How does the energy gap of an intrinsic semiconductor effectively change when doped with a (a) trivalent impur

Question

Q. How does the energy gap of an intrinsic semiconductor effectively change when doped with a

(a) trivalent impurity, and

(b) pentavalent impurity ? Justify your answer in each case.

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Doping introduces impurity energy levels inside the forbidden gap — an acceptor level just above the valence band (trivalent) or a donor level just below the conduction band (pentavalent). In both cases the energy gap is effectively reduced: free carriers can now be created with only $\sim 0{.}01$ – $0{.}05$ eV instead of the full gap $E_g$ ( $\approx 1{.}1$ eV for Si). The intrinsic band gap of the host crystal itself is not altered.

The Concept

The energy gap $E_g = E_c - E_v$ of an intrinsic semiconductor is the minimum energy needed to break a covalent bond and lift an electron from the valence band into the conduction band. In pure silicon this is about $1{.}1$ eV, so at room temperature only a tiny fraction of electrons make the jump — the pure crystal conducts poorly.

Doping does not rebuild the host lattice: the band edges $E_c$ and $E_v$ stay where they are. What doping does is insert new, localised energy levels inside the previously forbidden gap, very close to one band edge. Because carriers can now be created via these levels, the energy actually required drops from $E_g$ to the tiny impurity ionisation energy — the gap has effectively narrowed for carrier generation.

(a) Trivalent impurity (acceptor, p-type)

A trivalent atom (boron, aluminium, indium) substituting a silicon atom forms only three covalent bonds; the fourth bond is incomplete — a vacancy ready to accept an electron.
This introduces an acceptor energy level $E_a$ just above the valence band edge — about $0{.}01$ – $0{.}05$ eV above $E_v$ (for example $\approx 0{.}045$ eV for boron in silicon).
Even room-temperature thermal energy lets valence-band electrons jump into these acceptor levels, leaving mobile holes in the valence band.
Justification of the effective change: creating a hole now needs only $E_a - E_v \approx 0{.}01$ – $0{.}05$ eV instead of the full $E_g \approx 1{.}1$ eV. The energy gap has effectively decreased to the small acceptor ionisation energy.

(b) Pentavalent impurity (donor, n-type)

A pentavalent atom (phosphorus, arsenic, antimony) forms four covalent bonds with its silicon neighbours; the fifth valence electron is only weakly bound to the impurity ion.
This introduces a donor energy level $E_d$ just below the conduction band edge — about $0{.}01$ – $0{.}05$ eV below $E_c$ (for example $\approx 0{.}045$ eV for phosphorus in silicon).
Room-temperature thermal energy readily ionises the donors, releasing these electrons into the conduction band as free carriers. …

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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 46 on this concept.

CBSE 2026Set A1 markMCQ
$Q. The valency of impurity element to convert a germanium crystal into a p-type semiconductor is (A) 2 (B) 3 (C) 4 (D) 5$
›Reveal solutionSolution

Doping tetravalent germanium with a trivalent (valency-3) impurity creates holes, giving a p-type semiconductor.

Germanium is tetravalent (valency 4). To make it p-type, it is doped with a trivalent impurity — boron, aluminium, gallium or indium (valency 3). Such an atom forms only three covalent bonds with the surrounding germanium atoms, leaving one bond incomplete: a hole. …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Minority charge carriers in p-type material is:$
(a) Holes
(b) Electrons
(c) Both electrons and holes
(d) None of these
›Reveal solutionSolution

In a p-type semiconductor, holes are the majority carriers and electrons are the minority carriers.

A p-type semiconductor is formed by doping a tetravalent (Si/Ge) crystal with a trivalent impurity (like Boron), which creates an excess of holes. Holes therefore vastly outnumber the thermally generated free electrons, makin …

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CBSE 2026Set ANNUAL1 mark
$Q. Answer in one word/sentence: What is the controlled addition of impurities to intrinsic semiconductor is called?$
›Reveal solutionSolution

The deliberate, controlled addition of impurity atoms to a pure (intrinsic) semiconductor to alter its conductivity is called doping.

Adding a small, controlled amount of pentavalent impurity (like P, As) creates an n-type semiconductor, while adding a trivalent impurity (like B, In) creates a p-type semiconductor. This con …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): The electrical conductivity of a semiconductor increases on doping. Reason (R): Doping always increases the number of electrons in the semiconductor.$
(a) Both Assertion and Reason are true, and reason is the correct explanation
(b) Both Assertion and Reason are true, but the Reason is not the correct explanation
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
›Reveal solutionSolution

Doping does raise conductivity, but it does not always add electrons — trivalent (acceptor) doping adds holes instead.

