Level 2 — RecallMaterials & Atomic Structure

Materials & Atomic Structure

30 minutes40 marksprintable — key stays hidden on paper

Chapter: 1.3 Materials & Atomic Structure Level: 2 — Recall (definitions, standard textbook problems, short derivations) Time Limit: 30 minutes Total Marks: 40


Instructions: Answer all questions. Show working where numerical answers are required. Use ...... notation for any formulae.


Q1. State the Bohr model postulate for the energy of an electron in the nn-th shell of a hydrogen atom, and give the maximum number of electrons that any shell nn can hold. (4 marks)

Q2. Define valence electrons. How many valence electrons does a silicon atom have, and to which group of the periodic table does silicon belong? (4 marks)

Q3. Explain what a covalent bond is and describe how covalent bonding produces a stable outer shell for each atom in a pure silicon crystal. (4 marks)

Q4. Distinguish between an intrinsic and an extrinsic semiconductor. For an intrinsic semiconductor, state the relationship between the electron concentration nn and the hole concentration pp. (4 marks)

Q5. Define carrier mobility and state its SI unit. Given the drift-velocity relation vd=μEv_d = \mu E, calculate the drift velocity of an electron in silicon where μn=0.135 m2V1s1\mu_n = 0.135\ \text{m}^2\,\text{V}^{-1}\,\text{s}^{-1} and the applied field is E=500 V/mE = 500\ \text{V/m}. (5 marks)

Q6. Explain the process of electron–hole pair generation by thermal energy. State what happens to the intrinsic carrier concentration nin_i of silicon as temperature increases. (4 marks)

Q7. The conductivity of a semiconductor is given by σ=q(nμn+pμp).\sigma = q\,(n\,\mu_n + p\,\mu_p). For an intrinsic sample with n=p=ni=1.5×1016 m3n = p = n_i = 1.5\times10^{16}\ \text{m}^{-3}, μn=0.135 m2V1s1\mu_n = 0.135\ \text{m}^2\text{V}^{-1}\text{s}^{-1}, μp=0.048 m2V1s1\mu_p = 0.048\ \text{m}^2\text{V}^{-1}\text{s}^{-1} and q=1.6×1019 Cq = 1.6\times10^{-19}\ \text{C}, calculate the conductivity σ\sigma. (5 marks)

Q8. Give three reasons why silicon dominates over germanium in modern electronic devices. (3 marks)

Q9. Name three compound semiconductors and state one typical application of each. (3 marks)

Q10. State how the electrical conductivity of a metal and of an intrinsic semiconductor each change with rising temperature, and briefly explain why they behave differently. (4 marks)


End of Paper

Answer keyMark scheme & solutions

Q1. (4 marks)

  • Bohr postulate: the energy of the nn-th allowed orbit is En=13.6n2 eVE_n = -\dfrac{13.6}{n^2}\ \text{eV} (energy is quantised; electrons occupy fixed orbits without radiating). (2 marks) — the quantisation/fixed-orbit idea is the core of the model.
  • Maximum electrons in shell nn is 2n22n^2. (2 marks) — standard shell-capacity rule (e.g. K=2, L=8, M=18).

Q2. (4 marks)

  • Valence electrons = the electrons in the outermost shell of an atom, which participate in chemical bonding. (2 marks)
  • Silicon has 4 valence electrons (1 mark) and belongs to Group IV (14) of the periodic table (1 mark).

Q3. (4 marks)

  • A covalent bond is the sharing of a pair of electrons between two adjacent atoms. (2 marks)
  • Each silicon atom shares one electron with each of its 4 neighbours; combined with its own 4 valence electrons, this gives an effective 8 electrons in the outer shell — a stable (filled octet) configuration. (2 marks)

Q4. (4 marks)

  • Intrinsic: a pure semiconductor with no added impurities; conduction arises only from thermally generated electron–hole pairs. (1.5 marks)
  • Extrinsic: a semiconductor deliberately doped with impurity atoms (donors/acceptors) to increase carrier concentration and control conductivity. (1.5 marks)
  • For intrinsic material: n=p=nin = p = n_i. (1 mark)

Q5. (5 marks)

  • Mobility = drift velocity acquired by a carrier per unit applied electric field. (1.5 marks)
  • SI unit: m2V1s1\text{m}^2\,\text{V}^{-1}\,\text{s}^{-1}. (1 mark)
  • Calculation: vd=μnE=0.135×500=67.5 m/sv_d = \mu_n E = 0.135 \times 500 = 67.5\ \text{m/s}. (2.5 marks) — substitution (1) and correct result with unit (1.5).

Q6. (4 marks)

  • Thermal energy breaks a covalent bond, freeing an electron and leaving behind a vacancy (a hole); the freed electron and hole form an electron–hole pair, both able to conduct. (3 marks)
  • As temperature rises, more bonds break so nin_i increases (roughly exponentially with TT). (1 mark)

Q7. (5 marks)

  • σ=qni(μn+μp)\sigma = q\,n_i(\mu_n + \mu_p). (1 mark)
  • μn+μp=0.135+0.048=0.183\mu_n + \mu_p = 0.135 + 0.048 = 0.183. (1 mark)
  • σ=1.6×1019×1.5×1016×0.183\sigma = 1.6\times10^{-19} \times 1.5\times10^{16} \times 0.183. (1 mark)
  • =1.6×1.5×0.183×103=0.43920×1034.39×104 S/m= 1.6\times1.5\times0.183 \times 10^{-3} = 0.43920 \times 10^{-3} \approx 4.39\times10^{-4}\ \text{S/m}. (2 marks)

Q8. (3 marks) — any three, 1 mark each:

  • Silicon forms a stable, high-quality native oxide (SiO2\text{SiO}_2) essential for MOS technology.
  • Higher bandgap (~1.1 eV) → lower leakage current and operation at higher temperatures than germanium (~0.66 eV).
  • Abundant and cheap (from sand); mature, well-developed fabrication technology.

Q9. (3 marks) — any three, 1 mark each (semiconductor + valid application):

  • GaN — LEDs, high-power/high-frequency RF and power devices.
  • GaAs — high-speed microwave/RF circuits, laser diodes, solar cells.
  • SiC — high-power, high-temperature power electronics (e.g. EV inverters).

Q10. (4 marks)

  • Metal: conductivity decreases with rising temperature. (1 mark) — increased lattice vibrations scatter carriers, reducing mobility (carrier count is fixed). (1 mark)
  • Intrinsic semiconductor: conductivity increases with rising temperature. (1 mark) — the exponential rise in thermally generated carriers (nin_i) outweighs the fall in mobility. (1 mark)

[
  {"claim":"Q1 shell capacity 2n^2 gives L-shell (n=2) capacity of 8","code":"n=2; result = (2*n**2 == 8)"},
  {"claim":"Q5 drift velocity = mu*E = 0.135*500 = 67.5 m/s","code":"vd = 0.135*500; result = abs(vd - 67.5) < 1e-9"},
  {"claim":"Q7 conductivity approx 4.392e-4 S/m","code":"q=1.6e-19; ni=1.5e16; sigma=q*ni*(0.135+0.048); result = abs(sigma - 4.392e-4) < 1e-7"},
  {"claim":"Q7 mobility sum equals 0.183","code":"result = abs((0.135+0.048) - 0.183) < 1e-12"}
]