Two statements are given — one labelled Assertion (A) a — CBSE Class 12 Physics 2023 PYQ

Question

Q. Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (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) Assertion (A) is false and Reason (R) is also false. Assertion (A) : In an n-type semiconductor, the number density of electrons is greater than the number density of holes but the crystal maintains an overall charge neutrality. Reason (R) : The charge of electrons donated by donor atoms is just equal and opposite to that of the ionised donor.

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Concept understanding — N Type Semiconductor Doping

N-Type Semiconductor Doping: From Intuition to Precision

Imagine you have a pure silicon crystal. Silicon has four valence electrons, and in the crystal, every atom shares one electron with each of its four neighbours — forming perfect covalent bonds. Every electron is tied up in a bond. There are no free electrons to carry current. Pure silicon at room temperature is almost an insulator.

Now, what if you could sneak in an extra electron that has no bond to belong to? That extra electron would be free to wander through the crystal, carrying current. That is exactly what n-type doping does.

The Intuition: Adding a "Giver" Atom

Take a tiny amount of phosphorus — an element from Group V of the periodic table. Phosphorus has five valence electrons. When a phosphorus atom replaces a silicon atom in the crystal lattice, four of its electrons form normal bonds with the four neighbouring silicon atoms. The fifth electron has no partner. It is loosely held by the phosphorus nucleus, but at room temperature, thermal energy is enough to kick it free into the crystal's conduction band.

That freed electron can now move under an electric field. The phosphorus atom, having lost an electron, becomes a positively charged ion fixed in the lattice — it does not move. But the electron is mobile.

Note

The name "n-type" comes from negative — because the majority charge carriers are negatively charged electrons.

The Precise Statement

N-type semiconductor doping is the process of introducing impurity atoms from Group V (donors) into an intrinsic semiconductor (like silicon or germanium). Each donor atom contributes one extra electron to the crystal, creating a large number of free electrons that become the majority charge carriers. The donor atoms themselves become immobile positive ions.

The key result: the electron concentration $n$ becomes much larger than the hole concentration $p$ . In an n-type semiconductor at thermal equilibrium:

n \gg p

And if all donor atoms are ionised (which is true at room temperature for typical doping levels), the electron concentration is approximately equal to the donor concentration $N_D$ :

n \approx N_D

n \approx N_D \quad \text{(for n-type at room temperature)}

What Happens to Holes?

You might ask: if we add extra electrons, do holes still exist? Yes — but they are now the minority carriers. The law of mass action still holds:

n \cdot p = n_i^2

where $n_i$ is the intrinsic carrier concentration (about $1.5 \times 10^{10} \text{ cm}^{-3}$ for silicon at 300 K). So if $n \approx 10^{16} \text{ cm}^{-3}$ , then:

p = \frac{n_i^2}{n} \approx \frac{(1.5 \times 10^{10})^2}{10^{16}} = 2.25 \times 10^4 \text{ cm}^{-3}

That is a tiny number compared to the electron concentration. The material conducts almost entirely via electrons.

Common Donor Elements

Element	Group	Valence electrons	Notes
Phosphorus (P)	V	5	Most common for silicon

Why this formula?

Why N-Type Semiconductor Doping Works: The Physics Behind the Formula

When you dope a pure (intrinsic) semiconductor like silicon with a pentavalent impurity — an element from Group V of the periodic table, such as phosphorus, arsenic, or antimony — you create an n-type semiconductor. The "n" stands for negative, because the majority charge carriers are negatively charged electrons.

The key formula that governs n-type doping is:

n \approx N_D

where $n$ is the concentration of free electrons in the conduction band, and $N_D$ is the concentration of donor atoms introduced.

Let's understand why this simple relation holds, step by step.

Step 1: What happens at the atomic level?

Silicon has four valence electrons. It forms four covalent bonds with neighbouring silicon atoms, achieving a stable octet configuration. Now, introduce a phosphorus atom — it has five valence electrons.