Assertion: Doping a semiconductor with a suitable impurity increases the number of free charge carriers (electrons in n-type, holes in p-type), which increases conductivity $\sigma = n e \mu_e + p e \mu_h$ well beyond the intrinsic value. This is TRUE.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. For a p-type semiconductor, which of the following statement is true?$
(a) Holes are majority carriers and trivalent atoms are the dopants
(b) Holes are majority carriers and pentavalent atoms are the dopants
(c) Electrons are majority carriers and pentavalent atoms are the dopants
(d) Electrons are majority carriers and trivalent atoms are the dopants
›Reveal solutionSolution

p-type semiconductors are made by doping with trivalent (acceptor) atoms, which creates excess holes as majority carriers.

A p-type semiconductor is formed by doping a pure (intrinsic) semiconductor like Si or Ge with a trivalent impurity (Group 13 elements: B, Al, Ga, In). Each trivalent dopant atom has only 3 valence electrons, so when it bonds with 4 neighbouring Si/Ge atoms, one covalent bond is left incomplete — this vacancy is a HOLE, which behaves as a mobile positive charge carrier. Because these dopants "accept" an electron to complete the bond, they are called acceptor impurities. In the doped crystal, holes vastly outnumber the thermally generated electron …

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CBSE 2025Set 55/4/11 markMCQ
$Q. Assertion (A): An n-type semiconductor is not negatively charged. Reason (R): A neutral pentavalent impurity atom doped in an intrinsic (neutral) semiconductor donates its fifth unpaired electron to the crystal lattice and becomes a positive donor. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.$
›Reveal solutionSolution
An n-type semiconductor remains electrically neutral because every free electron donated by an impurity atom leaves behind a fixed positive ion in the lattice; the mobile electrons and immobile positive ions balance exactly. The correct option is (A).

Understanding Charge Neutrality in Doped Semiconductors

When we dope a pure semiconductor, we're adding impurity atoms that dramatically change the electrical properties—but does this make the material charged? The answer hinges on understanding what happens at the atomic level during doping.

An intrinsic (pure) semiconductor like silicon has a perfectly balanced number of protons and electrons. When we introduce pentavalent impurity atoms (like phosphorus or arsenic) into this crystal, each impurity atom brings five valence electrons but sits in a lattice position that requires only four for bonding.

Step-by-Step Analysis

Initial state of the dopant atom

A pentavalent impurity atom is electrically neutral when isolated: it has 15 protons (for phosphorus) and 15 electrons. The nuclear charge is $+15e$ and the total electronic charge is $-15e$ , giving zero net charge.

What happens during doping

When this atom is incorporated into the silicon lattice, four of its valence electrons participate in covalent bonds with neighboring silicon atoms. The fifth electron is loosely bound—so loosely that at room temperature, thermal energy is sufficient to free it into the conduction band.

The donor ion formation

Once the fifth electron is donated to the crystal, the impurity atom becomes a positive ion fixed in the lattice. This ion has $+15e$ of nuclear charge but now only $14e$ worth of electrons remaining bound to it, giving it a net charge of $+e$ . The Reason (R) correctly describes this process.

Tracking the overall charge balance

Here's the crucial point: the free electron that now roams the conduction band didn't appear from nowhere—it came from the donor atom. So for every free electron with charge $-e$ moving through the crystal, there's a corresponding immobile positive ion with charge $+e$ stuck in the lattice.

The charge accounting looks like:

$\text{Free electrons: } -Ne \quad + \quad \text{Donor ions: } +Ne \quad = \quad 0$
where $N$ is the number of dopant atoms.

Why the semiconductor stays neutral The n-type semiconductor has many more free electrons than an intrinsic semiconductor (that's why it conducts better), but it also has exactly the same number of positive donor ions. These charges cancel perfectly. The material as a whole remains electrically neutral.

Watch out

A common misconception is thinking "n-type means negative charge because of all those extra electrons." The "n" stands for negative charge carriers (electrons do the conducting), not for the material being negatively charged overall.

Evaluating Assertion (A) …
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CBSE 2025Set 55/5/11 markMCQ
$Q. Assertion (A): The impurities in p-type Si are not pentavalent atoms. Reason (R): The hole density in the valence band in a p-type semiconductor is almost equal to the acceptor density. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) Both A and R are false.$
›Reveal solutionSolution
The assertion is true (p-type Si uses trivalent, not pentavalent, impurities) and the reason is also true (hole density ≈ acceptor density in a p-type semiconductor), but the reason does not explain the assertion — they are independent facts. The correct option is (B).

Let’s unpack this step by step.

What makes a semiconductor p-type?

