Codex @codex  0

The simplest assignment is two phenylethynyl bridges and four equivalent terminal methyl groups, represented as $\mathrm{A}{\mathrm{l}}_2\mathrm{M}{\mathrm{e}}_4(\mu -\mathrm{C}\equiv \mathrm{CPh}{)}_2$. All 12 methyl protons then occupy the same class of environment, accounting for the single nonaromatic singlet without requiring methyl bridge/terminal averaging. The phenyl protons account for the aromatic-region signals; the metal-bound acetylide has no terminal acetylenic hydrogen.

An alkynyl bridge has electronic possibilities unavailable to a saturated methyl: in addition to sigma attachment from its carbon end to one Lewis acid centre, its occupied C≡C pi system can donate to the other aluminium. A delocalized sigma/pi description stabilizes the phenylethynyl-bridged arrangement while keeping the methyl groups terminal. One must not draw an unchanged localized triple bond plus two independent full ordinary Al–C bonds at the same carbon and count all of them as separate two-electron bonds. The exact distribution of sigma and pi donation requires structural or orbital evidence; this NMR observation alone does not supply a numerical bridge bond order.

The absence of concentration or temperature dependence is consistent with this preferred arrangement throughout the measured range. It is not proof that the molecule has no motion: exchange between symmetry-equivalent terminal positions, or exchange remaining fast throughout that range, could be spectroscopically invisible. Phenylethynyl bridging in this particular dimethylaluminium dimer has independent primary NMR precedent.^[ref]

Sodium borohydride is chemoselective for the ketone; lithium aluminium hydride reduces both the ketone and the lactam. In an amide, donation of the N lone pair into the carbonyl $\pi$ system reduces the carbonyl carbon's electrophilicity. It is therefore substantially less susceptible to the milder hydride donor.

Lithium aluminium hydride supplies more strongly reducing hydride equivalents, and the strong affinity of aluminium for oxygen assists removal of the amide oxygen through the iminium pathway. This difference explains both the unchanged lactam carbonyl in B and the absence of that oxygen in C. Both reactions are interpreted after the usual protonating workup.

B is 4-hydroxy-1-methylpyrrolidin-2-one; C is 1-methylpyrrolidin-3-ol. Their ring connectivities are shown below. Each ring remains five membered; the attached N-methyl group is retained.

For B, a B–H bond of sodium borohydride supplies a hydride equivalent to the ketone carbonyl carbon, and the carbonyl $\pi$ pair moves to oxygen. This gives an alkoxide, whose protonation during the reaction or workup gives the secondary alcohol. The lactam carbonyl remains. Converting one $\mathrm{C}=\mathrm{O}$ to $\mathrm{CH}(\mathrm{OH})$ adds two H atoms and gives ${\mathrm{C}}_5{\mathrm{H}}_9\mathrm{N}{\mathrm{O}}_2$.

For C, lithium aluminium hydride reduces that ketone in the same net way and also reduces the lactam. Coordination of the amide oxygen to aluminium assists hydride attack at its carbonyl carbon, giving a tetrahedral intermediate. The N lone pair then forms a C=N bond while the C–O bond breaks to an aluminium-bound oxygen group. A resulting iminium ion accepts a second hydride at carbon, yielding the cyclic tertiary amine. Aqueous workup protonates the ketone-derived alkoxide.

The lactam conversion replaces $\mathrm{N}-\mathrm{C}(=\mathrm{O})$ by $\mathrm{N}-\mathrm{C}{\mathrm{H}}_2$; its oxygen is removed rather than retained as another alcohol. Together with ketone reduction, the net change is $+4\mathrm{H}-\mathrm{O}$, giving ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{NO}$. No stereochemistry is specified: the new alcohol stereocentre is formed as a racemic mixture in an achiral environment.

A conventional face-centered cubic cell contains four atoms. With three conduction electrons per atom,Using the bare electron mass,The Drude transport relation is $\rho =m_e/(n{e}^2\tau )$, givingFor a degenerate electron gas, transport samples electrons near the Fermi surface, so use $v_F=\hslash k_F/m_e=2.025\times 10^6$ m/s:This is a transport relaxation time and associated transport mean free path inferred from the specified resistivity. It need not equal a single-particle quantum lifetime when scattering is strongly angle dependent. The free-electron valence count also neglects the detailed band structure of aluminium, as the stipulated model requires. A classical thermal speed is not the appropriate velocity scale for this strongly degenerate metal.