Four of phosphorus's electrons form normal covalent bonds with adjacent silicon atoms. The fifth electron has no place in the bonding structure. It is only very weakly bound to the phosphorus nucleus — the binding energy is tiny, about 0.045 eV for phosphorus in silicon (compared to the 1.1 eV band gap of silicon).

Note

This weak binding means that at room temperature (thermal energy ≈ 0.026 eV), almost all of these fifth electrons get enough energy to break free from their donor atoms and become free electrons in the conduction band.

Each phosphorus atom that loses its extra electron becomes a positively charged ion (fixed in the crystal lattice), but the freed electron is mobile and contributes to electrical conduction.

Step 2: Why $n \approx N_D$ and not exactly $N_D$ ?

The reasoning is straightforward:

Every donor atom contributes one free electron when ionised.
At room temperature, the ionisation is nearly complete — the donor energy level lies just below the conduction band edge (about 0.045 eV), so thermal energy easily kicks the electron into the conduction band.
Therefore, the number of free electrons $n$ is approximately equal to the number of donor atoms $N_D$ .

But why "approximately" and not exactly? Two reasons:

Intrinsic carriers still exist: Even in doped silicon, a small number of electron-hole pairs are thermally generated. The intrinsic carrier concentration $n_i$ (about $1.5 \times 10^{10} \text{ cm}^{-3}$ for silicon at 300 K) adds to the electron count. However, for typical doping levels ( $N_D \approx 10^{15}$ to $10^{18} \text{ cm}^{-3}$ ), $n_i$ is negligible — so $n \approx N_D$ is an excellent approximation.
Incomplete ionisation at very low temperatures: At extremely low temperatures (near 0 K), some donor atoms may not ionise. But for all practical operating temperatures of electronic devices, ionisation is essentially 100%.

Watch out

A common mistake is to think that $n = N_D$ exactly. The correct statement is $n \approx N_D$ because the intrinsic carrier concentration $n_i$ is always present, though negligible for moderate to heavy doping.

Step 3: What about the hole concentration?

In an n-type semiconductor, electrons are the majority carriers, and holes are the minority carriers. The product of electron and hole concentrations is always constant for a given semiconductor at a fixed temperature — this is the law of mass action:

n \cdot p = n_i^2

Since $n \approx N_D$ , we get:

$p \approx \frac{n_i^2}{N_D}$ …

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An n-type semiconductor remains electrically neutral because the extra electrons from donor atoms are balanced by an equal number of positively ionised donor ions fixed in the crystal lattice; the assertion is true but the reason, while true, doesn't explain why $n \gg p$ coexists with neutrality. The correct answer is (B).

The heart of this question lies in understanding what charge neutrality means in a doped semiconductor and why having more electrons than holes doesn't violate it.

Why n-type semiconductors stay neutral

When we dope a pure silicon crystal with pentavalent atoms (phosphorus, arsenic), each donor atom contributes one extra electron to the conduction band and becomes a positively charged ion ( $\text{P}^+$ , $\text{As}^+$ ) fixed in the lattice. The crystal as a whole must remain electrically neutral because we haven't added or removed net charge—we've only rearranged internal charges.

The neutrality condition for an n-type semiconductor is:

n + N_A^- = p + N_D^+

where $n$ is the electron density, $p$ is the hole density, $N_D^+$ is the density of ionised donors, and $N_A^-$ is the density of ionised acceptors (zero in pure n-type). This simplifies to:

n = p + N_D^+

At room temperature, nearly all donor atoms are ionised, so $N_D^+ \approx N_D$ (the doping concentration). Since $n \approx N_D$ and the intrinsic carrier product $np = n_i^2$ gives $p = n_i^2/n \ll n$ , we have many more electrons than holes—but the positive donor ions exactly balance the extra electrons.

Evaluating the statements

Assertion (A): Claims that $n > p$ while the crystal stays neutral. This is true. The majority carriers (electrons) far outnumber minority carriers (holes), yet overall neutrality holds because of the ionised donors.

