CUET UG Physics Booster Test 3- Extrinsic and Doping Concepts -

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CUET UG Physics Booster Test 3- Extrinsic and Doping Concepts

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 3- Extrinsic and Doping Concepts

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QUESTION 1 OF 20

The necessity to improve the conductivity of intrinsic semiconductors by adding impurities

Answer Review

1 The necessity to improve the conductivity of intrinsic semiconductors by adding impurities

A depends on specific properties of metals

B falls as 1/r² at large distances

C is dictated by the fact that the number of thermally generated carriers at room temperature is too low for electronic devices

D independent of the energy gap between conduction and valence bands

💡 Short Explanation

Intrinsic semiconductors have few charge carriers. Conductivity is low at room temperature. Doping increases carrier concentration.

✅ Correct Answer Explanation

The number of thermally generated electrons and holes in an intrinsic semiconductor is too small for efficient operation of electronic devices. Therefore, impurities are added to increase conductivity.

❓ Why Other Options Are Incorrect

Option A → Metal properties are not the reason.
Option B → Unrelated physical law.
Option D → Conductivity depends strongly on the energy gap.

🎯 Strategy

Used

Conceptual Understanding

Final Logic:

Low intrinsic carrier concentration necessitates doping.

🧠 Memory Trick

"Few Carriers → Add Dopants."

📚 Semiconductor Electronics – Need for Doping

2 For an intrinsic semiconductor properly doped to form an n-type material, if N_D is the donor concentration, the relation dictating the manifold increase in electrons will be

A nₑ ≈ N_D

B nₑ ≈ nᵢ

C nₑ ≈ nₕ

D nₑ − N_D ≈ nᵢ²

💡 Short Explanation

Donor atoms contribute free electrons. Electron concentration becomes approximately equal to donor concentration.

✅ Correct Answer Explanation

In an n-type semiconductor, each donor atom contributes approximately one free electron. Therefore, the electron concentration is nearly equal to the donor concentration: \(n_{e}\approx N_{D}\)

❓ Why Other Options Are Incorrect

Option B → True only for intrinsic semiconductors.
Option C → Majority and minority carriers are not equal.
Option D → Incorrect relation.

🎯 Strategy

Used

Formula Recall

Final Logic:

Donor concentration determines majority electrons.

🧠 Memory Trick

"One Donor → One Electron."

📚 Semiconductor Electronics – n-Type Semiconductor

3 Match List I with List II regarding dopant atomic substitution

List I	List II
P. Si/Ge Group in Periodic Table	1. Third Group
Q. Pentavalent dopants Group	2. Few parts per million (ppm)
R. Trivalent dopants Group	3. Fifth Group
S. Doping concentration typical limit	4. Fourth Group

A P-4, Q-3, R-1, S-2

B P-3, Q-4, R-2, S-1

C P-4, Q-1, R-3, S-2

D P-1, Q-2, R-4, S-3

💡 Short Explanation

Si and Ge → Group IV. Pentavalent → Group V. Trivalent → Group III. Doping concentration → ppm level.

✅ Correct Answer Explanation

Correct matching: P → Fourth Group (4) Q → Fifth Group (3) R → Third Group (1) S → Few ppm (2)

❓ Why Other Options Are Incorrect

Incorrect group assignments.

🎯 Strategy

Used

Matching Concepts

Final Logic:

Only Option A matches correctly.

🧠 Memory Trick

"IV Host, V Donor, III Acceptor."

📚 Semiconductor Electronics – Dopants

4 Incorrect statement about dopants in the lattice structure

A They must occupy a very few of the original semiconductor atom sites.

B A necessary condition is that sizes of dopant and semiconductor atoms should be nearly the same.

C A trivalent dopant atom provides four valence electrons to seamlessly form bonds with Si neighbors.

D The dopant must be such that it does not distort the original pure semiconductor lattice.

💡 Short Explanation

Trivalent atoms possess only three valence electrons. One bond remains incomplete.

✅ Correct Answer Explanation

Trivalent dopants such as Boron and Aluminum have only three valence electrons and therefore cannot provide four electrons for complete bonding.

❓ Why Other Options Are Incorrect

A, B, D → Correct requirements of doping.

🎯 Strategy

Used

Elimination

Final Logic:

Trivalent dopants cannot form four complete covalent bonds.