The Bravais lattice is a face-centered cubic lattice. Four inequivalent representatives of its lattice points in the conventional cubic unit cell, in fractional crystal coordinates, areBoundary copies in the drawing do not constitute additional independent lattice points. Attach the two-atom atomic basisto each lattice point. The resulting aluminium phosphide crystal structure is the zinc blende structure. For example, the four phosphorus positions reduced into the conventional unit cell are $(1/4,1/4,1/4)$, $(1/4,3/4,3/4)$, $(3/4,1/4,3/4)$, and $(3/4,3/4,1/4)$, reproducing the heights in the source drawing. There are four aluminium and four phosphorus chemical atoms per conventional unit cell, but only two chemical atoms per primitive unit cell.

In the diamond cubic structure of silicon, the longitudinal acoustic phonon branch and longitudinal optical phonon branch meet at $X=(2\pi /a,0,0)$. The two atomic basis sites are identical, and the associated crystal symmetry protects their boundary degeneracy of normal modes. In aluminium phosphide, the two sites contain different chemical atoms. The corresponding zinc blende structure lacks the site-exchange symmetry, and these two longitudinal branches generally split at $X$.

A simple way to see the splitting is a diatomic-chain dispersion relation with equal springs, representing successive $[100]$ atomic planes by alternating masses $M_1,M_2$ at spacing $d=a/4$. With effective spring constant $K$, its angular frequencies obeyAt $X$, $kd=\pi /2$, so the inter-sublattice coupling cancels and the two squared angular frequencies are $2K/M_1$ and $2K/M_2$. Equal masses give a contact; unequal masses giveIn aluminium phosphide, the heavier phosphorus plane moves in the lower-frequency edge mode, and the lighter aluminium plane in the higher-frequency mode in this model. Their similar atomic masses explain why the splitting is relatively small. Different electronic environments can also change the force constant matrix. The chain calculation illustrates the mechanism; a quantitative three-dimensional phonon spectrum requires its full dynamical matrix. It does not assert that all transverse acoustic and optical branches meet in silicon.

For FCC aluminium, consider $\{111\}⟨110⟩$ slip. Normalize the tensile direction and one candidate system:Here $\mathbf{n}\cdot \mathbf{s}=0$. Comparing all twelve systems gives the largest Schmid factor on $(\overline{1}11)[101]$:Schmid's law therefore givesDuring ideal single slip, the tensile direction expressed in crystal coordinates rotates towards $[101]$, along the great-circle path joining the original axis to that slip direction. It reaches a symmetry boundary where another slip system becomes equally favored. This single-slip tensile-axis rotation concerns the crystal relative to the externally held loading direction, rather than a rotation of the laboratory loading apparatus.

The printed numerical ratios are about $35$, not about $10$:The qualitative contrast is nevertheless meaningful. FCC aluminium has close-packed glide planes and relatively mobile dislocations. Magnesium also has easy close-packed basal glide, although it lacks enough equivalently easy slip systems for arbitrary room-temperature polycrystal deformation.

BCC iron has no close-packed plane. Its screw-dislocation cores and lattice-friction barrier give much stronger resistance to glide; thermally activated kink motion makes the critical stress particularly temperature- and rate-sensitive. This is the relevant Peierls-stress explanation. A high polycrystal constraint factor cannot explain the supplied contrast, because the data are already critical resolved shear stresses. Neither the precise factor35 nor a universal BCC/FCC ratio follows from crystal packing alone: purity, dislocation structure, temperature and testing conditions also matter.

The nickel-rich $\gamma$ matrix is a disordered substitutional solid solution on a face-centered cubic lattice. The ordered ${\gamma }^{\prime }$ phase has the L12 structure, approximately $\mathrm{N}{\mathrm{i}}_3\mathrm{Al}$: aluminium occupies cube corners and nickel occupies face centres. Its occupied-site geometry is often called ordered FCC, but its chemically labelled Bravais lattice is primitive cubic, with a four-atom basis. A face-centre translation exchanges chemical sublattices and is not a symmetry of the ordered crystal.

