JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
This MCQ module is based on: Energy Bands Classification
TOPIC 23 OF 27
Energy Bands Classification
🎓 Class 12
Physics
CBSE
Theory
Ch 14 – Semiconductor Electronics
⏱ ~14 min
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Energy Bands Classification
14.1 Introduction — From Vacuum Tubes to Solid-State Electronics
Electronic devices are simply ways to control the flow of electrons. Before 1948 this control was achieved with bulky vacuum tubes (valves) — diodes, triodes, tetrodes, pentodes — in which a heated cathode boiled off electrons that travelled through an evacuated glass envelope toward an anode. Vacuum tubes worked, but they were large, fragile, hot, slow and power-hungry.
The 1948 invention of the transistor at Bell Labs (Bardeen, Brattain & Shockley — Nobel 1956) launched the solid-state revolution. In a semiconductor, mobile electrons live inside the crystal itself; they are summoned by simple stimuli — light, heat, a small voltage — without any vacuum or hot filament. The result: tiny, robust, fast, low-power devices that today form the building blocks of every smartphone, computer, sensor and solar panel.
Why semiconductors? Resistivity in between metals and insulators (~10⁻⁵ to 10⁶ Ωm). The number and type of mobile carriers can be tuned by temperature, doping or light — giving us the diode, the transistor and the integrated circuit.
14.2 Classification — Conductors, Semiconductors, Insulators
By electrical resistivity (ρ) or conductivity (σ = 1/ρ)
| Class | Resistivity ρ (Ω·m) | Conductivity σ (S/m) | Examples |
|---|---|---|---|
| Metals (conductors) | 10⁻² – 10⁻⁸ | 10² – 10⁸ | Cu, Ag, Al, Fe |
| Semiconductors | 10⁻⁵ – 10⁶ | 10⁵ – 10⁻⁶ | Si, Ge, GaAs, CdS |
| Insulators | 10¹¹ – 10¹⁹ | 10⁻¹¹ – 10⁻¹⁹ | glass, mica, rubber |
Semiconductors of interest in this chapter:
- Elemental: Silicon (Si) and Germanium (Ge) — group IV elements with 4 valence electrons.
- Compound (inorganic): GaAs, CdS, InP, CdSe.
- Organic: anthracene, doped phthalocyanines, conducting polymers like polyaniline.
By energy bands
In an isolated atom, electrons occupy discrete energy levels. When N atoms come together to form a crystal, the outer-shell levels overlap with neighbours and split into N closely-spaced levels. With N ~ 10²³, this dense set of levels behaves like a continuous energy band.
The two bands of greatest importance:
- Valence band (VB): contains the valence electrons. Filled (or nearly filled) at T = 0 K.
- Conduction band (CB): the next-higher band. Empty (or nearly empty) at T = 0 K.
Between the top of the VB (energy E_V) and the bottom of the CB (energy E_C) lies a forbidden region called the energy band gap E_g = E_C − E_V. The size of this gap, plus the band's filling, determines whether the material conducts.
| Material | Band gap E_g (eV) | Conduction at 300 K | Class |
|---|---|---|---|
| Diamond (C) | 5.4 | Negligible | Insulator |
| Silicon (Si) | 1.12 | Moderate | Semiconductor |
| Germanium (Ge) | 0.72 | Moderate | Semiconductor |
| Gallium Arsenide (GaAs) | 1.43 | Moderate | Semiconductor |
| Tin (Sn — α form) | 0 | Strong | Metal |
Worked Example 14.1 (NCERT): C, Si and Ge all have the same diamond-like crystal structure and four valence electrons, yet C is an insulator while Si and Ge are semiconductors. Reason: the four bonding electrons sit in the 2nd, 3rd and 4th orbits respectively, so the ionisation/band-gap energy decreases from C → Si → Ge. The smaller the gap, the easier it is for thermal energy to promote electrons into the conduction band.
14.3 Intrinsic Semiconductors
An intrinsic semiconductor is pure Si or Ge with no impurities. Both have 4 valence electrons; in the crystal each atom forms covalent bonds with four neighbours, sharing electrons in pairs.
Electrons and holes
At T = 0 K, all bonds are intact and the crystal is an insulator. Above 0 K, thermal vibrations occasionally break a bond, releasing a free electron into the conduction band and leaving an electron vacancy behind. This vacancy is called a hole and behaves like a positive charge carrier with an effective charge +e.
Hole motion is just rearrangement of bound electrons: an electron from a neighbouring bond fills the hole, creating a new hole at its old location. The net effect is as if a positive particle moved.
For an intrinsic semiconductor:
n_e = n_h = n_i (intrinsic carrier concentration)
I = I_e + I_h (electron + hole currents in parallel)
At room temperature, n_i is small (~10¹⁶ /m³ for Si), so intrinsic Si conducts only weakly. To make useful devices we need many more carriers — that's what doping does.
Activity 14.1 — Heating a thermistor
A thermistor is a piece of semiconductor whose resistance depends strongly on temperature. Connect a small thermistor in series with a battery and an ammeter. Measure the current at room temperature, then warm the thermistor with your fingers or a hair-dryer.
Predict: does the current increase or decrease on heating?
Current increases. Heating provides extra thermal energy, freeing more electron-hole pairs across the band gap, so n_e and n_h grow exponentially with T. Conductivity rises and resistance falls — the opposite of metals (where R increases with T due to phonon scattering).