🧠 Memory Trick

"Trivalent Means Three."

📚 Semiconductor Electronics – Acceptor Impurities

5 In maintaining the lattice integrity of Si or Ge during doping

A each dopant has an equal charge for certain values of capacitances only

B each dopant must satisfy the necessary condition that its size should be nearly the same as the host atoms

C each impurity atom expands the lattice spacing to 5.43 Å inherently

D each dopant has different charge leading to macroscopic lattice fractures

💡 Short Explanation

Similar atomic size minimizes lattice distortion. Crystal structure remains intact.

✅ Correct Answer Explanation

The impurity atom should have a size close to that of the host semiconductor atom so that it can substitute into the lattice without causing major distortions.

❓ Why Other Options Are Incorrect

A, C, D → Not valid conditions for doping.

🎯 Strategy

Used

Direct Recall

Final Logic:

Size similarity preserves lattice integrity.

🧠 Memory Trick

"Same Size, Stable Lattice."

📚 Semiconductor Electronics – Doping Constraints

6 Suppose \(n_{i}=1.5\times {10}^{16} m^{-3}\)and a Si crystal is doped with pentavalent As such that \(N_{D}=5\times {10}^{22} m^{-3}\). What will be the value of minority carrier concentration \(\left(n_{h}\right)\)?

A \(4.5\times {10}^{9} m^{-3}\)

B \(2.25\times {10}^{32} m^{-3}\)

C \(3.0\times {10}^{16} m^{-3}\)

D \(7.5\times {10}^{22} m^{-3}\)

💡 Short Explanation

Use: \(n_{e}n_{h}=n_{i}^{2}\) and \(n_{e}\approx N_{D}\)

✅ Correct Answer Explanation

\(n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{{\left(1.5\times {10}^{16}\right)}^{2}}{5\times {10}^{22}}=\frac{2.25\times {10}^{32}}{5\times {10}^{22}}=4.5\times {10}^{9} m^{-3}\)

❓ Why Other Options Are Incorrect

B → Equals \(n_{i}^{2}\).
C → Not calculated value.
D → Donor concentration.

🎯 Strategy

Used

Mass Action Law

Final Logic:

Minority carriers decrease greatly in n-type semiconductors.

🧠 Memory Trick

"nₑnₕ = nᵢ²."

📚 Semiconductor Electronics – Mass Action Law

7 Correct statements about pentavalent donor impurities
1. They include Arsenic, Antimony, and Phosphorus.
2. The number of electrons made available depends strongly upon the doping level.
3. The number of electrons provided by dopants depends strongly on the ambient temperature.
4. They donate one extra electron for conduction.

A (A), (B), (D)

B (B), (C), (D)

C (A), (C), (D)

D (A), (B), (C), (D)

💡 Short Explanation

Pentavalent dopants donate electrons. More dopants mean more free electrons.

✅ Correct Answer Explanation

Arsenic, Antimony, and Phosphorus are common donor impurities. Each contributes one extra electron, and the total number of available electrons depends mainly on donor concentration. Statement C is incorrect because donor contribution depends primarily on doping level, not strongly on temperature at room temperature.

❓ Why Other Options Are Incorrect

Include incorrect statement C.

🎯 Strategy

Used

Statement Evaluation

Final Logic:

A, B, and D are correct.

🧠 Memory Trick

"As, Sb, P → Donate One Electron."

📚 Semiconductor Electronics – Donor Impurities

8 The ionization of the fifth electron is so easy because it is seen as part of the _____ and requires a very small _____:

A Valence band, Kinetic energy

B Covalent bond, Potential difference

C Effective core, Ionization energy

D Conduction band, Recombination rate

💡 Short Explanation

Fifth electron is weakly bound. Very small ionization energy is needed.

✅ Correct Answer Explanation

The fifth electron is only loosely attached to the donor atom and behaves as though it belongs to the effective core. Hence, only a small ionization energy is needed to free it.

❓ Why Other Options Are Incorrect

Incorrect physical interpretation.

🎯 Strategy

Used

Direct Recall

Final Logic:

Weak binding → low ionization energy.

🧠 Memory Trick

"Fifth Electron = Easily Freed."