Reason (R): States that the charge of donated electrons equals and opposes the charge of ionised donors. This is true as a standalone fact—each donor gives one electron and becomes $+e$ , so the charges are equal and opposite. …

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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 29 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): The majority charge carriers in n-type semiconductor are electrons. Reason (R): An n-type semiconductor is formed by doping an intrinsic semiconductor with trivalent impurity.$
(i) Both A and R are correct and R is correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) A is correct but R is incorrect.
(iv) Both A and R are incorrect.
›Reveal solutionSolution

Assertion is correct, but n-type is formed with PENTAVALENT (not trivalent) doping — Reason is false.

Assertion is correct: in an n-type semiconductor the majority charge carriers are indeed electrons. Reason is incorrect: an n-type semiconductor is formed by doping an intrinsic (pure) semiconductor with a pentavalent impurity (e.g. P, As, Sb), not a t …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In n-type semiconductor what kind of impurity is found?$
(a) Valency of three
(b) Valency of five
(c) Valency of two
(d) Valency of one
›Reveal solutionSolution

n-type doping uses pentavalent (5-valence-electron) impurity atoms, which contribute one extra free electron each.

Pure silicon or germanium is tetravalent (4 valence electrons), each atom forming 4 covalent bonds with its neighbours in the crystal lattice. To make an n-type semiconductor, the crystal is doped with a PENTAVALENT impurity (valency 5, e.g. phosphorus, arsenic, antimony). Four of the impurity atom's five valence electrons form covalent bonds with the surrounding silicon atoms, and the fifth electron is left loosely bound - easily freed to become a mobile cha …

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CBSE 2026Set ANNUAL1 mark
$Q. What are majority charge carriers in a n-type semiconductor?$
›Reveal solutionSolution

n-type doping (pentavalent impurity) contributes extra free electrons, which become the majority (dominant) charge carriers.

An n-type semiconductor is created by doping a tetravalent crystal (Si/Ge) with a pentavalent donor impurity (e.g. phosphorus). Each donor atom contributes one extra, loosely-bound electron that easily becomes a free (mobile) charge carrier, vastly outnumbering the small number of electron-hole pairs that arise from thermal generation in …

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CBSE 2026Set ANNUAL1 mark
$Q. A pure semiconductor when doped with a pentavalent impurity becomes a p-type semiconductor. (True/False)$
›Reveal solutionSolution

Doping with a pentavalent (group 15) impurity such as phosphorus or arsenic supplies extra free electrons and gives an n-type semiconductor; a p-type semiconductor needs a trivalent dopant instead.

A pentavalent atom (5 valence electrons) substituting for a tetravalent silicon or germanium atom in the crystal lattice uses 4 of its electrons to form covalent bonds with its four neighbouring lattice atoms; the 5th electron is only loosely bound and readily becomes a free (conduction) electron. Because it effectively 'donates' this extra electron, the pentavalent impurity is called a donor, and the resulting semiconductor has electrons as majority charge carriers — this is an n-type semiconductor.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Why is the conductivity of n-type semiconductor greater than that of p-type semiconductor even if both of these have the same level of doping?$
›Reveal solutionSolution

Electrons are inherently more mobile than holes in the semiconductor lattice, giving n-type higher conductivity at equal doping.

Conductivity in an extrinsic semiconductor is $\sigma = ne\mu_e$ (n-type, majority carriers are electrons) or $\sigma = pe\mu_h$ (p-type, majority carriers are holes), where $\mu_e$ and $\mu_h$ are the electron and hole mobilities respectively. Physically, electron conduction occurs by electrons moving relatively freely through the conduction band, while hole conduction occurs by valence electrons hopping from one bond to fill an adjacent vacancy (an indirect, effectively slower mechanism), so electron mobility is inherently greater than hole mobility ( $\mu_e > \mu_h$ ) in typical semiconductors like silicon and germanium. So even at the same doping level (s …

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CBSE 2026Set ANNUAL1 mark
$Q. Who are the majority charge carrier in P-type semiconductor ?$
›Reveal solutionSolution

Holes are the majority carriers in P-type material.

