Q.Doping a semiconductor results in

Showing the 12 most recent of 29 on this concept.

1. 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 trivalent one — doping with a trivalent impurity (e.g. B, Al, In) produces a p-type semiconductor instead.

   ✓Final answer

   (iii) A is correct but R is incorrect.

2. 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 charge carrier. This gives an excess of free electrons (negative charge carriers), hence the name 'n-type' (n for negative). Trivalent impurities (valency 3), by contrast, produce p-type semiconductors with excess holes.

   ✓Final answer

   (b) Valency of five.

3. 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 the intrinsic material. So in n-type material, free ELECTRONS are the majority charge carriers, while holes (present only from thermal generation) are the minority carriers.

   ✓Final answer

   Free electrons.

4. 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.

   A p-type semiconductor is instead obtained by doping with a trivalent (group 13) impurity such as boron, aluminium or indium, which creates a deficiency of one bonding electron (a 'hole') at each impurity site — these are called acceptors, and holes become the majority carriers.

   ✓Final answer

   False. A pentavalent dopant produces an n-type semiconductor (majority carriers = electrons), not a p-type one.

5. 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 (same carrier concentration $n=p$), an n-type semiconductor has higher conductivity than a p-type semiconductor, because $\mu_e>\mu_h$.

   ✓Final answer

   Because electron mobility ($\mu_e$) exceeds hole mobility ($\mu_h$) in a semiconductor, so at the same doping level, n-type conductivity ($ne\mu_e$) is greater than p-type conductivity ($pe\mu_h$).

6. 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 electrons.

   Therefore, in a P-type semiconductor, holes are the majority charge carriers and electrons are the minority carriers.

   ✓Final answer

   Holes.

7. 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.

   A small number of electron–hole pairs still form by thermal generation, producing a few holes. Since electrons dominate, these holes are the minority charge carriers.

   ✓Final answer

   (B) holes.

8. 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 donor atom becomes a fixed positive ion). Because of this, electrons vastly outnumber the thermally generated holes, so electrons are the majority charge carriers and holes are the minority charge carriers in an n-type semiconductor.

   ✓Final answer

   (i) Electrons.

9. 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. These donated free electrons vastly outnumber the thermally generated holes, so electrons are the majority carriers (holes are the minority carriers) in an n-type semiconductor.

   ✓Final answer

   (a) electrons.

10. 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 free (donor) electrons — an n-type semiconductor. (Doping with a trivalent atom instead would create holes, giving a p-type semiconductor; conductors and insulators are not "doped" onto an intrinsic semiconductor lattice at all.)

    ✓Final answer

    (d) pentavalent material

11. CBSE 2023Set 55/1/11 markMCQ
    Q.In the energy-band diagram of n-type Si, the gap between the bottom of the conduction band $E_C$ and the donor energy level $E_D$ is of the order of :

    (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.

    1. 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.

    2. 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.

    3. Plug in the numbers.

    $E_D \approx \frac{0.26}{(11.7)^2} \times 13.6\ \text{eV} \approx \frac{0.26}{136.9} \times 13.6\ \text{eV} \approx 0.0019 \times 13.6\ \text{eV} \approx 0.026\ \text{eV}$

    This is about $0.026\ \text{eV}$ — roughly $0.03\ \text{eV}$. For common donors like phosphorus, arsenic, and antimony in silicon, the measured ionisation energies are indeed in the range $0.044\ \text{eV}$ to $0.054\ \text{eV}$ (slightly higher due to details like central‑cell corrections). The order of magnitude is clearly $0.01\ \text{eV}$, not $0.1\ \text{eV}$ or larger.

    Watch out

    A common mistake is to confuse the donor ionisation energy with the band gap itself. The band gap of Si is $\sim 1.1\ \text{eV}$, but the donor level sits just below $E_C$ by only about $0.01\ \text{eV}$ — that’s two orders of magnitude smaller. Option (c) $0.1\ \text{eV}$ is still an order of magnitude too large for shallow donors in Si.

    4. Compare with the given options.
       • $10\ \text{eV}$ — far larger than the band gap; impossible.
       • $1\ \text{eV}$ — comparable to the band gap; that would be a deep level, not a shallow donor.
       • $0.1\ \text{eV}$ — still too large; this is more typical of donors in wider‑gap semiconductors like GaAs.
       • $0.01\ \text{eV}$ — matches the calculated and measured order of magnitude for shallow donors in silicon.

    Tip

    A quick memory aid: for silicon, shallow donor levels are about 0.01 eV below $E_C$, and shallow acceptor levels are about 0.01 eV above $E_V$. This tiny energy is why even at room temperature ($k_B T \approx 0.026\ \text{eV}$), most donors are ionised and contribute free electrons.

    ✓Final answer

    The gap is of the order of 0.01 eV, so the correct option is (d).

12. 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.

    But does (R) explain (A)? Here's the subtlety. Reason (R) tells us that electrons and donor ions balance each other, which is part of the neutrality equation. However, it doesn't address why we can have $n \gg p$ and still maintain neutrality—it doesn't mention that holes are minority carriers or that the donor ions are immobile and part of the lattice. The reason is a correct statement about local charge balance but misses the full picture needed to explain the assertion.

    Watch out

    A common mistake is thinking (R) fully explains (A) because both mention neutrality. The reason describes one piece of the charge-balance puzzle (electrons ↔ donors) but doesn't explain the coexistence of unequal carrier densities with overall neutrality—that requires understanding the role of immobile ionised donors in the lattice.

    ✓Final answer

    Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). The correct option is (B).