_o/(V_o+V_w)=0.10$ gives $V_o=V_w/9=1.111\times 10^{-4}{\mathrm{m}}^3$. Thus $A=333.3{\mathrm{m}}^2$ and $E_s=16.7\mathrm{J}$. This corresponds to about $2.65\times 10^{13}$ drops.

There is a wording convention worth making explicit: if the oil addition instead means $0.10$ times the initial water volume, then $V_o=100\mathrm{mL}$, $A=300{\mathrm{m}}^2$ and $E_s=15.0\mathrm{J}$. These differ only in the stated volume-fraction convention. The calculation uses the oil–water interface and neglects the container and air interfaces.

An amphiphile adsorbs at the oil–water interface, reducing the interfacial tension. At fixed drop radius and oil volume, $E_s(c)=A{\gamma }_{ow}(c)$ therefore falls with increasing concentration and tends toward a plateau once the interface is nearly saturated and further amphiphile mostly enters micelles. For an approximately ideal dilute nonionic surfactant, the Gibbs adsorption equation $d\gamma =-RT\Gamma d\ln a$ connects the fall in interfacial tension to positive surface excess $\Gamma$. Ionic systems require the appropriate counterion activities.

The critical micelle concentration gives the familiar break in a tension-versus-log-concentration curve, rather than a fall to zero interfacial energy. If mixing creates a different droplet size distribution as amphiphile is added, area also changes; a tension reduction alone does not determine the total interfacial energy. The plotted prediction holds area fixed.

A Pickering emulsion is stabilized by particles adsorbed at the liquid–liquid interface. A partially wetting sphere removes a patch of that interface. The smaller detachment barrier into either bulk phase is approximatelywhere $\theta$ is the three-phase contact angle. At approximately neutral wetting, $\theta =90^{\circ }$, the stated particle radius givesat $25^{\circ }\mathrm{C}$. Such a large attachment energy makes thermal detachment unlikely. A sufficiently covered interface resists drop coalescence, and a water-wet particle preference supports an oil-in-water dispersion. However, radius alone does not guarantee stabilization: completely water-wet or oil-wet particles have a vanishing attachment barrier in this model, and sufficient particle coverage is necessary. Sedimentation and droplet ripening are separate failure modes. The classical particle-stabilized-emulsion experiments provide the historical basis of the terminology.^[ref]

Balance upward surface-tension force $2\pi r\gamma \cos \theta$ against the weight $\rho g\pi r^{2}h$ of the liquid column. The capillary action isWith perfect wetting, $\cos \theta =1$, $D=10^{-3}\mathrm{m}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $h=2.94\times 10^{-2}\mathrm{m}=29.4\mathrm{mm}$. The diameter, not the radius, is $1\mathrm{mm}$. This neglects the small meniscus-volume correction, gas-density buoyancy and contact-angle hysteresis.

For two equal spheres, the Derjaguin approximation uses $R_{\mathrm{eff}}=R/2$. A planar nonretarded Van der Waals force interaction has energy per area $W(h)=-A_H/(12\pi h^2)$, soThe force is attractive, with magnitude $13.0\mu \mathrm{N}$, taking separation increase as positive. Using the sphere–plane value of $R_{\mathrm{eff}}$ would overestimate it by a factor of two. The gap is much smaller than either radius, as required. Treat the supplied Hamaker constant as the effective one for the intervening medium; roughness, solvation and molecular-scale contact can modify a literal $1\mathrm{nm}$ experiment.

The change from a positive surface potential to a negative potential outside the compact layer is consistent with specific phosphate adsorption that overcompensates the positive oxide charge. The negative compact-layer charge then has a diffuse excess of positive counterions. This charge reversal is an interfacial-chemistry interpretation, not something a single monotone double-layer profile with an unchanged positive boundary charge could generate by screening alone. Surface potential and zeta potential at a slip plane are distinct observables.

The Stern model separates the specifically adsorbed compact layer from a diffuse layer. In a spherical diffuse layer, a Poisson-Boltzmann equation description isAt sufficiently small $|z_{i}e\psi |/k_{B}T$, the Debye–Hückel approximation gives ${\nabla }^2\psi ={\kappa }^2\psi$. If $a_s$ is the outer compact-layer radius, its decaying solution isFor a locally flat surface, the geometric factor tends to unity. Thus the far-field negative potential approaches zero from below over the Debye–Hückel screening length. Neither $+120$ nor $-70\mathrm{mV}$ is safely in the low-potential linear regime at room temperature, so the simple exponential is a far-field approximation, not a globally valid fit. The phosphate layer requires adsorption chemistry or an equivalent specified boundary model.