A P-type semiconductor is formed by doping a pure semiconductor with a trivalent (acceptor) impurity such as boron. Each acceptor atom creates a hole (a vacancy that behaves as a positive charge carrier). This produces a large number of holes and only a few thermally-generated elect …

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CBSE 2025Set D1 markMCQ
$Q. In n-type semiconductor the minority charge carrier is/are (A) electrons (B) holes (C) electron and hole (D) none of these$
›Reveal solutionSolution

n-type doping (pentavalent) gives excess electrons as majority carriers; the few thermally generated holes are the minority carriers.

An n-type semiconductor is made by doping with a pentavalent element (e.g. phosphorus), which donates free electrons. Electrons are therefore the majority charge carriers.

…

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CBSE 2025Set A1 markMCQ
$Q. Majority charge carriers in n-type semiconductors are —$
(i) Electrons
(ii) Holes
(iii) Neutrons
(iv) Moving ion
›Reveal solutionSolution

In n-type semiconductors, electrons are majority carriers because pentavalent doping atoms donate free electrons.

An n-type semiconductor is formed by doping a pure (intrinsic) semiconductor such as Silicon or Germanium with a pentavalent impurity (like Phosphorus, Arsenic or Antimony), which has 5 valence electrons. Four of these electrons form covalent bonds with the neighbouring Si/Ge atoms, and the fifth electron is loosely bound and becomes free to move in the crystal even at room temperature. Each donor atom thus contributes one free electron without creating a corresponding hole (the d …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The majority charge carriers in n-type semiconductors are$
(a) electrons
(b) holes
(c) both electrons and holes
(d) ions
›Reveal solutionSolution

Doping a semiconductor with a pentavalent (group V) impurity donates one extra free electron per dopant atom, making electrons the majority charge carriers in n-type material.

When a tetravalent semiconductor (Si or Ge) is doped with a pentavalent impurity (e.g. P, As, Sb), four of the impurity atom's five valence electrons form covalent bonds with neighbouring silicon atoms, leaving the fifth electron loosely bound and free to conduct at room temperature. The …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. n-type semiconductor is obtained by doping the intrinsic semiconductor with$
(a) trivalent material
(b) conductor
(c) insulator
(d) pentavalent material
›Reveal solutionSolution

An n-type semiconductor has electrons as the majority charge carrier; this is achieved by doping a tetravalent crystal (Si/Ge) with an atom that has one more valence electron than the host, i.e. a pentavalent (Group V) impurity such as P, As or Sb.

Reasoning

Silicon/Germanium atoms are tetravalent — each forms 4 covalent bonds with its neighbours. When a pentavalent atom (5 valence electrons) replaces a host atom in the lattice, 4 of its electrons form the required covalent bonds and the 5th electron is left loosely bound, easily freed into the conduction band by thermal energy. This creates a large population of f …

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CBSE 2023Set 55/1/11 markMCQ
$E_C$
(a) $10\ \text{eV}$
(b) $1\ \text{eV}$
(c) $0.1\ \text{eV}$
(d) $0.01\ \text{eV}$
›Reveal solutionSolution
The donor level in n-type silicon lies very close to the conduction band edge — the ionisation energy is about 0.01 eV for typical donors like phosphorus, making option (d) correct.

The question asks about the energy gap between the bottom of the conduction band $E_C$ and the donor energy level $E_D$ in n-type silicon. This gap is the donor ionisation energy — the energy needed to free the extra electron from the donor atom so it can move into the conduction band and contribute to current.

Why is this gap so small? In an n-type semiconductor, we deliberately add impurity atoms (donors) that have one more valence electron than the host atoms. For silicon (group 14), a typical donor is phosphorus (group 15). The extra electron is only weakly bound to the donor nucleus because the silicon crystal’s high dielectric constant (about 11.7) screens the Coulomb attraction, and the electron’s effective mass is much smaller than the free‑electron mass. This makes the donor orbit large and its binding energy tiny — typically a few hundredths of an electronvolt.

