Materials & Atomic Structure

Question 1 (12 marks)

(a) Radius derivation (5)

Coulomb = centripetal: $\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2} = \frac{m_e v^2}{r} \quad(1)$ Quantisation: $m_e v r = n\hbar \Rightarrow v = \frac{n\hbar}{m_e r}$ (1)

Substitute $v$ into (1): $\frac{Ze^2}{4\pi\varepsilon_0 r} = m_e \frac{n^2\hbar^2}{m_e^2 r^2}$ (1)

Solve for $r$: $\boxed{r_n = \frac{4\pi\varepsilon_0 n^2\hbar^2}{m_e Z e^2}}$ (2)

(b) Energy derivation (5)

KE $= \tfrac12 m_e v^2 = \frac{1}{8\pi\varepsilon_0}\frac{Ze^2}{r}$ (from (1)) (1) PE $= -\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r}$ (1) Total $E = KE + PE = -\frac{1}{8\pi\varepsilon_0}\frac{Ze^2}{r}$ (1)

Insert $r_n$: $\boxed{E_n = -\frac{m_e Z^2 e^4}{8\varepsilon_0^2 h^2 n^2} = -\frac{m_e Z^2 e^4}{32\pi^2\varepsilon_0^2\hbar^2 n^2}}$ (2) (Standard result gives $-13.6\,Z^2/n^2$ eV.)

(c) Numerical (2)

$r_1 = \frac{4\pi(8.854\times10^{-12})(1.055\times10^{-34})^2}{(9.11\times10^{-31})(1.602\times10^{-19})^2}\approx 5.29\times10^{-11}\ \text{m}$ (1) = Bohr radius $a_0 \approx 0.529\ \text{Å}$ (1)

Question 2 (10 marks)

(a) Filling: $1s^2\,2s^2\,2p^6\,3s^2\,3p^2$ (or shells 2,8,4). (2) The outermost shell ($n=3$) contains $3s^2 3p^2$ = 4 electrons → 4 valence electrons. (1)

(b) Each Si atom shares one electron with each of its 4 neighbours, forming 4 covalent bonds. Each shared pair counts toward both atoms' outer shells, giving each atom an effective 8 outer electrons (stable octet). Sketch: central Si with 4 lines to 4 neighbours, each line = shared electron pair. (4) (2 for mechanism, 2 for octet completion / sketch)

(c) Diamond cubic lattice; coordination number = 4 (tetrahedral). (3) (2 lattice, 1 coordination)

Question 3 (12 marks)

(a) Creating one free electron simultaneously creates one hole; the energy $E_g$ is shared/the equilibrium involves both carrier types, and the Fermi level sits mid-gap, so the activation energy per carrier is $E_g/2$. (2)

(b) Exponent term: $\exp[-E_g/(2k_B)\,(1/350 - 1/300)]$. $-E_g/(2k_B) = -1.12/(2\times8.617\times10^{-5}) = -6498.8\ \text{K}$ $1/350 - 1/300 = 0.0028571 - 0.0033333 = -4.7619\times10^{-4}$ Exponent $= (-6498.8)(-4.7619\times10^{-4}) = 3.0947$ $\exp(3.0947) = 22.08$ (4) Prefactor: $(350/300)^{3/2} = (1.1667)^{1.5} = 1.260$ (1) Ratio $= 22.08 \times 1.260 \approx 27.8$ (1)

(c) In semiconductors, raising $T$ generates exponentially more electron–hole pairs (more carriers), so $\sigma$ rises despite reduced mobility. In metals the carrier count is already fixed/huge; heating only increases lattice vibrations (phonon scattering), lowering mobility and hence $\sigma$. (4)

Question 4 (10 marks)

(a) Intrinsic: pure semiconductor where carriers come only from thermal EHP generation ($n=p=n_i$). Extrinsic: deliberately doped with impurities so one carrier type dominates. (2)

(b) $J = q n v_{dn} + q p v_{dp}$ (charge × density × drift velocity, summed over both carriers). (1) Substitute $v_d = \mu E$: $J = q(n\mu_n + p\mu_p)E$. (2) Compare with $J = \sigma E$: $\sigma = q(n\mu_n + p\mu_p)$. (1) Symbols: $q$ elementary charge, $n,p$ electron/hole densities, $\mu$ mobilities, $E$ field. (1)

(c) $n \approx N_d = 10^{16}\ \text{cm}^{-3}$. $\sigma = q n \mu_n = (1.602\times10^{-19})(10^{16})(1350)$ $= 1.602\times10^{-19}\times1.35\times10^{19} = 2.163\ (\Omega\cdot\text{cm})^{-1}$ (3)

Question 5 (10 marks)

(a) Any three (1 each): native stable oxide $\text{SiO}_2$ (excellent insulator/gate dielectric); wider band gap → lower leakage & higher operating temperature than Ge; abundant/cheap (sand); higher melting point & mechanical robustness for processing. (3)

(b) (2 each — property + application)

• GaN: wide bandgap (~3.4 eV) → high breakdown voltage/high-frequency → power electronics / fast chargers / RF amplifiers.
• GaAs: high electron mobility & direct bandgap → high-speed RF / optoelectronics (LEDs, laser diodes, solar cells).
• SiC: very wide bandgap & high thermal conductivity → high-temperature, high-power devices (EV inverters). (6)

(c) More expensive / harder to manufacture / lacks a high-quality native oxide. (1)

Question 6 (6 marks)

(a) A valence electron absorbs thermal energy $\geq E_g$, breaking a covalent bond and jumping to the conduction band, leaving behind a vacancy (hole) in the valence band. Electron and hole are both mobile charge carriers. (3)

(b) Mobility $\mu = v_d/E$ — drift velocity per unit electric field. As $T$ rises: (1) increased lattice/phonon scattering and (2) more thermal agitation → shorter mean free time → lower $\mu$. (Ionised-impurity scattering dominates only at low T.) (3)

[
  {"claim":"Bohr ground-state radius ~5.29e-11 m","code":"eps0=8.854e-12; hbar=1.055e-34; me=9.11e-31; e=1.602e-19; r1=4*pi*eps0*hbar**2/(me*e**2); result = abs(float(r1)-5.29e-11) < 0.05e-11"},
  {"claim":"ni(350)/ni(300) ratio ~27.8","code":"Eg=1.12; kB=8.617e-5; import math; expterm=math.exp(-Eg/(2*kB)*(1/350-1/300)); pre=(350/300)**1.5; ratio=expterm*pre; result = abs(ratio-27.8) < 1.0"},
  {"claim":"n-type conductivity ~2.16 (ohm-cm)^-1","code":"q=1.602e-19; n=1e16; mun=1350; sigma=q*n*mun; result = abs(sigma-2.163) < 0.01"},
  {"claim":"T^3/2 prefactor at 350/300 ~1.26","code":"pre=(350/300)**1.5; result = abs(pre-1.260) < 0.005"}
]