Key Fact
Semiconductors
Materials with conductivity between conductors and insulators. Intrinsic (pure Si, Ge) have equal electrons and holes. Extrinsic: doped with donors (n-type) or acceptors (p-type).
-- NCERT Class 12 Physics, Ch. 14, p. 329Semiconductors
8 MCQs9-step worked example
Source: NCERT Electronic DevicesPYQ coverage: NEET 2021Official key: NTA-verifiedLast updated: 24 Sep 2026
Semiconductors, explained for NEET
The confusion that costs marks here is treating "semiconductor" as a fixed category of material rather than a position on a conductivity scale. A repeater who has memorised "metals conduct, insulators don't, semiconductors are in between" will still lose the mark when a stem asks about resistivity magnitude or about the sign of the temperature coefficient, because those are the two things actually tested at this level.
NCERT Class 12 Physics, Chapter 14 (page 324) classifies solids by resistivity ρ and conductivity σ. Metals sit at ρ ≈ 10⁻² to 10⁻⁸ Ω m. Insulators sit at ρ ≈ 10¹¹ to 10¹⁹ Ω m. Semiconductors occupy the band between, ρ ≈ 10⁻⁵ to 10⁶ Ω m — a span of eleven orders of magnitude, which is why a single remembered number is useless and the range is what gets examined.
The second classification runs on energy band gap E_g. A conductor has conduction and valence bands overlapping, so E_g is effectively zero. An insulator has E_g large — for diamond, about 6 eV — so at room temperature essentially no electron is promoted across it. A semiconductor has a small but non-zero gap: silicon about 1.1 eV, germanium about 0.7 eV. Small enough that thermal energy lifts a measurable number of electrons across; large enough that at 0 K the material is a perfect insulator.
That last point drives the behaviour NEET tests most often. Raise the temperature of a semiconductor and more electron-hole pairs are generated, so conductivity rises and resistivity falls — a negative temperature coefficient of resistance. A metal does the opposite: its carrier count is fixed, lattice vibrations increase, and resistivity rises.
Elemental semiconductors are Si and Ge. Compound semiconductors include GaAs, CdS, InP and organic solids like anthracene.
Watch out: "semiconductor at 0 K behaves as an insulator" is true and frequently examined. Do not mark it false because the material is called a semiconductor.
Can you answer these Semiconductors MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
According to the NCERT classification of solids on the basis of resistivity, the approximate resistivity range of a semiconductor is
Show answer and why every option is right or wrong
Answer: B. NCERT Class 12 Physics, Chapter 14 (page 324) gives the semiconductor resistivity band as roughly 10⁻⁵ Ω m to 10⁶ Ω m, sitting between metals and insulators.
Why A is wrong: A is wrong because 10⁻⁸ Ω m to 10⁻² Ω m is the metal range, not the semiconductor range.
Why C is wrong: C is wrong because 10¹¹ Ω m to 10¹⁹ Ω m is the insulator range.
Why D is wrong: D is wrong because it is an invented band above the semiconductor range; NCERT places the semiconductor upper bound at about 10⁶ Ω m, not its lower bound.
MCQ 2Easy RecallPractice
Which of the following is an elemental semiconductor?
Show answer and why every option is right or wrong
Answer: C. Germanium is an elemental semiconductor — a single element from group IV. NCERT Class 12 Physics, Chapter 14 (page 325) lists Si and Ge as the elemental semiconductors.
Why A is wrong: A is wrong because GaAs is a compound semiconductor formed from gallium and arsenic.
Why B is wrong: B is wrong because CdS is a compound semiconductor formed from cadmium and sulphur.
Why D is wrong: D is wrong because InP is a compound semiconductor formed from indium and phosphorus.
MCQ 3Easy RecallPractice
On the energy-band picture, a conductor is distinguished by which feature?
Show answer and why every option is right or wrong
Answer: A. In a conductor the conduction and valence bands overlap, leaving no forbidden gap, so electrons move into conducting states without any thermal promotion. NCERT Class 12 Physics, Chapter 14 (page 326) uses this overlap as the defining band-picture feature of a metal.
Why B is wrong: B is wrong because a gap of about 1.1 eV is silicon's — that is a semiconductor, not a conductor.
Why C is wrong: C is wrong because a gap above about 3 eV describes an insulator.
Why D is wrong: D is wrong because the valence band of a conductor is not empty; it is the overlap with the conduction band, not emptiness of the valence band, that makes conduction free.
