The energy levels of holes are: (A) in the valence band (B) in the conduction band (C) in the band gap but close to valence band (D) in the band gap but close to conduction band

Question

Accepted Answer

Concept understanding — Intrinsic and Extrinsic Semiconductors

A pure semiconductor -- pure silicon or pure germanium, with no impurity added -- is called INTRINSIC. Each silicon or germanium atom has a valence of four and forms four covalent bonds with its neighbours in a tetrahedral arrangement; at absolute zero, every bond is complete and no free carriers exist. At room temperature, thermal agitation breaks a small number of these bonds, promoting a valence electron into the conduction band and leaving behind a vacancy -- a HOLE -- in the valence band, which behaves as an effective positive charge. Because every broken bond produces one electron and one hole together, an intrinsic semiconductor always has equal carrier densities, $n_e=n_h=n_i$ (the intrinsic carrier density), and conduction happens through BOTH carrier types drifting in opposite directions when a field is applied -- electrons toward the positive terminal, holes (via a relay of neighbouring electron jumps) toward the negative terminal. Because so few of a crystal's roughly $5\times10^{22}$ atoms per $cm^3$ actually contribute a carrier this way, intrinsic conductivity is comparatively low and not easily controlled.

An EXTRINSIC semiconductor is an intrinsic semiconductor deliberately DOPED with a small, carefully chosen amount of impurity to raise and control its conductivity. Doping with a PENTAVALENT impurity (phosphorus, arsenic, antimony) gives an N-TYPE semiconductor: four of the dopant's five valence electrons bond normally, and the fifth remains only weakly bound (needing just 0.01-0.05 eV to free), donating a nearly-free electron per dopant atom, so $n_e\gg n_h$ and electrons are the majority carrier. Doping with a TRIVALENT impurity (boron, aluminium, indium) gives a P-TYPE semiconductor instead: the dopant's three electrons complete only three of the host's four bonds, leaving one incomplete bond -- a hole -- per dopant atom, so $n_h\gg n_e$ and holes are the majority carrier. In both cases, the crystal remains electrically NEUTRAL overall (every donated electron or accepted hole is balanced by an oppositely-charged, fixed impurity ion), and for ANY semiconductor, doped or not, the mass-action law $n_e n_h=n_i^2$ always holds, letting the minority carrier density be found once the (doping-set) majority density is known. Because electron mobility exceeds hole mobility (a hole's motion is really a relay of individual electron jumps, not one carrier moving freely), an n-type semiconductor conducts better than an equally-doped p-type one.

A hole is created when thermal energy breaks a covalent bond and promotes a valence electron into the conduction band; the vacancy this leaves behind, at the position the electron departed from, is the hole. Since that vacancy is a missing occupant of one of the valence band's own energy levels, the hole's energy level is, by construction, IN the valence band -- not in the conduction band and not in the band gap (those in-gap donor/acceptor levels described in Sections 14.5.1 and 14.5.2 are a separate, doping-related detail, not the general answer for where a hole 'lives'). [!ANSWER] (A) in the valence band

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set V11 markMCQ
$Q. Which of the following pairs are elemental semiconductors?$
(a) Silicon and aluminium
(b) Silicon and germanium
(c) Germanium and cadmium
(d) Aluminium and cadmium
›Reveal solutionSolution

Option (b) Silicon and germanium.

✓Final answer
$Option (b) Silicon and germanium.$

Silicon (Si) and germanium (Ge) are group-14 (tetravalent) elements that behave as intrinsic elemental semiconductors. Aluminium and cadmium are metals, so the other pairs are not elemental semiconductors.
CBSE 2026Set A1 markMCQ
$Q. Semiconductor is damaged by the strong current due to (A) lack of free electrons (B) excess of protons (C) excess of electrons (D) excess of neutrons$
›Reveal solutionSolution

A strong current means an excess of moving electrons, whose collisions dump heat that damages the semiconductor's crystal structure.

Unlike a metal, a semiconductor has relatively few free carriers, and its conductivity rises sharply with temperature. When a strong current flows, there is an excess of electrons moving through the material; their frequent collisions with the lattice generate intense Joule heating. This heating liberates still more carriers (thermal runaway), and the temperature rise can break the crystal bonds and permanently damage the device. So the damage is attributed to the excess of electrons (charge carriers) driven by the strong current.

✓Final answer
$(C) excess of electrons.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. When a semiconductor is heated, its resistance$
(a) decreases
(b) increases
(c) remains unchanged
(d) is uncertain
›Reveal solutionSolution
A semiconductor's resistance falls with rising temperature, because the number of charge carriers rises sharply (negative temperature coefficient of resistance).

In a semiconductor, conduction occurs through electrons and holes generated when covalent bonds break due to thermal energy. As temperature increases:

More covalent bonds break, generating more electron–hole pairs.

The number of free charge carriers (and hence conductivity $\sigma$ ) increases roughly exponentially with temperature.

Since resistance $R \propto 1/\sigma$ , the resistance decreases as temperature rises.