For an ideal dilute electrolyte with molar ionic strength $I=\frac{1}{2}{\sum }_i{c}_i{z}_i^2$, measured in $\mathrm{mol}{\mathrm{L}}^{-1}$, the Debye–Hückel screening length isTaking water's relative permittivity as approximately $78.5$ at $25^{\circ }\mathrm{C}$, $I=0.10\mathrm{M}$ for NaCl. Therefore ${\kappa }^{-1}\simeq 0.96\mathrm{nm}$. This is an exponential screening length, not a hard outer edge of the diffuse layer. The factor of $1000$ converts molarity into $\mathrm{mol}{\mathrm{m}}^{-3}$.

A $0.10\mathrm{M}$ formula-unit concentration of fully dissociated $\mathrm{AlC}{\mathrm{l}}_3$ gives $0.10\mathrm{M}$ trivalent cations and $0.30\mathrm{M}$ chloride:At unchanged temperature and permittivity, the Debye–Hückel screening length scales as $I^{-1/2}$, giving ${\kappa }^{-1}=0.96/\sqrt{6}\simeq 0.39\mathrm{nm}$. It falls by a factor $\sqrt{6}$, not by three. This is a formal screening estimate: at this ionic strength, trivalent-ion activity, finite ion size, hydrolysis and specific adsorption make the ideal dilute model quantitatively limited.

The oxygen-pressure exponents can be explained by a change in the dominant charge state of metal vacancies, with holes carrying the p-type current. For a divalent metal oxide, a doubly negative metal vacancy is $V_{\mathrm{M}}^{\prime \prime }$ in Kröger–Vink notation. A conventional oxygen-incorporation reaction isTreating regular oxygen-site activity as constant, the law of mass action gives $K_2=[V_{\mathrm{M}}^{\prime \prime }]p_h^2/p_{{\mathrm{O}}_2}^{1/2}$. If these are the compensating defects, charge neutrality gives $p_h=2[V_{\mathrm{M}}^{\prime \prime }]$, hence $p_h^3\propto p_{{\mathrm{O}}_2}^{1/2}$ and $\sigma \propto p_{{\mathrm{O}}_2}^{1/6}$, assuming nearly constant hole mobility.

For singly negative vacancies, useNow $p_h=[V_{\mathrm{M}}^{\prime }]$ and $p_h^2\propto p_{{\mathrm{O}}_2}^{1/2}$, so $\sigma \propto p_{{\mathrm{O}}_2}^{1/4}$. The charge-state equilibrium $V_{\mathrm{M}}^{\prime }\rightleftharpoons V_{\mathrm{M}}^{\prime \prime }+h^{\bullet }$ gives $[V_{\mathrm{M}}^{\prime \prime }]/[V_{\mathrm{M}}^{\prime }]=K_{\mathrm{ion}}/p_h$: the singly charged fraction can therefore dominate as oxygen pressure and hole concentration increase. This is a consistent dilute-defect model, not a unique experimental identification of every defect. Changes in mobility, defect association or other compensating defects require independent tests.

In a galvanic cell, a spontaneous oxidation-reduction reaction supplies electrical work: ${\Delta }_{r}G<0$ for discharge, the anode is negative, and the cathode positive. A discharging electric battery and a hydrogen–oxygen fuel cell are two practical examples. The first stores its reactants internally; the second is supplied continuously with fuel and oxidant.

An electrolytic cell consumes externally supplied electrical work to drive electrode reactions or material transfer under an applied potential. Its anode is positive and cathode negative. water electrolysis for hydrogen production and metal electroplating are two practical examples. oxidation always occurs at the anode and reduction at the cathode in both cell types; the electrode signs change. The electrolyte conducts ions, while electrons flow through the external circuit. A rechargeable battery can operate in either regime depending on whether it is discharging or being charged.

The reversible voltage follows from the two-electron hydrogen reaction:At the specified geometric current density, the magnitude of each activation loss follows its Tafel equation, $\eta =b{\log }_{10}(j/j_0)$:Both losses reduce the delivered fuel cell voltage. Therefore $V=0.36048\mathrm{V}$ and $P/A=jV=3.60\mathrm{W}{\mathrm{m}}^{-2}$. The very small oxygen exchange current is responsible for most of the activation loss. The calculation uses current densities on a common area basis, unit water activity and the stated gas conditions, and neglects the losses the problem excludes.

At unchanged intrinsic catalyst kinetics, increasing the electrochemically accessible true area by $100$ multiplies exchange current per geometric area by $100$. Equivalently, the local current density at a fixed total geometric current falls by $100$. Each Tafel equation activation loss decreases by $b{\log }_{10}100=2b$:Thus $V_{\mathrm{new}}=0.59948\mathrm{V}$ and $P_{\mathrm{new}}/A=5.99\mathrm{W}{\mathrm{m}}^{-2}$, an increase of $2.39\mathrm{W}{\mathrm{m}}^{-2}$, or about $66.3\%$. Power does not increase a hundredfold, because the voltage improvement is logarithmic and the specified geometric current is unchanged. This assumes both electrodes gain accessible active area, the original exchange currents use the original area basis, and the porous structure does not introduce additional transport or Ohmic losses.

