Level 5 — MasteryMaterials & Atomic Structure

Materials & Atomic Structure

90 minutes60 marksprintable — key stays hidden on paper

Level 5 (Mastery): Cross-domain — Physics · Mathematics · Coding · Proof Time limit: 90 minutes Total marks: 60

Instructions: Answer all THREE questions. Show full reasoning. Use ...... for inline and ...... for displayed mathematics. Physical constants where needed: kB=1.381×1023J/Kk_B = 1.381\times10^{-23}\,\text{J/K}, q=1.602×1019Cq = 1.602\times10^{-19}\,\text{C}, 1eV=1.602×1019J1\,\text{eV}=1.602\times10^{-19}\,\text{J}.


Question 1 — Bonding, lattice geometry & atomic accounting (20 marks)

Silicon crystallizes in the diamond cubic structure with lattice constant a=5.431A˚a = 5.431\,\text{Å}.

(a) Starting from the Bohr model, explain why the silicon atom (Z=14Z=14) has exactly 4 valence electrons, and connect this directly to the fact that each Si atom forms 4 covalent bonds in the crystal. (4)

(b) The diamond cubic cell contains 8 atoms per conventional cubic cell. Derive the atomic number density NN (atoms/m³) as a function of aa, then evaluate it numerically. (4)

(c) In the diamond structure the nearest-neighbour distance is d=34ad = \dfrac{\sqrt{3}}{4}a. Derive this from the geometry (a corner atom bonding to the atom at (14,14,14)a(\tfrac14,\tfrac14,\tfrac14)a) and evaluate dd in Å. (4)

(d) Compute the atomic packing fraction (APF) of the diamond cubic lattice, treating atoms as hard spheres of radius r=d/2r = d/2 that touch along the bond. Show your working and comment on why this value (~0.34) is low compared with FCC (~0.74). (5)

(e) Explain, in terms of valence electrons and bonding, why a perfect intrinsic silicon crystal at 0K0\,\text{K} behaves as an insulator. (3)


Question 2 — Intrinsic carriers, thermal generation & a proof (22 marks)

The intrinsic carrier concentration of a semiconductor obeys

ni(T)=CT3/2exp ⁣(Eg2kBT)n_i(T) = C\,T^{3/2}\exp\!\left(-\frac{E_g}{2k_B T}\right)

where CC is a material constant and EgE_g the bandgap.

(a) Prove that the logarithmic temperature sensitivity satisfies

d(lnni)dT=32T+Eg2kBT2.\frac{d(\ln n_i)}{dT} = \frac{3}{2T} + \frac{E_g}{2k_B T^2}.

State clearly which term dominates near room temperature for silicon (Eg=1.12eVE_g=1.12\,\text{eV}, T=300KT=300\,\text{K}) by computing the ratio of the two terms. (6)

(b) For intrinsic silicon ni(300K)1.0×1010cm3n_i(300\,\text{K}) \approx 1.0\times10^{10}\,\text{cm}^{-3}. Using the model above (ignore the weak T3/2T^{3/2} prefactor for this part, i.e. treat niexp(Eg/2kBT)n_i \propto \exp(-E_g/2k_BT)), estimate nin_i at T=400KT = 400\,\text{K}. Give the result to one significant figure. (6)

(c) Germanium has Eg=0.66eVE_g = 0.66\,\text{eV}. Using the same approximation, compute the ratio niGe/niSin_i^{Ge}/n_i^{Si} at 300K300\,\text{K} assuming equal prefactors CC. Hence argue quantitatively why silicon is preferred over germanium for devices operating at elevated temperature. (5)

(d) Sketch (describe) qualitatively how the conductivity σ=qni(μn+μp)\sigma = q n_i(\mu_n+\mu_p) of intrinsic Si varies with TT, noting that mobility falls roughly as μT3/2\mu \propto T^{-3/2} while nin_i rises exponentially. Which effect wins, and what does this imply for the temperature coefficient of resistance of intrinsic silicon? (5)


Question 3 — Extrinsic material, mobility & a small simulation (18 marks)

(a) Distinguish intrinsic from extrinsic semiconductors, then define carrier mobility μ\mu via drift velocity vd=μEv_d = \mu E. State the units of μ\mu. (4)

(b) A silicon sample is doped with ND=1×1016cm3N_D = 1\times10^{16}\,\text{cm}^{-3} donors (fully ionized). Using the mass-action law np=ni2np = n_i^2 with ni=1.0×1010cm3n_i=1.0\times10^{10}\,\text{cm}^{-3}, compute the equilibrium electron (nn) and hole (pp) concentrations. (4)

(c) For this sample μn=1350cm2/V⋅s\mu_n = 1350\,\text{cm}^2/\text{V·s} and μp=480cm2/V⋅s\mu_p = 480\,\text{cm}^2/\text{V·s}. Compute the resistivity ρ=1/σ\rho = 1/\sigma where σ=q(nμn+pμp)\sigma = q(n\mu_n + p\mu_p). Comment on which carrier dominates conduction and why. (5)