This is opposite to the behaviour of a metallic conductor, whose resistance increases with temperature (as electron-lattice collisions increase while carrier density stays essentially fixed).

✓Final answer
$(a) decreases.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which one in the following is a semiconductor-$
(a) Phosphorus
(b) Aluminium
(c) Polyaniline
(d) Copper
›Reveal solutionSolution

Among the four choices, polyaniline is a conducting/semiconducting organic polymer; the others are an insulating-side element, a metal, and a metal.

Phosphorus (a non-metal element, used as a dopant but not itself classed as a semiconductor here) and copper (a good metallic conductor) and aluminium (a metal, a conductor) are not semiconductors. Polyaniline is a conjugated conducting polymer whose electrical conductivity can be tuned over a wide range (from insulating to semiconducting to metallic-like) by doping — it behaves as an organic semiconductor and is used in semiconductor devices such as organic transistors and sensors.

✓Final answer
$(c) Polyaniline$
CBSE 2025Set ANNUAL1 mark
$Q. The resistivity of semiconductors is between ____________ ohm-m to ____________ ohm-m.$
›Reveal solutionSolution
Semiconductors occupy the middle ground between conductors and insulators, with resistivity roughly between $10^{-5}$ and $10^{6}$ ohm-metre.

Comparing typical resistivity ranges:

Conductors (metals): $\rho \sim 10^{-8}$ to $10^{-2}\ \Omega$ m

Semiconductors: $\rho \sim 10^{-5}$ to $10^{6}\ \Omega$ m

Insulators: $\rho \sim 10^{11}$ to $10^{19}\ \Omega$ m

This intermediate resistivity (and correspondingly intermediate conductivity) is the defining property that gives semiconductors their name and their usefulness in electronic devices, since it can be further controlled by doping and temperature.

✓Final answer
$10^{-5}$
CBSE 2025Set ANNUAL1 mark
$Q. Semi-conductor is a device that obeys Ohm's law. (T/F)$
›Reveal solutionSolution

This statement is False — a semiconductor (and semiconductor devices such as diodes) generally do NOT obey Ohm's law; their V-I relationship is non-linear.

Ohm's law ( $V = IR$ with constant $R$ ) holds for an ohmic conductor whose resistance stays constant regardless of the applied voltage or current direction. A semiconductor's conductivity strongly depends on temperature and carrier concentration, and semiconductor junction devices (like a p-n junction diode) show a markedly non-linear V-I characteristic (near-zero current in reverse bias, exponentially rising current in forward bias beyond the threshold voltage). This is precisely why semiconductor devices are classified as non-ohmic.

✓Final answer
$False — a semiconductor generally does not obey Ohm's law.$
CBSE 2024Set ANNUAL1 mark
$Q. What happens to the electrical resistance of a semiconductor when its temperature is raised?$
›Reveal solutionSolution

Unlike metals, a semiconductor's resistance falls when it is heated, because more charge carriers become available for conduction.

In a pure (intrinsic) semiconductor at low temperature, very few electrons have enough thermal energy to jump across the energy gap from the valence band to the conduction band, so the number of free charge carriers (electrons and holes) is small and the resistance is relatively high.

As the temperature is raised, more valence electrons acquire enough thermal energy to break their covalent bonds and jump into the conduction band, creating additional electron–hole pairs. The number of free charge carriers therefore increases rapidly (roughly exponentially) with temperature.

Since conductivity is proportional to the number of free carriers, the conductivity increases and hence the resistance DECREASES with rising temperature. This is opposite to the behaviour of a metal, whose resistance increases with temperature (due to increased lattice vibration/scattering, not a change in carrier density) — semiconductors are said to have a negative temperature coefficient of resistance.

✓Final answer
$Its resistance decreases — more electron-hole pairs are thermally generated, increasing conductivity (negative temperature coefficient of resistance).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Example of inorganic semiconductor is -$
(a) Ge
(b) CdS
(c) anthracene
(d) polyaniline
›Reveal solutionSolution
Semiconductors are classed as inorganic (elemental/compound, e.g. Ge, Si, CdS, GaAs) or organic (e.g. anthracene, polyaniline); CdS is the inorganic compound example here.

Semiconducting materials are broadly grouped as:

Elemental inorganic semiconductors: Silicon (Si), Germanium (Ge).

Compound inorganic semiconductors: CdS, GaAs, CdSe, InP, etc.

Organic semiconductors: anthracene, doped phthalocyanines, polyaniline, etc.

Among the given options, anthracene and polyaniline are organic materials, so they are ruled out. Between Ge and CdS, CdS is the textbook example specifically cited for an inorganic compound semiconductor, distinguishing it from the elemental semiconductor Ge.

✓Final answer
$(b) CdS.$
CBSE 2023Set 55/1/11 markMCQ
$Q. When an intrinsic semiconductor is doped with a small amount of trivalent impurity, then :$
(a) its resistance increases.
(b) it becomes a p-type semiconductor.
(c) there will be more free electrons than holes in the semiconductor.
(d) dopant atoms become donor atoms.
›Reveal solutionSolution

Doping an intrinsic semiconductor with a trivalent impurity (like boron) creates p-type material, where holes are the majority carriers and resistance decreases — the correct option is (b).