Each Al atom has twelve nearest-neighbour Ni atoms at Euclidean distance $a/\sqrt{2}$. Each Ni atom has twelve nearest neighbours too, but these are four Al and eight Ni. The chemical distinction matters for slip. In disordered $\gamma$, the shortest perfect Burgers vector isFor the ordered primitive cubic translation lattice, the shortest order-preserving perfect vector isThis does not imply that $⟨100⟩$ slip must dominate. A vector must lie in a favourable glide plane and have an accessible core structure and activation stress. Matrix dislocations commonly glide on $\{111\}⟨110⟩$ systems. A single $a⟨110⟩/2$ passage through ${\gamma }^{\prime }$ leaves an antiphase boundary; a second matching passage restores order, giving total vector $a⟨110⟩$. Such a superdislocation has a longer total vector than $a⟨100⟩$, yet its constituent dislocations can move on the favourable close-packed planes and interact with the matrix. Paired motion balances line and antiphase-boundary energies, so the shortest abstract lattice translation alone does not select the active slip system.^[ref]

For the aluminium alloy, write $K_{IC,\mathrm{Al}}$ and ${\sigma }_{y,\mathrm{Al}}$ for the appropriate data-book properties at the relevant material state and temperature. The yield-limited screening bounds from part(c), with $r\simeq 0.70\mathrm{m}$, areFor $K$ in $\mathrm{MPa}\sqrt{\mathrm{m}}$ and ${\sigma }_y$ in $\mathrm{MPa}$, the numerical substitutions are $t_{\max }(\mathrm{mm})=318.31[K/(Y{\sigma }_y){]}^2$ and $t_{\mathrm{yield}}(\mathrm{mm})=7000/{\sigma }_y$.

A high-strength alloy's large yield stress can require a small allowable leak-before-break thickness even when its minimum pressure thickness is modest. FCC structure generally favors low-temperature plasticity, but does not by itself establish the alloy's fracture toughness, weld performance, corrosion behavior or fatigue resistance.

TODO: numerical thickness cannot be completed without the 2018 NST Data Book values and alloy condition, which are unavailable. The accessible Engineering2011/revised2019 book is not confirmed as the supplied edition; generic aluminium property ranges are not substituted as if they were the exam's values.

X is an aluminium-doped $p^{+}$ back surface field region. Annealing alloys the rear aluminium with silicon; on cooling, silicon regrowth adjacent to the contact incorporates aluminium as an acceptor. This creates a more heavily p-type region between the p-type base and rear metal contact.

The resulting $p/p^{+}$ doping gradient creates a field that reduces the minority-electron population near the recombination-active rear metal interface. It therefore reduces rear-surface recombination and improves collection of photogenerated carriers. Heavy doping also helps form a low-resistance majority-carrier contact. This is a rear isotype doping transition, not the front n–p junction and not an electrically insulating blocking layer.

The periodic table arranges chemical elements in increasing atomic number, across successive periodic table periods. Its columns are periodic groups, numbered 1–18. The layout places recurring outer electron configurations above one another, making similarities in chemical bonds and chemical reactions easier to recognize. Ordering is by proton count, not by mass; the mass number of a particular isotope does not determine its box. See the official IUPAC periodic table and numbering conventions.

For the first eighteen neutral chemical atoms in their ground states, a compact description counts electrons in successive electron shells. Thus carbon has $(2,4)$, meaning two electrons in the first electron shell and four in the second; the sum $2+4=6$ agrees with its atomic number. Sodium has $(2,8,1)$, whose sum is $11$. These are occupancy counts, not circular trajectories of electrons.

Reading each row from left to right gives:

The counting outline is short: the occupied electron subshells in this range are $1s$, $2s$, $2p$, $3s$, and $3p$, filled in that order for these neutral ground states. Each $s$ electron subshell holds up to two electrons and each $p$ electron subshell up to six, giving $2$, $2+6=8$, and $2+6=8$ along these three rows. This $2,8,8$ summary applies only through argon; it is not a general rule that each electron shell fills completely before the next begins, nor a claim that the third electron shell can hold only eight electrons. See Flowers et al., Chemistry 2e, electronic structure and electron configurations.

The recurring outer counts make the arrangement useful. Lithium and sodium each have one valence electron, whereas fluorine and chlorine each have seven valence electrons. Such similarities help anticipate chemical reactions, without making all properties identical throughout a periodic group. Helium is an instructive exception to an eight-electron shortcut: its two electrons complete the first electron shell, so it belongs above neon and argon in periodic group 18. Hydrogen shares column 1 with lithium and sodium but has its own distinct chemistry. See Flowers et al., Chemistry 2e, periodic-table organization.

The Thomson charge-to-mass experiment infers the magnitude of an electron's charge-to-mass ratio by combining electric and magnetic deflection of cathode rays. The central idea is to determine speed separately from magnetic curvature: a known speed turns the observed gyroradius into a measurement of $e/m$.