📚 Semiconductor Electronics – Donor Electron

9 Majority electrons in n-type semiconductor statements
1. Their total number is due to electrons contributed by donors and intrinsically generated ones.
2. Their presence increases the rate of recombination of holes.
3. \(n_{e}>>n_{h}\)describes their dominance.
4. They are primarily generated through thermal agitation of the host lattice at absolute zero.

A (A), (B), (C)

B (A), (C), (D)

C (B), (C), (D)

D (A), (B), (C), (D)

💡 Short Explanation

Donors provide most electrons. Electrons dominate conduction. Recombination reduces holes.

✅ Correct Answer Explanation

Statements A, B, and C are correct. Statement D is incorrect because thermal generation does not occur at absolute zero.

❓ Why Other Options Are Incorrect

Include incorrect statement D.

🎯 Strategy

Used

Option Grouping

Final Logic:

A, B, and C are correct.

🧠 Memory Trick

"n-Type = Electron Dominance."

📚 Semiconductor Electronics – Majority Carriers

10 The dynamic controlling the reduction of minority holes in n-type material

A depends on the angle between r and p

B depends on the ionization energy of the pentavalent core

C depends entirely on the fixed donor core dropping into the valence band

D is that the rate of recombination of holes increases due to the increase in the number of electrons

💡 Short Explanation

More electrons increase recombination. Minority holes are destroyed more frequently.

✅ Correct Answer Explanation

In n-type semiconductors, donor impurities provide a large number of electrons. These electrons recombine with holes more often, reducing the minority hole concentration.

❓ Why Other Options Are Incorrect

A, B, C → Not responsible for minority-carrier reduction.

🎯 Strategy

Used

Cause-and-Effect Analysis

Final Logic:

More majority electrons → more recombination → fewer holes.

🧠 Memory Trick

"Many Electrons, Few Holes."

📚 Semiconductor Electronics – Recombination and Minority Carriers

11 Match List I with List II regarding trivalent dopant characteristics

List I	List II
P. Indium (In) or Boron (B)	1. Three valence electrons
Q. Valence electrons of dopant	2. Hole
R. Vacancy in the fourth bond	3. Effectively negatively charged
S. Dopant atom ionization state	4. Trivalent impurity

A P-4, Q-1, R-2, S-3

B P-1, Q-4, R-3, S-2

C P-4, Q-2, R-1, S-3

D P-2, Q-1, R-4, S-3

💡 Short Explanation

Boron and Indium are trivalent impurities. They possess three valence electrons. The fourth bond contains a hole.

✅ Correct Answer Explanation

Correct matching: P → Trivalent impurity (4) Q → 3 valence electrons (1) R → Hole (2) S → Effectively negatively charged (3)

❓ Why Other Options Are Incorrect

Incorrect assignment of hole, valence electrons, and ionization state.

🎯 Strategy

Used

Option Grouping

Application:

Match the properties of trivalent acceptor impurities.

Final Logic:

Only Option A correctly matches all characteristics.

🧠 Memory Trick

"Trivalent → 3 Electrons → 1 Hole."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Acceptor Impurities

12 If a trivalent Aluminum atom in a Silicon lattice assumes an effective negative charge −q after bonding, and the associated hole has an effective charge +q, the net crystal contribution before recombination will be

A (−q + q)

B (q + q)

C (−q − q)

D (q − q)

💡 Short Explanation

Acceptor ion carries −q. Hole carries +q. Net charge remains zero.

✅ Correct Answer Explanation

When Aluminum accepts an electron, it becomes negatively ionized (−q). The missing electron in the bond appears as a hole (+q). Their charges balance each other. \((-q)+(+q)=0\)

❓ Why Other Options Are Incorrect

Option B → Net positive charge.
Option C → Net negative charge.
Option D → Represents incorrect charge assignment.

🎯 Strategy

Used

Charge Conservation

Application:

Add the charges of acceptor ion and hole.

Final Logic:

Total contribution remains electrically neutral.

🧠 Memory Trick

"Negative Acceptor + Positive Hole = Neutral Crystal."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Charge Neutrality in p-Type Semiconductors

13 Incorrect statement about p-type majority carrier generation

A The dopant atom of a p-type material can be treated as a core of one negative charge with its associated hole.

B One acceptor atom provides exactly one hole to the lattice.

C The number of conduction electrons acts as the majority carrier.