A thermoelectric generator uses the Seebeck effect and needs a spatial temperature gradient: a couple develops $V\simeq (S_p-S_n)\Delta T$ between hot and cold junctions. It can deliver continuous power under a steady thermal gradient and has no moving parts, but requires sustained heat flow and adequate hot- and cold-side thermal contacts. Thermal conduction, internal electrical resistance and the material's thermoelectric figure of merit limit conversion.

A pyroelectric generator responds to a time-dependent spontaneous electric polarization. Under approximately uniform heating or cooling, $I=pAdT/dt$ when the electrical boundary condition allows charge flow. It can harvest temporal thermal fluctuations without a persistent spatial gradient. However, at a constant uniform temperature its ordinary pyroelectric current stops; practical extraction needs cycling, low-leakage electronics and a suitable load. A large open-circuit voltage is not evidence of a large available average power. Temperature-driven nanowire measurements provide a primary experimental example of this distinction.^[ref]

For electrode area $A$, the change in spontaneous polarization density supplies a charge scale $\Delta Q=pA\Delta T$, while the film's capacitance is $C=\epsilon A/d$. The voltage magnitude is therefore $|\Delta V|=|p|d|\Delta T|/\epsilon$. More explicitly, for an initially compensated film with no change in free electrode charge, $\Delta D=\epsilon \Delta E+p\Delta T=0$. Taking $z$ from the lower electrode to the upper one gives $\Delta V_{\mathrm{upper}-\mathrm{lower}}=-d\Delta E=pd\Delta T/\epsilon$.The measurable magnitude is $0.339\mathrm{V}$, with polarity reversed by reversing the poling direction or the temperature change. This is an ideal open-circuit estimate: leakage, the measuring instrument's input capacitance and finite charge-relaxation time can reduce or reshape the observed voltage.

The secondary pyroelectric effect is the change of polarization density produced when thermal strain acts through piezoelectricity. It is additional to the primary change at fixed strain. A linear constitutive description along the polar axis iswhere $e_{3j}$ are piezoelectric stress coefficients and ${\alpha }_j$ the freely developing thermal strains per kelvin. Thermal-expansion mismatch with a support or a deliberately chosen compliant structure can change the secondary contribution. It enhances the total response only if its sign reinforces the primary response; mechanical boundary conditions and thermal cycling must be characterized rather than assuming every restraint is beneficial.

Preventing axial expansion removes the free axial-strain contribution, but does not automatically eliminate every secondary contribution. In a one-dimensional idealization with only ${\epsilon }_3$ coupling and perfect axial clamping, that strain-mediated term vanishes. A real rod may still expand transversely, and nonzero $e_{31}$ or $e_{32}$ can change $P_3$ through those strains. In a stress-based constitutive equation, the same restraint appears as thermal stress and a piezoelectric term; do not add it a second time to a coefficient already specified for the constrained condition. Full suppression of all relevant thermal strains leaves the primary coefficient at fixed strain. The answer therefore depends on which degrees of freedom are constrained, not simply on the word “nanorod”.

A wireless sensor node combines a sensor, signal conditioning and conversion, a processor, a radio and a power subsystem. It can use a battery, a wired supply where practical, or local harvesting such as light, vibration or a thermal gradient, usually with a storage capacitor, supercapacitor or rechargeable battery to bridge interruptions. Harvested power must meet the time-averaged demand and storage must supply peak radio current.

Reduce power by duty cycling the sensor, processor and radio; use a low-leakage sleep mode and switch off unused peripherals. Buffer and batch data, transmit fewer unnecessary bits, use event-triggered acquisition when it meets the application, and reduce retries with a reliable link and an appropriate transmit-power level. Include wake-up, sensing, processing and communication energy in the budget: reducing payload alone may not reduce a radio dominated by startup overhead.

The duty cycle $D=t_{\mathrm{active}}/(t_{\mathrm{active}}+t_{\mathrm{sleep}})$ is the fraction of time spent active. At a constant supply voltage and with switching energy neglected,For $D=0.01$, the average current is $0.01(25\mathrm{mA})+0.99(6.1\mu \mathrm{A})=256.039\mu \mathrm{A}$. Thus $\bar{P}=512.078\mu \mathrm{W}\simeq 0.512\mathrm{mW}$. Multiplying only the active power by the duty cycle would omit the sleep contribution.

For $D=0.001$, the same duty cycle model gives $\bar{I}=25\mu \mathrm{A}+0.999(6.1\mu \mathrm{A})=31.0939\mu \mathrm{A}$. Thus $\bar{P}=62.1878\mu \mathrm{W}$. The tenfold reduction in active fraction reduces average power by only about $8.23$ because the sleep current remains. The ideal limit as $D\to 0$ is $V{I}_s=12.2\mu \mathrm{W}$, before leakage elsewhere or wake-up energy. Active and sleep contributions become comparable at a duty cycle of roughly $I_s/(I_a+I_s)=0.0244\%$; below that scale lowering sleep current is particularly valuable.