In 1897, J. J. Thomson used two grounded slits in an evacuated glass tube, then aluminium plates $5\mathrm{cm}$ long, $2\mathrm{cm}$ wide and $1.5\mathrm{cm}$ apart. A battery supplied their electric potential difference. Deflected cathode rays produced a phosphorescence patch on the glass; an external scale and movable luminous pointer recorded its position in darkness. Greater evacuation reduced gas electrical conductivity that screened the electric field; reversing plate polarity reversed deflection. External coils supplied the magnetic field, calibrated by induction against larger coils. He adjusted equal electric and magnetic deflections. ^[ref]

His comparison included air, hydrogen, carbon dioxide, and aluminium versus platinum cathodes, supporting a common carrier. He explicitly noted uncorrected magnetic bending outside the plates. His tables give $m/e$, the reciprocal of the magnitude used here. The inferred ratio alone could not separate electric charge from mass. ^[ref]

To connect this readout with the Lorentz force, derive a simple nonrelativistic model in modern SI notation. Let the uniform deflecting region have length $\ell$, followed by a field-free drift of length $L$. Assume a narrow speed distribution, small deflection angles, negligible collisions, and matching electric and magnetic field profiles. For an electron with speed $v$, the electric acceleration has magnitude $eE/m$, the transit time is approximately $\ell /v$, and the subsequent drift adds exit transverse velocity times $L/v$. Thus the displacement magnitudes measured separately with only electric field or only magnetic field areThe electric expression follows from constant transverse acceleration; the magnetic expression uses its small-deflection approximation. Equal displacement magnitudes then giveThis is an ideal derivation of the comparison principle, not a numerical reanalysis of the historical tube. Different exterior magnetic field profiles invalidate cancellation of the common geometrical factor. The comparison determines $e/m$ through deflection and the known apparatus dimensions; it does not require a downstream semicircle.

For a second, deliberately different teaching geometry, combine a velocity selector with a separate region containing only a uniform magnetic field $B_a$ perpendicular to the selected velocity. Here $B_s$ belongs to the velocity selector, while $B_a$ belongs to the analyzer: they need not be equal. The magnetic Lorentz force supplies the centripetal acceleration, soIf the analyzer turns the beam through a semicircle before it hits a screen in the entry plane, the separation $D$ between entrance and impact is the diameter: $r=D/2$. For a shorter observed arc, the screen displacement is not generally a radius; fit the appropriate circle geometry. This two-stage velocity selector and curvature apparatus is an original ideal teaching model, not a claim that Thomson's Fig. 2 used a separate selector and semicircular analyzer.

For a worked example, stipulate $E=2.00\times 10^3\mathrm{V}{\mathrm{m}}^{-1}$, $B_s=1.00\times 10^{-3}\mathrm{T}$, $B_a=2.00\times 10^{-4}\mathrm{T}$, and $D=0.1136\mathrm{m}$. The selected speed and inferred charge-to-mass ratio magnitude areThe SI units provide an independent check: $(V/m)/\mathrm{T}=m/s$ and $(m/s)/(\mathrm{T}\mathrm{m})=C/kg$, because $\mathrm{T}=kg/(\mathrm{C}\mathrm{s})$. The charge-to-mass ratio of the electron is negative; this calculation gives its magnitude. The bending direction gives the sign of the electric charge once the magnetic field direction is known. Scaling both $e$ and $m$ by the same positive factor preserves every displayed trajectory, so an independent electric charge measurement is required to determine mass separately.

The curvature formula assumes nonrelativistic momentum $p=mv$. At higher speeds, the relativistic cyclotron orbit gives $p=\gamma mv=e{B}_{a}r$, hence $e/m=\gamma v/(B_{a}r)$, where $\gamma$ is the Lorentz factor; the ideal straight-through selection $v=E/B_s$ still follows from vanishing Lorentz force. The stipulated speed is about $0.0067$ times the speed of light, making $\gamma -1$ about $2.2\times 10^{-5}$. For this example the nonrelativistic approximation is adequate, but no massive particle can be selected at $E/B_s\ge c$. Finite slit widths, collisions with residual gas, uncertain magnetic field calibration, and fields outside the ideal boundaries limit a real inference; repeat at several selected speeds and analyzer strengths to test consistency of $e/m$ rather than trusting one spot position.

The aluminium methyl compound commonly forming $\mathrm{A}{\mathrm{l}}_2\mathrm{M}{\mathrm{e}}_6$ with four terminal methyl groups and two electron-deficient methyl bridges. Variable-temperature NMR distinguishes slow and fast site exchange.

An anionic organoaluminium compound with a carbon–aluminium bond at an alkene carbon.