D Holes are the majority carriers and electrons are the minority carriers.

💡 Short Explanation

Holes dominate conduction in p-type material. Electrons are minority carriers.

✅ Correct Answer Explanation

In p-type semiconductors, acceptor impurities create holes, making holes the majority carriers. Electrons exist only in smaller numbers and act as minority carriers.

❓ Why Other Options Are Incorrect

Option A → Correct.
Option B → Correct.
Option D → Correct.

🎯 Strategy

Used

Conceptual Elimination

Application:

Identify the statement inconsistent with p-type behavior.

Final Logic:

Majority carriers in p-type semiconductors are holes, not electrons.

🧠 Memory Trick

"p-Type = Plenty of Holes."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: p-Type Semiconductor

14 When a p-type semiconductor is in thermal equilibrium

A each minority electron has an equal charge for certain values of impurity doping only

B the minority carriers produced thermally have a significantly increased chance of meeting majority holes and getting destroyed

C each electron has a different charge depending on its orbital level

D each donor atom suppresses the generation of minority electrons

💡 Short Explanation

Majority holes are abundant. Minority electrons recombine easily. Electron concentration decreases.

✅ Correct Answer Explanation

In p-type semiconductors, thermally generated electrons frequently encounter majority holes. This increases recombination and reduces the minority electron concentration.

❓ Why Other Options Are Incorrect

Option A → Electron charge is constant.
Option C → Electron charge does not vary.
Option D → Donor atoms are not present in p-type materials.

🎯 Strategy

Used

Carrier Recombination Analysis

Application:

Understand minority-carrier behavior.

Final Logic:

Abundant holes increase recombination probability.

🧠 Memory Trick

"More Holes → Fewer Electrons."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Minority Carriers in p-Type Semiconductors

15 Correct statements regarding the donor energy level (E_D)
1. E_D exists slightly below the bottom of the conduction band (E_C).
2. Electrons move from E_D into the conduction band with very small energy supply.
3. At room temperature, very few (~10¹²) atoms of Si get ionized compared to donor atoms.
4. E_D is positioned directly in the middle of the energy band gap.

A (A), (B), (C)

B (A), (C), (D)

C (B), (C), (D)

D (A), (B), (D)

💡 Short Explanation

Donor levels lie close to the conduction band. Small energy ionizes donor electrons. Most donor atoms ionize at room temperature.

✅ Correct Answer Explanation

Donor energy levels are located just below the conduction band. A small thermal energy is sufficient to move donor electrons into the conduction band. Intrinsic ionization remains much smaller compared to donor ionization.

❓ Why Other Options Are Incorrect

Statement D → Donor levels are not located at the middle of the gap.

🎯 Strategy

Used

Energy Band Analysis

Application:

Locate donor levels and their behavior.

Final Logic:

A, B, and C are correct.

🧠 Memory Trick

"Donor Level Near Conduction Band."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Donor Energy Levels

16 With a very small supply of energy, an electron from the _____ can jump to the acceptor level E_A and leave behind a _____ in the valence band.

A Conduction band, Hole

B Valence band, Hole

C Effective core, Electron

D Donor level, Cation

💡 Short Explanation

Acceptor level lies near the valence band. Electrons move from valence band to E_A. A hole remains behind.

✅ Correct Answer Explanation

The acceptor level lies slightly above the valence band. Therefore, a valence electron can easily move to the acceptor level, leaving a hole in the valence band.

❓ Why Other Options Are Incorrect

Option A → Electron originates from valence band.
Option C → Incorrect source and carrier.
Option D → Not applicable.

🎯 Strategy

Used

Energy Level Reasoning

Application:

Determine the origin of acceptor electrons.

Final Logic:

Valence-band electrons fill acceptor states.