Let’s work through the reasoning step by step.

Recall the scale of the band gap in silicon.

The fundamental band gap of silicon (between the valence band $E_V$ and the conduction band $E_C$ ) is about $1.1\ \text{eV}$ at room temperature. That’s the energy needed to break a covalent bond and create an electron‑hole pair. Options (a) $10\ \text{eV}$ and (b) $1\ \text{eV}$ are comparable to or larger than the band gap itself — far too large for a shallow donor level.

Understand what a donor level is.

A donor atom replaces a silicon atom in the lattice. Its fifth electron is not needed for bonding and orbits the donor ion, much like the electron in a hydrogen atom — but inside a crystal. The binding energy of this “hydrogen‑like” state is given by a modified Rydberg formula:

$E_D = \frac{m^* e^4}{2 (4\pi \epsilon_0 \epsilon_r)^2 \hbar^2} \cdot \frac{1}{n^2}$
For the ground state ( $n=1$ ), this becomes:
$E_D = \frac{m^*}{m_0} \cdot \frac{1}{\epsilon_r^2} \times 13.6\ \text{eV}$
where $13.6\ \text{eV}$ is the hydrogen ground‑state energy, $m^*$ is the electron’s effective mass (about $0.26\,m_0$ for silicon), and $\epsilon_r \approx 11.7$ for silicon.

Plug in the numbers. $E_D \approx \frac{0.26}{(11.7)^2} \times 13.6\ \text{eV} \approx \frac{0.26}{136.9} \times 13.6\ \text{eV} \approx 0.0019 \times 13.6\ \text{eV} \approx 0.026\ \text{eV}$ …
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CBSE 2023Set 55/4/11 markMCQ
$Q. Two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false. Assertion (A) : In an n-type semiconductor, the number density of electrons is greater than the number density of holes but the crystal maintains an overall charge neutrality. Reason (R) : The charge of electrons donated by donor atoms is just equal and opposite to that of the ionised donor.$
›Reveal solutionSolution

An n-type semiconductor remains electrically neutral because the extra electrons from donor atoms are balanced by an equal number of positively ionised donor ions fixed in the crystal lattice; the assertion is true but the reason, while true, doesn't explain why $n \gg p$ coexists with neutrality. The correct answer is (B).

The heart of this question lies in understanding what charge neutrality means in a doped semiconductor and why having more electrons than holes doesn't violate it.

Why n-type semiconductors stay neutral

When we dope a pure silicon crystal with pentavalent atoms (phosphorus, arsenic), each donor atom contributes one extra electron to the conduction band and becomes a positively charged ion ( $\text{P}^+$ , $\text{As}^+$ ) fixed in the lattice. The crystal as a whole must remain electrically neutral because we haven't added or removed net charge—we've only rearranged internal charges.

The neutrality condition for an n-type semiconductor is:
$n + N_A^- = p + N_D^+$
where $n$ is the electron density, $p$ is the hole density, $N_D^+$ is the density of ionised donors, and $N_A^-$ is the density of ionised acceptors (zero in pure n-type). This simplifies to:
$n = p + N_D^+$
At room temperature, nearly all donor atoms are ionised, so $N_D^+ \approx N_D$ (the doping concentration). Since $n \approx N_D$ and the intrinsic carrier product $np = n_i^2$ gives $p = n_i^2/n \ll n$ , we have many more electrons than holes—but the positive donor ions exactly balance the extra electrons.

Evaluating the statements

Assertion (A): Claims that $n > p$ while the crystal stays neutral. This is true. The majority carriers (electrons) far outnumber minority carriers (holes), yet overall neutrality holds because of the ionised donors.

Reason (R): States that the charge of donated electrons equals and opposes the charge of ionised donors. This is true as a standalone fact—each donor gives one electron and becomes $+e$ , so the charges are equal and opposite. …
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Why n-type semiconductors stay neutral

Evaluating the statements

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