MCQ 4Direct ApplicationPractice
A specimen of pure silicon is cooled from 300 K towards 0 K. As the temperature falls, its electrical conductivity
Show answer and why every option is right or wrong
Answer: B. In a semiconductor the carriers are created by thermal excitation across the gap, so cooling removes carriers and conductivity falls; at 0 K silicon behaves as an insulator. NCERT Class 12 Physics, Chapter 14 (page 326) states that at 0 K a semiconductor's valence band is full and its conduction band empty.
Why A is wrong: A is wrong because it applies the metallic mechanism — in a metal the carrier count is fixed and reduced lattice scattering raises conductivity on cooling. In a semiconductor the carrier count itself collapses, and that dominates.
Why C is wrong: C is wrong because conductivity depends on the number of free carriers, not on the number of atoms; the atoms remain but their bonded electrons are no longer promoted.
Why D is wrong: D is wrong because there is no reversal — the loss of thermally generated pairs is monotonic as T falls to 0 K.
MCQ 5Direct ApplicationPractice
Four solids are found to have band gaps of 0.0 eV, 0.7 eV, 1.1 eV and 6.0 eV. Which one is most likely to be an insulator at room temperature?
Show answer and why every option is right or wrong
Answer: D. An insulator has a band gap far above the available thermal energy at room temperature; 6.0 eV is diamond-like and blocks promotion of electrons entirely. NCERT Class 12 Physics, Chapter 14 (page 326) uses E_g > 3 eV as the insulator marker.
Why A is wrong: A is wrong because a zero gap means the bands overlap — that is a conductor.
Why B is wrong: B is wrong because 0.7 eV is germanium's gap, which is small enough for appreciable thermal generation; that is a semiconductor.
Why C is wrong: C is wrong because 1.1 eV is silicon's gap — a semiconductor, the very case the question contrasts against.
MCQ 6Direct ApplicationPractice
A sample of germanium and a sample of copper are each heated from 300 K to 400 K. Which statement correctly describes the change in their resistances?
Show answer and why every option is right or wrong
Answer: C. Germanium is a semiconductor with a negative temperature coefficient — heating generates more electron-hole pairs, so resistance falls. Copper is a metal with a fixed carrier count and increased lattice scattering, so resistance rises. NCERT Class 12 Physics, Chapter 14 (page 326) contrasts the two behaviours.
Why A is wrong: A is wrong because it applies the metallic (positive coefficient) rule to germanium as well; a semiconductor's resistance falls on heating.
Why B is wrong: B is wrong because it applies the semiconductor rule to copper; a metal's resistance rises on heating.
Why D is wrong: D is wrong because it reverses both materials — it assigns the positive coefficient to germanium and the negative one to copper.
MCQ 7Concept TrapPractice
A student argues: "Silicon is called a semiconductor, so it must conduct at least a little at every temperature." Which response is correct?
Show answer and why every option is right or wrong
Answer: B. "Semiconductor" names the size of the band gap, not a guarantee of conduction; at 0 K there is no thermal energy to promote electrons, so pure silicon behaves as a perfect insulator. NCERT Class 12 Physics, Chapter 14 (page 326) states this explicitly.
Why A is wrong: A is wrong because it treats the label as a property of the material at every temperature; the label describes only the gap size, and conduction requires carriers that must be thermally created.
Why C is wrong: C is wrong because silicon conducts appreciably at ordinary room temperature, far below its melting point — this overstates the threshold enormously.
Why D is wrong: D is wrong because the band gap does not close on cooling; it is the thermal energy available to cross the fixed gap that vanishes.
MCQ 8CalculationPractice
Three solids X, Y and Z are tested. X has resistivity 4 × 10⁻⁷ Ω m and its resistance rises when heated. Y has resistivity 2 × 10² Ω m and its resistance falls when heated. Z has resistivity 5 × 10¹⁵ Ω m. Identify X, Y and Z in that order.
Show answer and why every option is right or wrong
Answer: C. X at 4 × 10⁻⁷ Ω m lies in the metal band (10⁻⁸ to 10⁻² Ω m) and has a positive temperature coefficient — a conductor. Y at 2 × 10² Ω m lies in the semiconductor band (10⁻⁵ to 10⁶ Ω m) with a negative coefficient. Z at 5 × 10¹⁵ Ω m lies in the insulator band (10¹¹ to 10¹⁹ Ω m). NCERT Class 12 Physics, Chapter 14 (page 324) supplies all three ranges.