(d) Coding task. Write a short Python function n_i(T, Eg=1.12, C=<your choice>) that returns intrinsic carrier concentration using the full model of Q2. Then, in pseudocode or Python, describe a loop that finds (by scanning TT) the temperature at which the thermally generated nin_i becomes comparable (within 10%) to the doping ND=1016cm3N_D=10^{16}\,\text{cm}^{-3} — the onset of loss of extrinsic behaviour. Explain the device significance of this temperature. (5)

Answer keyMark scheme & solutions

Question 1

(a) Bohr shells fill as 2,8,18,2, 8, 18,\dots (2n22n^2). For Z=14Z=14: shell 1 holds 2, shell 2 holds 8, leaving 1410=414-10=4 in shell 3 (the outermost/valence shell). Why: the 4 outermost electrons are the valence electrons; each is shared with a neighbour to complete an octet, giving 4 covalent bonds per Si atom. (2 for shell filling, 2 for the bonding link)

(b) Volume of conventional cell =a3=a^3, 8 atoms per cell: N=8a3,a=5.431×1010mN = \frac{8}{a^3}, \quad a=5.431\times10^{-10}\,\text{m} a3=1.602×1028m3,N=81.602×10284.99×1028m3a^3 = 1.602\times10^{-28}\,\text{m}^3,\quad N = \frac{8}{1.602\times10^{-28}} \approx 4.99\times10^{28}\,\text{m}^{-3} (5.0×1022cm3)(\approx 5.0\times10^{22}\,\text{cm}^{-3}). (2 for formula, 2 for value)

(c) Corner atom at (0,0,0)(0,0,0), tetrahedral partner at (14,14,14)a(\tfrac14,\tfrac14,\tfrac14)a: d=(a4)2×3=34ad = \sqrt{\left(\tfrac{a}{4}\right)^2\times 3} = \frac{\sqrt3}{4}a d=1.7324(5.431)=2.352A˚d = \frac{1.732}{4}(5.431) = 2.352\,\text{Å} (2 derivation, 2 value)

(d) Hard spheres touch along bond so r=d/2=38ar=d/2 = \frac{\sqrt3}{8}a. Volume of 8 spheres =843πr3= 8\cdot\frac{4}{3}\pi r^3. APF:

= \frac{32\pi}{3}\cdot\frac{3\sqrt3}{512} = \frac{\pi\sqrt3}{16}\approx 0.340$$ **Comment:** diamond cubic is an *open* structure — the directional tetrahedral covalent bonds require large empty interstitial space, unlike close-packed metallic FCC (0.74) where bonding is non-directional. **(3 derivation, 2 value+comment)** **(e)** At $0\,\text{K}$ every valence electron is locked in a covalent bond; there are no free electrons and no holes (no broken bonds). With zero mobile carriers, conductivity is zero → insulator. **(3)** --- ## Question 2 **(a)** $\ln n_i = \ln C + \tfrac32\ln T - \dfrac{E_g}{2k_B T}$. Differentiate: $$\frac{d(\ln n_i)}{dT} = \frac{3}{2T} + \frac{E_g}{2k_B T^2}\quad\checkmark$$ (since $\frac{d}{dT}(-\frac{E_g}{2k_B}T^{-1}) = +\frac{E_g}{2k_B}T^{-2}$.) **Ratio of exponential term to prefactor term:** $$\frac{E_g/(2k_BT^2)}{3/(2T)} = \frac{E_g}{3k_B T}$$ $= \dfrac{1.12\times1.602\times10^{-19}}{3\times1.381\times10^{-23}\times300} \approx 14.4$. The exponential (bandgap) term dominates by a factor ~14. **(3 proof, 3 dominance calc)** **(b)** $\dfrac{n_i(400)}{n_i(300)} = \exp\!\left[\dfrac{E_g}{2k_B}\left(\dfrac1{300}-\dfrac1{400}\right)\right]$. $\dfrac{E_g}{2k_B} = \dfrac{1.12\times1.602\times10^{-19}}{2\times1.381\times10^{-23}} = 6496\,\text{K}$. $\left(\tfrac1{300}-\tfrac1{400}\right)=8.333\times10^{-4}$. Exponent $=6496\times8.333\times10^{-4}=5.41 \Rightarrow e^{5.41}\approx 224$. $$n_i(400) \approx 224\times10^{10} \approx 2\times10^{12}\,\text{cm}^{-3}$$ **(3 exponent, 3 result)** **(c)** $\dfrac{n_i^{Ge}}{n_i^{Si}} = \exp\!\left[\dfrac{E_g^{Si}-E_g^{Ge}}{2k_BT}\right]$ (smaller $E_g$ → larger $n_i$). $\Delta E_g = 1.12-0.66 = 0.46\,\text{eV}$. $\dfrac{0.46\times1.602\times10^{-19}}{2\times1.381\times10^{-23}\times300} = \dfrac{7.37\times10^{-20}}{8.286\times10^{-21}} = 8.89$. Ratio $=e^{8.89}\approx 7.3\times10^{3}$. Ge has ~7000× more intrinsic carriers, so leakage/thermal generation is far higher and it loses extrinsic behaviour at much lower $T$ → Si preferred for high-temperature operation. **(3 calc, 2 argument)** **(d)** $\sigma\propto n_i(T)\,\mu(T) \propto T^{3/2}e^{-E_g/2k_BT}\cdot T^{-3/2} = e^{-E_g/2k_BT}$ (the $T^{3/2}$ cancels). The exponential rise of $n_i$ overwhelmingly beats the mild mobility decrease → conductivity *increases* with temperature. Hence intrinsic silicon has a **negative temperature coefficient of resistance** (resistance drops as T rises), opposite to metals. **(3 competition analysis, 2 conclusion)** --- ## Question 3 **(a)** *Intrinsic:* pure semiconductor, carriers only from thermal electron–hole pair generation, $n=p=n_i$. *Extrinsic:* deliberately doped; majority carriers set by dopant (donors→n-type, acceptors→p-type). Mobility: $v_d=\mu E$, so $\mu = v_d/E$; units $\text{cm}^2/(\text{V·s})$ (or m²/V·s). **(2 distinction, 2 mobility)** **(b)** Donor-dominated: $n\approx N_D = 1\times10^{16}\,\text{cm}^{-3}$. $p = n_i^2/n = (10^{10})^2/10^{16} = 10^{20}/10^{16} = 1\times10^{4}\,\text{cm}^{-3}$. **(2 for n, 2 for p)** **(c)** Electron term $=n\mu_n = 10^{16}\times1350 = 1.35\times10^{19}$; hole term $=p\mu_p=10^4\times480=4.8\times10^6$ (negligible). $\sigma = q(n\mu_n+p\mu_p) = 1.602\times10^{-19}\times1.35\times10^{19} = 2.163\,(\Omega\text{·cm})^{-1}$. $$\rho = 1/\sigma \approx 0.462\,\Omega\text{·cm}$$ Electrons (majority carriers) dominate — their concentration exceeds holes by $10^{12}$, so hole contribution is utterly negligible. **(3 σ, 1 ρ, 1 comment)** **(d)** ```python import numpy as np def n_i(T, Eg=1.12, C=3.9e16): # C tuned so n_i(300)=1e10 cm^-3 kB_eV = 8.617e-5 # eV/K return C * T**1.5 * np.exp(-Eg/(2*kB_eV*T)) ND = 1e16 for T in range(300, 1000): if abs(n_i(T) - ND)/ND < 0.10: print("onset T =", T); break ``` The scan finds the temperature where thermally generated $n_i \approx N_D$; above it, intrinsic carriers swamp the dopants and the device loses its designed n-type behaviour (junctions leak, transistors fail). This sets the **maximum operating temperature** — and because Si's larger $E_g$ pushes this onset higher than Ge's, Si tolerates hotter environments. **(2 code, 1 loop, 2 significance)** ```verify [ {"claim":"Silicon atomic density N=8/a^3 ~ 5.0e28 /m^3", "code":"a=5.431e-10; N=8/a**3; result = abs(N-4.99e28)/4.99e28 < 0.02"}, {"claim":"Nearest-neighbour distance d=sqrt(3)/4*a = 2.352 Angstrom", "code":"a=5.431; d=sqrt(3)/4*a; result = abs(float(d)-2.352) < 0.01"}, {"claim":"Diamond APF = pi*sqrt(3)/16 ~ 0.340", "code":"apf=pi*sqrt(3)/16; result = abs(float(apf)-0.340) < 0.005"}, {"claim":"n_i(400)/n_i(300) ~ 224 giving ~2e12 cm^-3", "code":"Eg=1.12*1.602e-19; kB=1.381e-23; r=exp(Eg/(2*kB)*(1/300-1/400)); result = abs(float(r)-224)/224 < 0.05"}, {"claim":"Ge/Si intrinsic ratio ~ 7.3e3 at 300K", "code":"dE=0.46*1.602e-19; kB=1.381e-23; r=exp(dE/(2*kB*300)); result = abs(float(r)-7.3e3)/7.3e3 < 0.10"}, {"claim":"n-type Si resistivity ~ 0.462 ohm-cm", "code":"q=1.602e-19; n=1e16; p=1e4; sig=q*(n*1350+p*480); rho=1/sig; result = abs(rho-0.462) < 0.01"} ] ```