The Core Idea: What Doping Actually Does

An intrinsic (pure) semiconductor has equal numbers of electrons and holes — both generated by thermal energy. When you add a trivalent impurity (3 valence electrons, e.g., boron, aluminium, gallium), each impurity atom replaces a silicon atom in the crystal lattice. Silicon has 4 valence electrons; the trivalent atom has only 3. That missing bond creates a hole — a vacancy that can accept an electron.

This is the key: trivalent impurities are acceptor atoms, not donors. They introduce extra holes into the crystal, making holes the majority carriers. The material becomes p-type (p for positive charge carriers).

Let’s walk through each option carefully.

Step-by-Step Analysis

1. Option (a): “its resistance increases”

Doping always adds charge carriers to the semiconductor. In an intrinsic semiconductor, the number of carriers is very small (on the order of $10^{10} \text{ cm}^{-3}$ for silicon at room temperature). Doping with even 1 part per million of a trivalent impurity increases the hole concentration dramatically — to about $10^{16} \text{ cm}^{-3}$ or more.

Conductivity $\sigma$ is given by:
$\sigma = q (n \mu_n + p \mu_p)$
where $q$ is the electronic charge, $n$ and $p$ are electron and hole concentrations, and $\mu_n$ , $\mu_p$ are their mobilities. Adding more carriers (holes) increases conductivity, so resistance decreases. Option (a) is false.

Watch out

A common mistake is to think that “impurities block current” — but in semiconductors, controlled impurities (dopants) are what make them conduct. Pure silicon is almost an insulator; doped silicon is a useful conductor.

2. Option (b): “it becomes a p-type semiconductor”

Exactly. With trivalent doping, holes outnumber electrons. The material is p-type. This is the definition. Option (b) is true.

3. Option (c): “there will be more free electrons than holes”

No. Trivalent atoms accept electrons, creating holes. The majority carriers are holes; electrons are minority carriers. In a p-type semiconductor at thermal equilibrium:
$p \gg n$
The product $n \cdot p = n_i^2$ (where $n_i$ is the intrinsic carrier concentration) still holds, but $p$ is large and $n$ is correspondingly small. Option (c) is false.

4. Option (d): “dopant atoms become donor atoms”

Donor atoms are pentavalent (5 valence electrons, e.g., phosphorus, arsenic) — they donate an extra electron. Trivalent atoms are acceptors — they accept an electron (creating a hole). Calling a trivalent impurity a “donor” is a direct contradiction of terminology. Option (d) is false.

Tip

A simple mnemonic: Trivalent → Three electrons → Acceptor (starts with A, like “add a hole”). Pentavalent → Provides an electron → Donor (starts with D, like “donates”).

Final Answer

✓Final answer
$The correct option is (b) — doping an intrinsic semiconductor with a trivalent impurity makes it a p-type semiconductor.$
CBSE 2023Set F1 markMCQ
$Q. The temperature coefficient of a semi-conductor is (A) positive (B) Negative (C) Zero (D) Infinity$
›Reveal solutionSolution

For semiconductors, resistance decreases with rising temperature → negative temperature coefficient of resistance.

In a semiconductor, raising the temperature releases more charge carriers (electrons and holes) by breaking covalent bonds. This large increase in carrier concentration outweighs the increased scattering, so the resistivity (and resistance) decreases as temperature increases. Since the temperature coefficient of resistance α is defined by $R = R_0(1+\alpha\,\Delta T)$ , a decrease of R with rising T means α is negative.

✓Final answer
$(B) Negative.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. As temperature increases the resistance of a semiconductor$
(1) increases
(2) decreases
(3) remains constant
(4) none of these
›Reveal solutionSolution

A semiconductor has a NEGATIVE temperature coefficient of resistance — its resistance falls as temperature rises.

In a semiconductor, raising the temperature breaks more covalent bonds, generating more free electrons and holes (charge carriers). Since conductivity depends on the number of available carriers, more carriers mean higher conductivity and hence LOWER resistance — opposite to the behaviour of a metallic conductor.

✓Final answer
$(2) decreases.$
CBSE 2023Set ANNUAL1 mark
$Q. What is the effect of rise in temperature on the conductivity of a semiconductor?$
›Reveal solutionSolution

A rise in temperature breaks more covalent bonds in the semiconductor, generating additional free electrons and holes, which raises conductivity.

As temperature increases, thermal energy breaks more covalent bonds, creating more electron-hole pairs. Since conductivity is proportional to the number of free charge carriers, this INCREASES the conductivity of the semiconductor (equivalently, its resistance decreases) — the opposite of the behaviour of a metal, whose conductivity decreases with temperature due to increased lattice scattering.

✓Final answer
$The conductivity of a semiconductor increases with a rise in temperature.$

Q.The energy levels of holes are: (A) in the valence band (B) in the conduction band (C) in the band gap but close to valence band (D) in the band gap but close to conduction band

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