21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, December 2016 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Seven questions of 20 marks each (Roman numerals I–VII); the rubric asks for any five, with only the first five in the answer book marked. All seven are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and an error-function table are provided in the exam's own appendix (reproduced where used below).
Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only Question VI (dislocation theory) is genuinely deformation/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — electron structure, bonding, crystal structure, crystallographic directions/planes, solid solubility, XRD, microscopy and phase diagrams — and is answered as such below.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
a. Aufbau principle. ("building-up") States that, in the ground state, electrons fill available atomic orbitals starting from the lowest available energy level and proceeding to successively higher levels, subject to the sub-level filling order set by $(n+l)$ (and, for ties, the lower $n$ first) — e.g. $4s$ (n+l=4) fills before $3d$ (n+l=5) even though $n=3\lt4$. Example: potassium ($Z=19$) fills $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$, placing the 19th electron in $4s$ rather than the energetically higher $3d$.
b. Pauli's exclusion principle. No two electrons in the same atom may share all four quantum numbers ($n,l,m_l,m_s$); equivalently, each orbital (fixed $n,l,m_l$) can hold at most two electrons, and only if their spins are opposed ($m_s=+\tfrac12,-\tfrac12$). Example: the $1s$ orbital of helium holds exactly two electrons, $1s^2$, with opposite spin — a third electron cannot enter $1s$ and must begin the $2s$ level instead (as in Li, $1s^22s^1$).
c. Electronegativity. A dimensionless number (Pauling scale, roughly 0.7–4.0) describing an atom's relative ability to attract shared bonding electrons toward itself. It increases left-to-right across a period and decreases top-to-bottom down a group. Example: F ($\chi=4.0$) is the most electronegative element; in HF the bonding electrons sit much closer to F than to H, giving the molecule a polar-covalent bond.
Given. Hydrogen atom ($Z=1$), transition $n_i=4\to n_f=3$; appendix constant $R_E=13.61$ eV.
Find. $\Delta E$, and whether the energy is absorbed or emitted.
Approach. The appendix's one-electron energy levels are $E_n=-R_E/n^2$; the transition energy is $\Delta E=E_f-E_i$.
The negative sign means the electron loses energy dropping from the $n=4$ to the (lower, more tightly bound) $n=3$ state, so the $0.662$ eV is emitted as a photon (an $n=4\to3$ line in the Paschen series of the hydrogen spectrum).
Given. $f=1$ Hz; $h=6.63\times10^{-34}$ J·s; $c=3\times10^{8}$ m/s.
Find. Energy of a single emitted quantum, $E$; its wavelength, $\lambda$.
Approach. A single photon's energy is $E=hf$; its wavelength follows from $c=f\lambda$.
Check. A real heated tungsten filament radiates thermally with a spectrum peaking in the infrared/visible ($f\sim10^{14}$ Hz), never at $f=1$ Hz — a 1 Hz "photon" would have a wavelength of $3\times10^{8}$ m (about the Earth–Moon distance), far outside any electromagnetic emission a filament could produce. The question's own numbers are used exactly as given (mechanically exercising $E=hf$ and $c=f\lambda$); the physically unrealistic frequency is a feature of the question, not an error introduced here.
de Broglie's hypothesis. Every moving particle of momentum $p=mv$ has an associated wavelength $\lambda=h/p$ (the same relation photons obey, $p=h/\lambda$, extended to matter). Particles — electrons in particular — therefore exhibit wave-like behaviour (diffraction, interference) alongside their particle behaviour, the basis of electron microscopy and diffraction.
Given. Electron speed $v=c/6$; position uncertainty $\Delta x=1\%$ of the electron's own de Broglie wavelength (the natural length scale the question supplies, since it introduces de Broglie's hypothesis in the same part — stated here as the governing assumption, per the check note below); $m_e=9.11\times10^{-31}$ kg.
Find. Uncertainty in speed, $\Delta v$.
Approach. Compute the de Broglie wavelength $\lambda=h/(m_ev)$, take $\Delta x=0.01\lambda$, apply the Heisenberg relation $\Delta x\,\Delta p\ge h/(4\pi)$ from the appendix, then $\Delta v=\Delta p/m_e$.
Check — assumption and a surprising result. The question gives "the uncertainty in knowing its position is one percent" without stating one percent of what; taking it as 1% of the electron's own de Broglie wavelength is the only length scale the problem itself supplies (and pairs naturally with the de Broglie hypothesis asked for in the same part). Note the result: $\Delta v\approx3.98\times10^{8}$ m/s exceeds the speed of light. This is not an arithmetic error — it is the literal, non-relativistic Heisenberg-uncertainty answer for a position uncertainty this small ($1.46\times10^{-13}$ m, roughly a tenth of a picometre); it simply illustrates that the non-relativistic uncertainty formula breaks down as a physically meaningful velocity bound at this length scale, which is itself worth stating explicitly rather than silently rounding the answer down to $c$.