14.4 Extrinsic Semiconductors — Doping
The number of charge carriers in pure Si or Ge can be enormously increased by adding tiny, controlled amounts (parts-per-million) of selected impurities. This process is called doping; the impurities are dopants; the result is an extrinsic (or impurity) semiconductor. To preserve the lattice structure, the dopant atoms must be roughly the size of Si/Ge.
Two categories matter for tetravalent Si/Ge:
- Pentavalent (group V): As, Sb, P — they have 5 valence electrons → donor.
- Trivalent (group III): B, Al, In, Ga — they have 3 valence electrons → acceptor.
n-type (donor doping)
When a pentavalent atom replaces a Si atom, four of its electrons form covalent bonds with neighbours; the 5th is loosely bound and easily liberated (ionisation energy only ~0.05 eV for P in Si — far less than E_g ≈ 1.1 eV). Each donor contributes one electron to the conduction band:
n_e ≈ N_D >> n_h (electrons = majority, holes = minority)
p-type (acceptor doping)
A trivalent dopant has only 3 valence electrons. Its 4th bond with a neighbouring Si atom has a missing electron — i.e., a hole. A small thermal kick lets a nearby valence electron jump in, leaving a hole somewhere else. Each acceptor contributes one hole to the valence band:
n_h ≈ N_A >> n_e (holes = majority, electrons = minority)
Mass-action law: In thermal equilibrium, regardless of doping,
\[ n_e \cdot n_h = n_i^2 \]
where n_i is the intrinsic concentration at the given temperature. Crystal as a whole remains electrically neutral.
Worked Example 14.2 (NCERT)
A pure Si crystal has 5 × 10²⁸ atoms/m³. It is doped with 1 ppm pentavalent As. Find n_e and n_h. Given n_i = 1.5 × 10¹⁶ /m³.
Donor concentration: N_D = 1 ppm × 5 × 10²⁸ = 5 × 10²² /m³.
Since N_D >> n_i, essentially every donor is ionised: n_e ≈ N_D = 5 × 10²² /m³.
Using mass action n_e · n_h = n_i²:
\[ n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = \mathbf{4.5 \times 10^{9}\ /m^3} \]Doping has multiplied n_e by ~10⁶ over n_i, while reducing n_h by the same factor — typical of an extrinsic semiconductor.
Interactive — Band-Gap Explorer
How does band gap classify materials?
Pick a material and see its band gap, conduction-band electron population at 300 K, and the resulting class (insulator / semiconductor / metal).
300 K
E_g: 1.12 eV
Class: Semiconductor
n_i (relative): 1×
Competency-Based Questions
Q1 (MCQ). In an n-type silicon, the dopants are:
(b) Pentavalent atoms donate one extra electron per impurity, making electrons the majority carriers.
Q2 (MCQ). The order of band gaps is:
(a) C: 5.4 eV (insulator), Si: 1.12 eV, Ge: 0.72 eV.
Q3 (Fill-in-the-blank). In a p-type semiconductor, ____ are the majority carriers and ____ are the minority carriers.
Holes are the majority; electrons are the minority.
Q4 (Numerical). Pure Ge has n_i = 2.4 × 10¹⁹ /m³ at 300 K. If it is doped with 10¹⁵ /m³ acceptor atoms, find n_h and n_e.
Note here N_A = 10¹⁵ << n_i, so doping has minor effect. Both n_h and n_e remain ~ n_i ≈ 2.4 × 10¹⁹ /m³ to leading order. (Doping must exceed n_i to dominate.)
Q5 (HOTS). Why does an intrinsic semiconductor's conductivity increase with temperature, while a metal's decreases?
Semiconductor: n_i ∝ exp(−E_g/2kT). Heating exponentially raises the number of free carriers. This dominates over the slight reduction in mobility from phonon scattering, so σ increases. Metal: the carrier density is fixed and very high. Heating only increases lattice vibrations, scattering carriers more frequently and reducing mobility — so σ falls.
Assertion–Reason Questions
Options: (A) Both true, R correct explanation. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: A hole behaves like a positive charge carrier in semiconductors.
Reason: A hole represents the absence of an electron in a covalent bond, which an external field causes neighbouring electrons to fill — making the vacancy appear to drift in the direction of the field.
(A) Both correct, and the reason explains the assertion.
Assertion: Doping always increases the total number of charge carriers in a semiconductor.
Reason: n_e × n_h = n_i² stays constant in equilibrium.
(B) Both true, but the reason does NOT explain the assertion. The total carriers increase because the majority concentration rises far more than the minority falls. Mass-action holds independently.
Assertion: A semiconductor is a metal at very high temperatures.
Reason: Thermal agitation eventually fills the conduction band like a metal.
(D) The first claim is false (semiconductors and metals differ in their band structure, not just population). The second is partially true but not the right reason. So (D)-equivalent: A is false; R is true (in spirit).
Frequently Asked Questions - Energy Bands Classification
What is the main concept covered in Energy Bands Classification?
In NCERT Class 12 Physics Chapter 14 (Semiconductor Electronics), "Energy Bands Classification" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Energy Bands Classification useful in real-life applications?
Real-life applications of "Energy Bands Classification" from NCERT Class 12 Physics Chapter 14 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Energy Bands Classification?
Key formulas in "Energy Bands Classification" (NCERT Class 12 Physics Chapter 14 Semiconductor Electronics) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 14?
NCERT Class 12 Physics Chapter 14 (Semiconductor Electronics) is structured so each part builds on the previous one. "Energy Bands Classification" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Energy Bands Classification?
CBSE board questions from "Energy Bands Classification" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Energy Bands Classification" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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