🧠 Memory Trick

"Electron Leaves Valence Band → Hole Appears."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Acceptor Energy Levels

17 If \(n_{i}=1.0\times {10}^{16} m^{-3}\)and \(n_{h}=2.0\times {10}^{22} m^{-3}\), what is the electron concentration \(n_{e}\)?
\(n_{e}n_{h}=n_{i}^{2}\)

A \(5.0\times {10}^{9} m^{-3}\)

B \(1.0\times {10}^{32} m^{-3}\)

C \(2.0\times {10}^{6} m^{-3}\)

D \(2.0\times {10}^{38} m^{-3}\)

💡 Short Explanation

Use the mass action law: \(n_{e}=\frac{n_{i}^{2}}{n_{h}}\)

✅ Correct Answer Explanation

\(n_{e}=\frac{{\left(1.0\times {10}^{16}\right)}^{2}}{2.0\times {10}^{22}}=\frac{1.0\times {10}^{32}}{2.0\times {10}^{22}}=5.0\times {10}^{9} m^{-3}\)

❓ Why Other Options Are Incorrect

Option B → Equals \(n_{i}^{2}\).
Option C → Incorrect calculation.
Option D → Unphysical value.

🎯 Strategy

Used

Formula Substitution

Application:

Apply the mass action law.

Final Logic:

\(n_{e}=5.0\times {10}^{9} m^{-3}\)

🧠 Memory Trick

"Product of Carriers = \(n_{i}^{2}\)."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Mass Action Law

18 Statements regarding ionization and band gap energies
1. Energy required to jump the forbidden band at room temperature is ~1.1 eV for Si.
2. The energy gap E_g for diamond (C) is 5.4 eV.
3. The ionization energy for a pentavalent fifth electron in Si is ~0.05 eV.
4. The ionization energy for a donor is roughly equal to the intrinsic band gap.

A (A), (C), (D)

B (A), (B), (C)

C (B), (C), (D)

D (A), (B), (D)

💡 Short Explanation

Silicon band gap ≈ 1.1 eV. Diamond band gap ≈ 5.4 eV. Donor ionization energy ≈ 0.05 eV.

✅ Correct Answer Explanation

Statements A, B, and C are correct. Donor ionization energies are much smaller than intrinsic band gaps.

❓ Why Other Options Are Incorrect

Statement D → Donor ionization energy is not equal to the band gap.

🎯 Strategy

Used

Numerical Recall

Application:

Compare standard semiconductor energy values.

Final Logic:

A, B, and C are correct.

🧠 Memory Trick

"0.05 eV vs 1.1 eV vs 5.4 eV."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Donor Ionization Energy and Band Gap

19 Match List I with List II regarding charges in a p-type semiconductor

List I	List II
P. Acceptor atom core	1. Hole (effective positive charge)
Q. Associated majority carrier	2. Electron (effective negative charge)
R. Minority carrier	3. Effectively negatively charged
S. Overall crystal structure	4. Charge neutrality maintained

A P-3, Q-1, R-2, S-4

B P-1, Q-2, R-3, S-4

C P-4, Q-3, R-1, S-2

D P-3, Q-4, R-2, S-1

💡 Short Explanation

Acceptor core becomes negatively charged. Holes are majority carriers. Electrons are minority carriers.

✅ Correct Answer Explanation

Correct matching: P → Effectively negatively charged (3) Q → Hole (1) R → Electron (2) S → Charge neutrality maintained (4)

❓ Why Other Options Are Incorrect

Incorrect carrier and charge assignments.

🎯 Strategy

Used

Matching Concepts

Application:

Associate charges and carrier types.

Final Logic:

Only Option A matches correctly.

🧠 Memory Trick

"Negative Acceptor, Positive Hole."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Charge Neutrality in p-Type Semiconductors

20 The mechanism by which the dopant indirectly helps to reduce the intrinsic concentration of minority carriers

A depends on the exact atomic mass of the intrinsic semiconductor

B depends on the angle between r and p in the lattice structure

C falls as 1/r² at large distances

D is due to the abundance of majority current carriers giving minority carriers more chance of meeting them and getting destroyed

💡 Short Explanation

Majority carriers become abundant. Recombination increases. Minority carrier concentration decreases.

✅ Correct Answer Explanation

Doping introduces a large number of majority carriers. These majority carriers increase the probability of recombination with minority carriers, reducing their concentration.

❓ Why Other Options Are Incorrect

A, B, C → Unrelated to recombination dynamics.

🎯 Strategy

Used

Cause-and-Effect Analysis

Application:

Relate doping to carrier recombination.

Final Logic:

More majority carriers → more recombination → fewer minority carriers.

🧠 Memory Trick

"Majority Up, Minority Down."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Carrier Recombination in Extrinsic Semiconductors

CUET UG Physics Booster Test 3- Extrinsic and Doping Concepts

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