In silicon, the intrinsic carrier concentration is very low at room temperature. To create a p-type semiconductor, we deliberately introduce impurity atoms that have three valence electrons (trivalent), such as boron, aluminium, or gallium. These atoms are called acceptors because they accept an electron from the silicon lattice, leaving behind a hole in the valence band.

Pentavalent atoms (like phosphorus, arsenic) have five valence electrons and donate an extra electron — they make the material n-type, not p-type. So the assertion — “The impurities in p-type Si are not pentavalent atoms” — is absolutely correct.

What does the reason say?

The reason states: “The hole density in the valence band in a p-type semiconductor is almost equal to the acceptor density.”

This is a standard result from charge neutrality. In a p-type semiconductor at moderate temperatures (where intrinsic carriers are negligible compared to dopants), the number of holes $p$ is approximately equal to the acceptor concentration $N_A$ (assuming all acceptors are ionised). More precisely, $p \approx N_A$ when $N_A \gg n_i$ , where $n_i$ is the intrinsic carrier concentration. So the reason is also true.

Does the reason explain the assertion? …
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CBSE 2025Set ANNUAL1 mark
$8\times10^{13}\ \text{cm}^{-3}$
›Reveal solutionSolution

Electron concentration exceeds hole concentration, so electrons are the majority carriers — this defines an n-type semiconductor.

Given: electron concentration $n_e = 8\times10^{13}\ \text{cm}^{-3}$ , hole concentration $n_h = 5\times10^{12}\ \text{cm}^{-3}$ . Since $n_e \gg n_h$ (electrons are far more numerous than holes), electrons are the majority charge carriers and holes are the minority carriers. A semiconductor in which electrons are the majority carriers (typically produced by doping with a pentavalent impurity) is called an n-type s …

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CBSE 2025Set D1 markMCQ
$Q. When boron is mixed as impurity in silicon, then resultant matter is (A) n-type semiconductor (B) p-type semiconductor (C) n-type conductor (D) p-type conductor$
›Reveal solutionSolution

A trivalent dopant (boron) provides acceptor levels and creates holes, making silicon a p-type semiconductor.

Silicon is tetravalent. Boron has only three valence electrons, so when a boron atom replaces a silicon atom, one of the four covalent bonds is left short of an electron — creating a hole (an acceptor site).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Silicon is a semi-conductor. By adding a small amount of arsenic to it, its conductivity$
(a) will increase
(b) will decrease
(c) will remain unchanged
(d) will become zero
›Reveal solutionSolution

Arsenic has 5 valence electrons versus silicon's 4, so each arsenic atom substituted into the silicon lattice contributes one extra, loosely-bound electron that becomes a free charge carrier, sharply increasing conductivity (n-type doping).

Pure (intrinsic) silicon is a poor conductor at room temperature because it has very few free charge carriers.

When a small amount of arsenic (a pentavalent, Group 15 element) is added, each As atom forms four covalent bonds with neighbouring Si atoms (using 4 of its 5 valence electrons); the 5th electron is left loosely bound and easily becomes a free conduction electron at room temperature.

…

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$Q. Which type of semi-conductor is made by mixing Indium with Germanium ?$
(a) n-type
(b) p-type
(c) both (A) and (B)
(d) none of these
›Reveal solutionSolution

Indium is a trivalent (group-13) impurity, so doping it into germanium creates holes — a p-type semiconductor.

Germanium is a tetravalent (group-14) element. Doping it with a trivalent element such as indium (which has only 3 valence electrons) means each indium atom can form only 3 covalent bonds with its germanium neighbours; the 4th bond is left with a missing electron — a hole. These holes act as majority charge carriers, ma …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. On doping the conductivity of semi-conductor :$
(a) increases.
(b) remains same.
(c) decreases.
(d) none of these.
›Reveal solutionSolution

Doping adds impurity atoms that contribute extra free charge carriers (electrons or holes), which increases the conductivity of a semiconductor by several orders of magnitude.

A pure (intrinsic) semiconductor has very few free charge carriers at room temperature, so its conductivity is low. Doping introduces pentavalent (donor) or trivalent (acceptor) impurity atoms into the crystal lattice, creating extra free electrons (n-type) or holes (p-type). These additional charge carriers dramatically increase the number of car …

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Q.How does the energy gap of an intrinsic semiconductor effectively change when doped with a

Concept understanding — Extrinsic Semiconductor (Doping)

The Concept

(a) Trivalent impurity (acceptor, p-type)

(b) Pentavalent impurity (donor, n-type)

Understanding Charge Neutrality in Doped Semiconductors

Step-by-Step Analysis

Q.How does the energy gap of an intrinsic semiconductor effectively change when doped with a

The Concept

(a) Trivalent impurity (acceptor, p-type)

(b) Pentavalent impurity (donor, n-type)

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