Level 1 — RecognitionBand Theory & Carrier Physics

Band Theory & Carrier Physics

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Chapter: 2.1 Band Theory & Carrier Physics Level: 1 (Recognition) Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Choose the single best answer.

Q1. The energy band occupied by the outermost electrons at absolute zero is called the: (a) conduction band (b) valence band (c) forbidden band (d) Fermi band (1 mark)

Q2. For an intrinsic semiconductor, the Fermi level lies approximately: (a) at the top of the conduction band (b) at the bottom of the valence band (c) near the middle of the band gap (d) above the conduction band (1 mark)

Q3. Which material has essentially overlapping valence and conduction bands (zero band gap)? (a) insulator (b) semiconductor (c) conductor (d) intrinsic silicon (1 mark)

Q4. The mass action law for a semiconductor in thermal equilibrium is: (a) $n + p = n_i$ (b) $np = n_i^2$ (c) $n - p = n_i$ (d) $n/p = n_i$ (1 mark)

Q5. In GaAs, an electron can recombine directly with a hole emitting a photon because it is a: (a) indirect band-gap material (b) direct band-gap material (c) conductor (d) insulator (1 mark)

Q6. Drift current arises due to: (a) a concentration gradient (b) an applied electric field (c) temperature alone (d) photon emission (1 mark)

Q7. The Einstein relation connects diffusion coefficient $D$ and mobility $\mu$ as: (a) $D = \mu q / kT$ (b) $D/\mu = kT/q$ (c) $D\mu = kT/q$ (d) $D = \mu q T$ (1 mark)

Q8. In an n-type semiconductor, the majority carriers are: (a) holes (b) electrons (c) photons (d) phonons (1 mark)

Q9. As temperature increases, the intrinsic carrier concentration $n_i$ : (a) decreases (b) stays constant (c) increases (d) becomes zero (1 mark)

Q10. The Fermi–Dirac distribution $f(E)$ gives: (a) the number of energy bands (b) the probability an energy state is occupied by an electron (c) the band gap width (d) the drift velocity (1 mark)

Section B — Matching (1 mark each, 5 marks)

Q11. Match each material/quantity in Column X to its correct description in Column Y. (5 marks)

Column X		Column Y
(i) Conductor		(A) Band gap ≈ 1.1 eV, controllable conduction
(ii) Semiconductor (Si)		(B) Very large band gap (> ~5 eV), few free carriers
(iii) Insulator		(C) No/overlapping band gap, many free carriers
(iv) Diffusion current		(D) Driven by carrier concentration gradient
(v) Recombination		(E) An electron–hole pair is annihilated

Section C — True/False with Justification (1 mark T/F + 1 mark reason = 2 marks each, 30 total but capped)

State True or False and give a one-line justification. (Marks: 1 for correct T/F, 1 for correct justification.)

Q12. "At $E = E_F$ , the Fermi–Dirac occupation probability equals 0.5 at any non-zero temperature." (2 marks)

Q13. "Silicon has a direct band gap, which makes it ideal for making LEDs." (2 marks)

Q14. "In a p-type semiconductor, electrons are the minority carriers." (2 marks)

Q15. "A larger band gap generally means the material is a better electrical conductor at room temperature." (2 marks)

Q16. "Diffusion current can flow even when no external electric field is applied." (2 marks)

Q17. "The generation of electron–hole pairs increases with temperature." (2 marks)

End of paper. Total = 10 (MCQ) + 5 (Matching) + 12 (T/F) = 27… allocate remaining 3 marks equally weighting; scale to 30. Note: Sections A (10) + B (5) + C (12) = 27 marks; a 3-mark presentation/clarity allowance is added → Total 30.

Answer keyMark scheme & solutions

Section A — MCQ (1 mark each)

Q1. (b) valence band — Valence band = highest band filled with valence electrons at 0 K; conduction band lies above it. (1)

Q2. (c) near the middle of the band gap — For intrinsic material electron and hole concentrations are equal, forcing $E_F$ close to mid-gap. (1)

Q3. (c) conductor — Metals have overlapping/zero gap so electrons move freely into empty states. (1)

Q4. (b) $np = n_i^2$ — In equilibrium the product of electron and hole concentrations is a constant equal to $n_i^2$ . (1)

Q5. (b) direct band-gap material — In direct-gap GaAs the conduction-band minimum and valence-band maximum share the same crystal momentum $k$ , so radiative recombination needs no phonon. (1)

Q6. (b) an applied electric field — Drift velocity $v_d = \mu E$ ; current $J_{drift}=q n \mu E$ . (1)

Q7. (b) $D/\mu = kT/q$ — Einstein relation; equivalently $D = \mu kT/q$ . (1)

Q8. (b) electrons — n-type doping (donors) supplies excess electrons as majority carriers. (1)

Q9. (c) increases — $n_i \propto T^{3/2}\exp(-E_g/2kT)$ , a strongly increasing function of $T$ . (1)

Q10. (b) the probability an energy state is occupied by an electron — Definition of $f(E)=1/(1+e^{(E-E_F)/kT})$ . (1)

Section B — Matching (Q11, 5 marks, 1 each)

(i) → C, (ii) → A, (iii) → B, (iv) → D, (v) → E. Reasoning: conductors have overlapping bands (many free carriers); Si gap ≈1.1 eV; insulators >5 eV; diffusion driven by gradient; recombination destroys an e–h pair.

Section C — True/False + Justification (2 marks each)

Q12. TRUE. Substituting $E=E_F$ : $f=1/(1+e^0)=1/2$ . (T/F 1 + reason 1)

Q13. FALSE. Silicon has an indirect band gap; poor radiative efficiency, so unsuitable for LEDs. Direct-gap materials (GaAs) are used. (1+1)

Q14. TRUE. In p-type, holes are majority carriers; electrons are the minority carriers. (1+1)

Q15. FALSE. Larger band gap → fewer carriers thermally excited → poorer conduction (insulators have large gaps). (1+1)

Q16. TRUE. Diffusion current is driven by a concentration gradient $J=qD\,dn/dx$ , independent of external field. (1+1)

Q17. TRUE. Higher $T$ gives more thermal energy to break bonds, raising generation rate and $n_i$ . (1+1)

Verification of key numeric/derivation claims

[
  {"claim":"Fermi-Dirac gives f=1/2 at E=E_F (Q12)","code":"E,EF,k,T=symbols('E EF k T',positive=True); f=1/(1+exp((E-EF)/(k*T))); val=f.subs(E,EF); result = simplify(val-Rational(1,2))==0"},
  {"claim":"Einstein relation D/mu = kT/q (Q7)","code":"D,mu,k,T,q=symbols('D mu k T q',positive=True); expr=Eq(D/mu, k*T/q); Dexpr=solve(expr,D)[0]; result = simplify(Dexpr - mu*k*T/q)==0"},
  {"claim":"Mass action np=ni^2 => n=ni^2/p consistent (Q4)","code":"n,p,ni=symbols('n p ni',positive=True); eq=Eq(n*p,ni**2); nsol=solve(eq,n)[0]; result = simplify(nsol - ni**2/p)==0"},
  {"claim":"ni increases with T: derivative of T**(3/2)*exp(-Eg/(2kT)) wrt T positive (Q9)","code":"T,Eg,k=symbols('T Eg k',positive=True); ni=T**Rational(3,2)*exp(-Eg/(2*k*T)); d=diff(ni,T); val=d.subs({Eg:1.1,k:8.617e-5,T:300}); result = val>0"}
]