Why A is wrong: A is wrong because 4 × 10⁻⁷ Ω m is nine orders of magnitude below the semiconductor lower bound; X cannot be a semiconductor, and it also shows the metallic rising-resistance behaviour.
Why B is wrong: B is wrong because 2 × 10² Ω m is well inside the semiconductor band and nowhere near the insulator band starting at 10¹¹ Ω m; Y's falling resistance on heating confirms semiconductor.
Why D is wrong: D is wrong because it inverts the resistivity scale entirely — it labels the lowest-resistivity sample an insulator and the highest a conductor.
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How do you solve a Semiconductors question? A worked example
- 1
Given.
A solid sample of uniform cross-section has length L = 2.0 × 10⁻² m and cross-sectional area A = 4.0 × 10⁻⁶ m². Its resistance measured at 300 K is R = 1.0 × 10³ Ω. When the sample is warmed to 350 K, its measured resistance drops to 6.0 × 10² Ω.
- 2
Required.
Calculate the resistivity at 300 K, and classify the solid as conductor, semiconductor or insulator using both the resistivity value and the observed temperature behaviour.
- 3
Concept.
Two independent classification handles exist for a solid. First, the magnitude of resistivity ρ against the NCERT bands: metals 10⁻⁸–10⁻² Ω m, semiconductors 10⁻⁵–10⁶ Ω m, insulators 10¹¹–10¹⁹ Ω m. Second, the sign of the temperature coefficient: a metal's resistance rises on heating (fixed carriers, more lattice scattering), a semiconductor's falls (more electron-hole pairs generated). A sound answer uses both handles and checks they agree.
- 4
Formula.
ρ = RA / L.
- 5
Substitution.
ρ = (1.0 × 10³ Ω)(4.0 × 10⁻⁶ m²) / (2.0 × 10⁻² m).
- 6
Calculation.
Numerator: (1.0 × 10³)(4.0 × 10⁻⁶) = 4.0 × 10⁻³ Ω m².
Divide: (4.0 × 10⁻³) / (2.0 × 10⁻²) = 2.0 × 10⁻¹ Ω m.
All three given quantities carry two significant figures, so the result is quoted to two. No exact constants enter this calculation — every number here is a measured quantity, and each one therefore does limit the significant-figure count. (Contrast a case such as a cube's S = 6a², where the 6 is a counting integer and would not limit it.) - 7
Final answer.
ρ = 2.0 × 10⁻¹ Ω m at 300 K.
This value lies inside the semiconductor band (10⁻⁵ to 10⁶ Ω m) and is about one order of magnitude above the metal ceiling of 10⁻² Ω m. The temperature check agrees: resistance fell from 1.0 × 10³ Ω to 6.0 × 10² Ω on heating, a negative temperature coefficient, which is semiconductor behaviour and rules out a metal. The solid is a semiconductor. - 8
Common trap.
Reading 2.0 × 10⁻¹ Ω m as "small, therefore metallic." It is small compared with everyday insulators but it is ten times larger than the metal upper bound of 10⁻² Ω m. The bands are logarithmic, so an order-of-magnitude comparison — not an intuitive sense of "small" — decides the classification. The second trap is skipping the temperature check: resistivity alone near a band edge can be ambiguous, and the sign of the coefficient resolves it cleanly. Here both handles point the same way, which is what makes the conclusion safe.
- 9
Similar NEET-style question.
A wire of length 5.0 × 10⁻² m and cross-sectional area 2.0 × 10⁻⁷ m² has resistance 1.5 × 10⁻² Ω at 300 K, and this resistance increases to 1.8 × 10⁻² Ω at 400 K. Find its resistivity and classify the material. *(Answer: ρ = 6.0 × 10⁻⁸ Ω m, inside the metal band, and the resistance rises on heating — a conductor.)*
What to remember before solving Semiconductors questions
More in Electronic Devices: 3 exam traps and mistakes · 3 formulas · 3 question patterns from its other lessons.
Semiconductors questions from past NEET papers
1 question from NEET 2021. Answers verified against NTA official keys. — click to collapse
NEET 2021
1No current will flow in p-type, current will only flow in n-type
2Current in n-type = current in p-type
3Current in p-type > current in n-type
4Current in n-type > current in p-type.
NTA Answer: Option 4(final)
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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